Class 10 Maths - Surface Areas And Volumes

Consider a frustum of cone with larger and smaller radius equal to $r_1 \ and\ r_2$ , respectively. Then, the total surface area of frustum of a cone is $\pi l(r_1+r_2)+\pi (r_{1}^{2}+r_{2}^{2})$ . if correct enter 1 else enter 0

Total surface area of frustum of a cone is

$\pi l(r_1+r_2)+\pi (r_{1}^{2}+r_{2}^{2})$

State True(1) or False(0).

The volume of the frustum of a cone is $\frac{1}{3}\pi h\left [ r_{1}^{2}r_{2}^{2}-r_{1}r_{2} \right ],$ where h is vertical
height of the frustum and $r_1, r_2$ are the radii of the ends.

From the Volume of frustum cone

$=\dfrac{\pi h}{3} (R^2+Rr+r^2)$ .. False

State True(1) or False(0).

An open metallic bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The surface area of the metallic sheet used is equal to curved surface area of frustum of a cone + area of circular base + curved surface area of cylinder

Surface area of the used metallic sheet = curved surface area of frustum of a cone + area of circular base + curved surface area of cylinder.
Hence true.

surface area of a frustum of a cone is $\pi l (r_1+r_2)$ . If true enter 1 else 0.

Curved surface area of a frustum of a cone is

$\pi l(r_{1}+r_{2})$

A spherical ball of radius $3\ cm$ is melted and recast into three spherical balls. The radii of the two of the balls are $1.5\ cm$ and $2\ cm$ respectively. Find the radius of the third ball.

$V_{total}=V_1+V_2+V_3$

$\dfrac43\pi R^3=\dfrac43\pi(r_1^3 +r_2^3+r_3^3)$

$R^3=(r_1^3 +r_2^3+r_3^3)$

$R^3-r_1^3 -r_2^3=r_3^3$

$r_3^3=R^3 -r_1^3-r_2^3$

$\Rightarrow r_3^3=3^3-1.5^3-2^3=15.625$

Hence

$r_3=2.5\ cm$

So diameter

$= 5\ cm$

A fraction clutch is in the form of a frustum of a cone, the diameter of the ends being $32\ cm$ and $20\ cm$ and length $8\ cm$ . Find its bearing surface and volume.[Take $\pi=3.14$ ]

Diameter of ends are

$32\ cm$ and

$20\ cm$

${r}_{1}=16cm$ ,

${r}_{2}=10cm$ and

$h=8cm$

$l=\sqrt{{h}^{2}+{({r}_{1}-{r}_{2})}^{2}}$

$=\sqrt{{8}^{2}+{(16-10)}^{2}}$

$=\sqrt{100}=10cm$

Total surface area

$=\pi [({r}_{1}+{r}_{2})l+{{r}_{1}}^{2}+{{r}_{2}}^{2}]$

$=\pi [{(16+10)10+{16}^{2}+{10}^{2}}]$

$=616\pi{cm}^{2}$

and volume

$=\dfrac{\pi}{3}[{{r}_{1}}^{2}+{r}_{1}{r}_{2}+{{r}_{2}}^{2}]h$

$=\pi[256{cm}^{2}+160{cm}^{2}+100{cm}^{2}]8cm$

$=1376\pi{cm}^{3}$

The decorative block shown in Fig. 13.7 is made of two solids - a cube and a hemisphere. The base of the block is a cube with edge 5 cm. and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block. $\left(Take \pi =\dfrac { 22 }{ 7 } \right)$

The total surface area is

$=$ area of cube

$+$ curved area of hemisphere

$-$ base area of hemisphere

Area of the cube

$A=6(a)^2=6(5)^2=150cm^2$

Curved area of hemisphere

$B=2\pi r=$

$2\pi\left(\dfrac{4.2}{2}\right)$

$=27.72cm^2$

Base area of hemisphere

$C=\pi r^2=\pi \dfrac{4.2}{2}=13.86cm^2$

Total surface area

$=A+B-C=150+27.72-13.86=163.86cm^2$

How many spherical bullets each of $5cm$ in diameter can be cast from a rectangular block of metal $11dm\times 1m\times 5dm$ ?

$1 dm = 10 cm$

$1 m =100 cm$

Hence total volume of block =

$L\times B\times H$

$=110\times 100 \times 50 cm^3=5,50,000cm^3$

No. of bullets cast =

$\dfrac{550000}{\dfrac43\pi (2.5)^3}=8403$ bullets

A $16\text{ m}$ deep well with diameter $3.5\text{ m}$ is dug up and the earth from it is spread evenly to form a platform $27.5\text{ m}$ by $7\text{ m}$ . Find the height of the platform.

For cylindrical well:

$R=\dfrac{3.5}{2}=1.75\text{ m}$

$=16\text{ m}$

For cuboidal platform:

$l=27.5\text{ m}$

$b=7\text{ m}$

$h=?$

Earth obtained from the digging well will be completely used to form the platform. Hence,

Volume of earth from well

$=$ Volume of earth from cuboidal platform

$\begin{aligned}{}\pi {R^2}H& = l \times b \times h\\\frac{{22}}{7} \times {\left( {1.75} \right)^2} \times 16& = 27.5 \times 7 \times h\\154 &= 192.5 \times h\\h &= 0.8\text{ m}\end{aligned}$

Hence, the height of the platform is equal to

$0.8\text{ m}$ .

A well of diameter $2\ m$ is dug $14\ m$ deep. The earth taken out of it is spread evenly all around it to form an embankment of height $40\ cm$ . Find the width of the embankment.

Well is a cylinder of

$Depth ,L' = 14 m$ and radius

$R=1m$

Embankment is a hollow cyllinder with inner and outer radii :

$R_i = 1m \ R_e =? \ \ Height, L =0.4m$

Width of embankment

$=R_e-R_i$

$\pi (R_e^2-R_i^2)L=\pi R^2L'$

$R_e^2=\dfrac{R^2L'}{L} + R_i^2=35+1=36$

$\Rightarrow R_e=6 m$

Hence width of embankment

$=5m$

Rain water, which falls on a flat rectangular surface of length $6m$ and breadth $4m$ is transferred into a cylindrical vessel of internal radius $20cm$ . What will be the height of water in the cylindrical vessel if a rainfall of $1cm$ has fallen? (Use $\pi =\cfrac { 22 }{ 7 }$ )

$L\times B\times H=\pi r^2h$

$6\times 4 \times 0.01 = \pi 0.2^2h$

$h= 1.9098 m = 190.98cm$

Find the volume of the largest right circular cone that can be cut out of a cube whose edge is $9cm$ .

Largest right circular cone will have same height and diameter as edge of cube.

$Volume = \dfrac13 \pi r^2h$

$=\dfrac13 \pi (4.5)^29 = 190.85cm^3$

$50$ circular plates each of diameter $14cm$ and thickness $0.5cm$ are placed one above the other to form a right circular cylinder. Find its total surface area.

Given object is cylinder of radius

$7$ cm and height,

$l = 50\times 0.5=25cm$

$T.S.A= 2\pi r(r+l)=1407.43 cm^2$

A cylindrical tub of radius $12cm$ contains water to a depth of $20cm$ . A spherical form ball of radius $9cm$ is dropped into the tub and thus the level of water is raised by $hcm$ . What is the value of $h$ ?

Since rise in water level is solely caused by ball,

$\pi\times (12^2)\times h=\dfrac43 \pi\times (9^3)$

$\Rightarrow h =6.75 cm$

A conical flask is full of water. The flask has base-radius $r$ and height $h$ . The water is poured into a cylindrical flask of base-radius $mr$ . Find the height of water in the cylindrical flask.

$\dfrac13 \pi r^2h=\pi (mr)^2 H$

$H=\dfrac{h}{3m^2}$

A hollow sphere of internal and external radii $2\ cm$ and $4\ cm$ respectively is melted into a cone of base radius $4\ cm$ . Find the height and slant height of the cone.

Volume of sphere

$=V_s=\dfrac43 \pi (R_e^3-R_i^3)$

$V_c= \dfrac13\pi r^2h$

$V_c=V_s\Rightarrow \dfrac13 \pi 4^2 h = \dfrac43 \pi (4^3-2^3)$

$\Rightarrow h =14\ cm$

$l =\sqrt{14^2+4^2}=14.56\ cm$

A hollow sphere of internal and external diameters $4cm$ and $8cm$ respectively is melted into a cone of base diameter $8cm$ . Calculate the height of the cone.

$\dfrac43 \pi (R_e^3-R_i^3)=\dfrac13 \pi r^2h$

$\Rightarrow h = \dfrac{4(4^3-2^3)}{4^2}=14cm$

$25$ circular plates, each of radius $10.5cm$ and thickness $1.6cm$ , are placed one above the other to form a solid circular cylinder. Find the curved surface area and the volume of the cylinder so formed.

Given object is cylinder of radius

$10.5$ cm and height,

$l = 25\times 1.6=40cm$

$C.S.A=2\pi RL = 2\pi \times 10.5\times 40 =2638.93 cm^3$

$Volume = \pi R^2L=\pi \times (10.5)^2\times 40=13854.42cm^3$

If $h$ , $s$ , $V$ be the height, curved surface area and volume of a cone respectively, then $\left( {3\pi V{h^3} + 9{V^2} - {s^2}{h^2}} \right) = ?$

$3\pi v{h^3} + 9{v^2} - {5^2}{h^2}$

$3\pi \left( {\dfrac{1}{3}\pi {r^2}h} \right){x^3} + 9{\left( {\dfrac{1}{3}\pi {r^2}h} \right)^2} - {\left( {\pi rl} \right)^2}{h^2}$

${\pi ^2}{r^2}{h^4} + {\pi ^2}{r^4}{h^2} - {\pi ^2}{r^2}{l^2}{h^2}$

${\pi ^2}{r^2}{h^4} + {\pi ^2}{r^4}{h^2} - {\pi ^2}\left( {{h^2} + {r^2}} \right){h^2}$

${\pi ^2}{r^2}{h^4} + {\pi ^2}{h^2}{r^4} - {\pi ^2}{r^2}{h^4} - {\pi ^2}{r^4}{h^2}$ \

$= 0$

The largest sphere is to be carved out of a right circular cylinder of radius $7cm$ and height $14cm$ . Find the volume of the sphere.

The largest sphere is obviously of radius

$7cm$

Volume

$=\dfrac43\times \pi\times 7^3=1436.75 cm^3$

A vessel in the shape of a cuboid contains some water. If three identical spheres are immersed in the water, the level of water is increased by $2cm$ . If the area of the base of the cuboid is $160{cm}^{2}$ and its height $12cm$ , determine the radius of any of the spheres.

Let the radius of each sphere be

$r$

$cm$

Volume of a sphere

$= (\dfrac43 \pi r^3)$

$cm^3$

Increase in volume of the cuboid

$=162\times2$

$cm^2$

It is known that increase in volume of the cuboid will be equal to the volume of all three spheres.

$160\times2 = 3(\dfrac43 \pi r^3)$

$320=3(\dfrac43\times\dfrac{22}{7}\times r^3)$

$r^3=25.45$

$\Rightarrow r =2.94$

$cm$

The diameter of a metallic sphere is equal to $9cm$ . It is melted and drawn into a long wire of diameter $2mm$ having a uniform cross-section. Find the length of the wire.

$\dfrac43 \pi r^3 = \pi r'^2h'$

$\Rightarrow h' =\dfrac{4(4.5)^3}{3(0.1)^2}=12150cm=121.5m$

A copper sphere of radius $3\ cm$ is melted and recast into a right circular cone of height $3\ cm$ . Find the radius of the base of the cone.

Let the radius of the right circular cone be

$r.$

Given:

Radius of sphere,

$R_s=3\ cm$

Height of the cone,

$h=3\ cm$

So,

Volume of copper sphere

$=$ Volume of the right circular cone

$\dfrac43\pi R_s^3 = \dfrac13 \pi r^2h$

$\Rightarrow \dfrac43\times \pi\times 3^3 = \dfrac13\times \pi\times r^2\times3$

$\Rightarrow 4\times 3^2 = r^2$

$\Rightarrow r = 2\times 3$

$\Rightarrow r=6\ cm$

A self help group wants to manufacture jokes caps (conical shapes) of $3 \text cm$ radius and $4cm$ height. If the available colour paper sheets is $1000 \text cm^{2}$ , then how many caps can be manufactured from that paper sheet?

$\Rightarrow$ We have given,

$Height(h)=4\,cm$ and

$radius(r)=3\,cm$

$\Rightarrow$ Let slant height be

$l$ .

We know that,

$l^2=h^2+r^2$

$l^2=(4)^2+(3)^2$

$l^2=16+9$

$l^2=25$

$\therefore$

$l=5\,cm$

$\Rightarrow$ Curve surface area of

$1$ joker cap

$=\pi r l$

$=\dfrac{22}{7}\times 3\times 5$

$=47.14\,cm^2$

$\Rightarrow$ Available color paper sheet

$=1000\,cm^2$

$\Rightarrow$ Number of caps manufacture from

$1000\,cm^2$ paper sheet

$=\dfrac{1000}{47.14}=21.21$

994679_1062208_ans_4eeab1d8168d4d12aec7f47fff88f1e1.png

A spherical ball of radius $3\ cm$ is melted and recast into three spherical balls. The radii of two of the balls are $1.5\ cm$ and $2\ cm$ . Find the diameter of the third ball.

Let

$r$ be the radius of the third ball. Then,

Given:

$V_{total}=V_1+V_2+V_3$

$\Rightarrow \dfrac43\pi R^3=\dfrac43\pi(r_1^3 +r_2^3+r_3^3)$

$\Rightarrow \cfrac { 4 }{ 3 } \pi \times { 3 }^{ 3 }=\cfrac { 4 }{ 3 } \pi \times { (1.5) }^{ 3 }+\cfrac { 4 }{ 3 } \pi \times { 2 }^{ 3 }+\cfrac { 4 }{ 3 } \pi { r }^{ 3 }\quad$

$\Rightarrow 27=\cfrac { 27 }{ 8 } +8+{ r }^{ 3 }$

$\Rightarrow { r }^{ 3 }=\cfrac { 125 }{ 8 }$

$\Rightarrow r=\cfrac { 5 }{ 2 }$

$=2.5\ cm$

Metal spheres, each of radius $2cm$ , are packed into a rectangular box of internal dimension $16cm\times 8cm\times 8cm$ when $16$ spheres are packed the box is filled with preservative liquid. Find the volume of this liquid. (Use $\pi =\dfrac{669}{213}$ )

Volume of liquid

$=$ Volume of box

$-$ volume of spheres

Hence

$V= 16\times8\times8-16(\dfrac43\times\pi\times 2^3)$

$=16(64-\dfrac{32\pi}{3})=487.96 cm^2$

A washing tub in the shape of a frustum of a cone has height $21\ cm$ . The radii of the circular top and bottom are $20\ cm$ and $15\ cm$ respectively. What is the capacity of the tub ? $(\pi=\dfrac{22}{7})$

Given that,

Height

$h = 21 cm$

Top radius

$R = 20 cm$

Bottom radius

$r = 15 cm$

Capacity of the tub = Volume of frustum of the cone

$= \dfrac{1}{3} \pi h (R^2 + r^2 + Rr)$

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 21 \times (20^2 + 15^2 + 20\times 15)$

$= 22 \times (400 + 225 + 300)$

$= 22 \times 925$

$= 20350 cu.cm$

$= 20.350 l$ (Ans)

The radius of a solid sphere is $7$ cm. If the solid sphere is melt to form small spheres of radius $3.5$ cm. Then how many small spheres can be formed?

Given the radius of the solid sphere is

$7$ cm.

Then its volume is

$\dfrac{4}{3}\times \pi\times 7^3$ cm

$^3$ .

Also given the radius of the small sphere is

$3.5$ cm.

Then the volume of the small sphere is

$\dfrac{4}{3}\times \pi\times \left(\dfrac{7}{2}\right)^3$ cm

$^3$ .

Let there be

$n$ small spheres.

Then equating the volumes of

$n$ spheres to the bigger spheres we get,

$\dfrac{4}{3}\times \pi\times 7^3=$

$n\times\dfrac{4}{3}\times \pi\times \left(\dfrac{7}{2}\right)^3$

or,

$\dfrac{4}{3}\times \pi\times 7^3=$

$n\times\dfrac{4}{3}\times \pi\times \dfrac{7^3}{8}$

or,

$n=8$ .

Given : $ED = 14 cm$

R.E.F image

Area of shaded region

$ED=14cm$ (Given)

$ar(\triangle EBD)=ar(\triangle ACDE)-ar(\triangle ABE)-ar(\triangle BED)$ ...(1)

$ED=AC=14cm=2AB$ (Since

$AB=BC$ )

$\Rightarrow AB=BC=7cm=AE=CD$

$\therefore ar(ACDE)=AC\times CD=14\times 7\Rightarrow 98cm^2$

$ar(\triangle ABE )=\dfrac{1}{2}\times AB\times BC=\dfrac{1}{2}\times 7\times 7=\dfrac{49}{2}cm^2$

$ar(\triangle BCD)=\dfrac{1}{2}\times BC\times CD=\dfrac{1}{2}\times 7\times 7=\dfrac{49}{2}cm^2$

$\therefore \boxed{ae(\triangle EBD)=98-\dfrac{49}{2}-\dfrac{49}{2}=49cm^2}$

$ar(semi \,circle )\dfrac{\pi r^2}{2}=\dfrac{\pi}{2}(\dfrac{7}{2})^2=\dfrac{\pi}{2}\times (\dfrac{14}{2})^2$

$=\dfrac{22}{7}\times \dfrac{1}{2}\times 7\times 7$

$\boxed{area(semicircle)=77cm^2)}$

$\therefore$ area of shaded region

$=(1)+(2)$

$\therefore 77+49$

$\therefore 126cm^2$

1074065_1115863_ans_d3ffd69098cb4cebb93ad0e66e5ce216.png

The slant height of a cone with hemispherical base is $5$ cm. If the total surface area of the article is $103.62 cm^2$ , find its height.

Given,

slant height

$l=5cm$

total surface area

$A=103.62\text{cm}^{2}$

Formulae:

$l=\sqrt{h^2+r^2}$

$A=\pi r (r+l)$

$\Rightarrow r = \dfrac{1}{2}\sqrt{l^2+ 4 \dfrac{A}{\pi}}-\dfrac{l}{2}$

$\Rightarrow r = \dfrac{1}{2}\sqrt{5^2+ 4 \dfrac{103.62}{\pi}}-\dfrac{5}{2}$

$\Rightarrow r = 3.76$

$l^2 = h^2+r^2$

$\Rightarrow h^2 = l^2 - r^2$

$\Rightarrow h^2 = 5^2 - 3.76^2$

$\Rightarrow h^2 = 10.8624$

Height,

$h=3.29\text{ cm}$

A rectangular reservoir is 120 m long and 75 m wide. At what speed per hour must water flow into it through a square pipe of 20 cm wide so that the water rises by 2.4 m in 18 hours ?

Volume of water accumulated in the reservoir in 18 hours =

$(120 \times 75 \times 2.4) \ m^3$ ........(1)

Let the speed of the water be v km/hr.

Water in the pipe forms a cuboid of width

$\dfrac{20}{100}=\dfrac{1}{5} \ m, height =\dfrac{20}{100}=\dfrac{1}{5} \ m$

Length of the water cuboid formed in 18 hours = 18 v km = 18000 v metres

Therefore,

Volume of water accumulated in the reservoir in 18 hours

$=18000 v \times \dfrac{1}{5} \times \dfrac{1}{5} \ m^3=720v \ m^3$ .......(2)

From 1 and 2, we get,

$720v=120 \times 75 \times 2.4$

$v=30 \ km/hr$

Hence, water must flow at the speed of 30 km/hr into the reservoir.

Threee cubes of the same metals, whose edges are 6,8,10 cm are melted and formed into a single cube. find the diagonal of these single cube.

As we know that the volume of cube

$\left( V \right)$ is given by,

$V = {a}^{3}$

where as,

$a$ is the side of cube

Volume of cube of edge

$6 \; cm, \left( {V}_{1} \right) = {6}^{3} = 216 \; {cm}^{3}$

Volume of cube of edge

$8 \; cm, \left( {V}_{2} \right) = {8}^{3} = 512 \; {cm}^{3}$

Volume of cube of edge

$10 \; cm, \left( {V}_{3} \right) = {\left( 10 \right)}^{3} = 100 \; {cm}^{3}$

Total volume of the single cube formed

$\left( V' \right) = {V}_{1} + {V}_{2} + {V}_{3} = 216 + 512 + 1000 = 1728 \; {cm}^{3}$

Let

$a'$ be the edge of single cube formed.

$\therefore V' = {\left( a' \right)}^{3}$

$\Rightarrow a' = \sqrt[3]{1728} = 12 \; cm$

Now as we know that the length of diagonal

$\left( d \right)$ of a cube of edge

$a$ is given as-

$d = \sqrt{3} a$

$\therefore$ Length of diagonal of the single cube formed

$= \sqrt{3} \times 12 = 20.784 \; cm$

Hence the correct answer is

$20.784 \; cm$ .

From a solid cylinder whose height is $2.4cm$ and diameter $1.4cm$ a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest ${cm}^{2}$

We have

Height of solid cylinder

$=2.4$ cm

diameter of solid cylinder

$= 1.4$ cm

$l=\sqrt{r^{2}+h^{2}}=\sqrt{(0.7)^{2}+(2.4)^{2}}$

$=2.5$

clearly,

Total surface area of the remaining solid = Curved surface area of the cylinder + Area of the base of the cylinder + Curved surface area of the cone.

Surface Area

$=(\pi r^{2})+(2\pi rh)+(\pi rl)$

$=\pi ((0.7)^{2}+(2)(0.7)(2.4)+(0.7)(2.5))$

$=\pi (5.6)$

$=\dfrac{22}{7}(5.6)$

$=17.6 cm^{2}$

1062405_1091080_ans_c2938e0f10204e07a9798adefa05d7c8.png

Find the volume of the largest circular cone that can be cut out of a cube whose edge is $9\ cm. \left (\pi = \dfrac {22}{7}\right )$ .

Since the edge of cube is 9 cm this will be the altitude of the cone.
The radius of circular cone will be 9:2=4.5 cm.
The volume of cone is given by formula: V=%281%2F3%29%2A%28pi%29%2AR%5E2%2AH

, where R is the radius and H the altitude of cone. Since we are asked for the largest cone, its volume must be equal or less than the volume of cube. We write:

+%281%2F3%29%2A%28pi%29%2A%284.5%29%5E2%2A%289%29

=
V=

A hemispherical bowl of internal radius 9 cm is full of liquid. The liquid is to be filled into cylindrical shaped small bottles, each of diameter 3 cm and height 4 cm. How many bottles are needed to empty the bowl ?

Radius of hemispherical bowl = 9 cm

Volume of liquid in the bowl =

$\dfrac{2}{3}\pi r^3 =(\dfrac{2}{3}\pi\times 9\times 9 \times 9)=486 \pi \ cm^3$

Radius of each cylindrical bottle =

$\dfrac{3}{2} \ cm$

And, height =

$4 \ cm$

Volume of each cylindrical bottle =

$\pi R^2 h=\pi \times \dfrac{3}{2} \times \dfrac{3}{2}\times 4 =9\pi \ cm^3$

Therefore, required number of bottles =

$\dfrac{486\pi}{9\pi}=54$

A cylindrical container with diameter of base $42\ cm$ contains sufficient water to submerge a rectangular solid of iron with dimensions $22\ cm\times 14\ cm \times 10.5\ cm$ . Find the rise in level of the water when the solid is sub-merged.

1300884_1129110_ans_ca4ab19b4be54cd8934f0ed7eabc7213.jpg

The height of a cone is $40\ cm$ A Small cone is cut off at the top by a plane parallel to its base. If its volume be $\dfrac {1}{64}$ of the volume of the given cone at what height above the base is the section cut?

we have been given the volume scale factor which is 1/64.

According to this ratio, the formula for volume scale factor will be:

Volume of the smaller cone divide by volume of larger cone.

V. S. F=1/64

We get the linear scale factor:

(l.s.f)^3=V.S.F

L.S.F=1/4

L.s.f=height of small cone/height of large cone.

1/4=x/64

64=4x

X=16cm

Height above base where section was made:

40-16=24cm

How many silver coins, $1.75cm$ in diameter and of thickness $2mm$ , must be melted to form a cuboid of dimensions $5.5cm\times 10cm\times 3.5cm$ ?

Let the number of coin be

$x$

$x\times$ volume of 1 coin

$=$ volume of box

$x\times \pi r^{2}h=l\times b\times h$

$x\times \dfrac{22}{7}\times \dfrac{175}{100\times 2}\times\dfrac{175}{100\times 2}\times \dfrac{2}{10} = \dfrac{55}{10}\times 10\times \dfrac{35}{10}$

$x=\dfrac{55\times 35\times 100\times 100\times 2}{22\times 25\times 175}$

$= 20\times 20$

$=400 coins$

In figure, the object is made of two solids a cube and a hemisphere. The base of the figure is a cube with edges of 5 cm and the hemisphere fixed on the top has a diameter of 4.2 cm . Find the total surface area of the object (take pi = 22/7 ) .

1112862_1205853_ans_510fec724bdc4561b4b1f201001f4e9e.jpg

A spherical glass vessel has a cylindrical neck $8\;cm$ long $2\;cm$ in diameter, the diameter of a spherical part is $8.5\;cm$ . By measuring the amount of water it holds, a child finds its volume to be $345\;cm^3$ . Check whether she is correct, taking the above as the inside measurements, and $\pi=3.14$

Diameter of spherical glass vessel = 8.5 cm

$\therefore r = 4.25$ cm
Length of cylindrical neck = 8 cm
Diameter = 2 cm

$\therefore$ Radius, r = 1 cm
Volume of water she found = 345 cm

$^3$
(i) Volume of cylinder neck, V

$= \pi r^2 h$

$= \pi \times (1)^2 \times 8$

$= 8 \pi cm^3$
(ii) Volume of cylindrical vessel

$= \frac{4}{3} \pi r^3$

$= \frac{4}{3} \times \pi \times (4.25)^3$

$\therefore$ Total volume of water, V

$= 8 \pi + \frac{4}{3} \pi \times (4.25)^3$

$= 8 \times 3.14 + \frac{4}{3} \times 3.14 \times (4.25)^3$

$= 25.12 + 321.38$

$= 346.5 cm^3$

$\therefore$ Her answer is not correct.
1840898_1206323_ans_c7fdbbaa3d6549f59418a0802f733f48.png

The radii of circular ends of a bucket of height 24 cm. Find the area of its curved surface.

REF. Image.

Since it is a bucket. Therefore, it can act as a frustum of

$h =24 cm$

$r = 5 cm$

$R = 15 cm$

Slant height =

$l = \sqrt{h^{2} + (R - r)^{2}}$

$l = \sqrt{(24)^{2}+(15-5)^{2}} = \sqrt{(24)^{2}+(10)^{2}}$

$l = \sqrt{576+100}$

$= \sqrt{676}$

$= 26cm$

C.S.A

$= \pi (R+r)l.$

$= \frac{22}{7} (15 + 5) 26$

$= \frac{22}{7}\times 20\times 26$

$= \frac{11440}{7}$

$cm^{2}$

$= 1634.28571 cm^2$

1208708_1349027_ans_0a1f340896ef478fb6de56ce9e3aa7d0.png

The slant height of the frustum of a cone is $5\ cm$ . If the difference between the radii of its two circular ends is $4cm$ . Calculate the height of the frustum.

Given

$\rightarrow$ slant height = 5 cm and the difference between the radii to of two circular ends = 4 cm

To find : height of frustum

It is given by the relation:

$(slant\, Height)^{2} = (Height)^{2}+(Difference \,b/w\, the\, radii \, to\, ends)^{2}$

$(5)^{2} = (Height)^{2}+(4)^{2}$

$(Height)^{2} = (25-16)cm$

$(Height)^{2} = 9 cm$

Height

$= 3 cm$

Therefore, Height of the frustum

$= 3 cm$

The slant height of the frustum of a cone is $4cm$ . If the perimeters of its circular bases be $18cm$ and $6cm$ , find the curved surface area of the frustum and also find the cost of painting its total surface at the rate of Rs. $12.50$ per $100{cm}^{2}$

R.E.F image

$2\pi R=18$

$2\pi r=4$

$\pi R=9$

$\pi r=2$

$R=\dfrac{9}{\pi}$

$r=\dfrac{2}{\pi}$

Curved surface area

$=\pi (R+2)l$

$=\left(\dfrac{9}{\pi}+\dfrac{2}{\pi}\right)4$

$=44cm^2$

Total cost

$=\dfrac{12.50}{100}\times 44$

$=5.5 Rs$

1084253_1173139_ans_9d1fd44428374ab092a9cf0446675773.png

Fill in the blanks:
A rectangular pyramid has____ faces.

A rectangular pyramid has

$5$ faces.

If the total surface area of a solid hemisphere is $46c{m^2}$ , find its volume

Total surface area of solid hemisphere

$=3\pi r^2=46$

$r=2.21$

Volume

$=\dfrac{2}{3}\pi r^3=22.615 cm^3$

What length of tarpaulin 3 m wide will be required to make a conical tent of height 8 m and base radius 6 m ? Assume that the extra length of material will be required for stitching margins and wastage in cutting is approximately 20 cm.

h = Height of tent = 8 m

r = Base radius = 6 m

Therefore,

$l=\sqrt{r^2+h^2}=10$ $m$

Curved surface area of the cone = $\pi rl=3.14 \times 6 \times 10$ $m^{2}$

Therefore, length of 3 m wide tarpaulin required = $\dfrac{3.14 \times 6 \times 10}{3}=62.8$ $m$

Extra length required = $0.2$ $m$

Therefore, total length required = $63$ $m$

A sphere is inscribed in a cylinder than find the ratio of volume of sphere to volume of cylinder.

A sphere is inscribed in a cylinder

here height of cylinder

$h = 2r$

Volume of sphere

$= \dfrac{4}{3}\pi {r^3}$

and Volume of cylinder

$= \pi {r^2}h$

$\begin{array}{l} =\pi { r^{ 2 } }\left( { 2r } \right) \\ =2\pi { r^{ 3 } } \end{array}$

Ratio of volume of sphere to volume of cylinder

$= \dfrac{{4\pi {r^3}}}{{3 \times 2\pi {r^3}}} = 2:3$

1210997_1316542_ans_49f9b74413074799b1524e57e39adb45.PNG

A bucket is in the form of a frustum of a cone of height is $30\ cm$ with radii of its lower and upper ends as $10\ cm$ and $20\ cm$ respectively. Find the capacity of the bucket.

Radius of the upper end of the frustum of cone

$= R = 20\ cm$
radius of the lower end of the frustum of cone

$= r = 10 \ cm$
Height

$h = 30\ cm$

Volume

$= \dfrac{1}{3}πh[R^2 + r^2 + R\times r]$

$= \dfrac{1}{3}\times \dfrac{22}{7}\times 30\times [20^2 + 10^2 + 20\times 10]$

$= \dfrac{660}{21}\times [400 + 100 + 200]$

$= \dfrac{660\times 700}{21}$

$=22000 \ cm ^3$

capacity of cone

$=$ Volume of cone

$= 22\ liters$

Write the formula for finding volume of a frustum of a cone.

Let

$h$ be the height,

$R$ the radius of the lower base, and

$r$ the radius of the upper base.

Volume of a frustum of a cone

$= \cfrac{\pi}{3} h \left( {R}^{2} + {r}^{2} + Rr \right)$

1199114_1507951_ans_86ed9e235f9c4aa79276195111520506.gif

The silant height and base diametors of a conical to mb we 25m and 14 m qupertiuity. Find the cost of washisg its surface surface at the rate of RS 210 per 100 $m^{ 2 }$

1323530_1491342_ans_2789c066a8b54730a772d5854fb4ce0c.JPG

A metallic bucket, open at the top, of height $24cm$ is in the form of the frustum of a cone, the radii of whose lower and upper circular ends are $7cm$ and $14cm$ respectively. Find
the volume of water which can completely fill the bucket;
the area of the metal sheet used to make the bucket.

Let the slant height of the frustum of the cone be 'l' and the radius of the upper end be 'R' and the radius of the lower end be 'r' respectively.
Therefore,
l² = √(R - r)² + h²
l² = √(14 - 7)² + 24²
l² = √7² + 24²
l² = √49 + 576
l² = √625
l = 25 cm
So, slant height is 25 cm

Volume = 1/3πh(R² + r² + R*r)
⇒ 1/3*22/7*24*(14² + 7² + 14*7)
⇒ 1/3*22/7*24*(196 + 49 + 98)
⇒ 1/3*22/7*24*343
⇒ 181104/21
⇒ 8624 cm³

Curved surface area = πl(R + r)
⇒ 22/7*25*(14 + 7)
⇒ 22/7*25*21
⇒ 11550/21
⇒ 1650 cm²

Area of the base (lower end) of the bucket = πr²
⇒ 22/7*7*7
⇒ 154 cm²

Area of the metal sheet used to make the bucket = Curved surface area + Area of the base (lower end)
= 1650 + 154
= 1804 cm²

A sphere of diameter $18\text{ cm}$ is dropped into a cylindrical vessel of diameter $36\text{ cm},$ partly filled with water. If the sphere is completely submerged, then the increase in water level ( $\text{in cm}$ ) will be

Given

$\Rightarrow$

$r_c=18\text{ cm}$ and

$r_s=9\text{ cm}$ .

Let the rise in the water level be equal to

$h$ .

Volume of water level rises in cylinder will be equal to volume of sphere. So,

$\begin{aligned}{}\pi r_c^2h &= \frac{4}{3}\pi r_s^3\\\pi {\left( {\frac{{36}}{2}} \right)^2}h &= \frac{4}{3}\pi {\left( {\frac{{18}}{2}} \right)^3}\\{\left( {18} \right)^2}h &= \frac{4}{3} \times \frac{{18 \times 18 \times 18}}{{2 \times 2 \times 2}}\\h& = \frac{{18}}{6}\\ &= 3\text{ cm}\end{aligned}$

So, the rise in the water level will be equal to

$3\text{ cm}$ .

In the above figure, area of a square $ABCD$ is $S$ and area of each shaded portion is $P$ and area of inner circle is $Q$ . Then the area of remaining part of a square is

Area of each shaded region is P and there are four shaded regions.

Let R be the area of remaining region.

Area of square ABCD = Area of shaded region x 4 + area of inner circle + area of remaining region

therefore,

$S = 4 \times P + Q + R$

thus area of remaining region ( R) = S - 4P -Q

Find the ratio of the volumes of a cone and of a cylinder whose base diameter and heights are equal.

Given based and hegight of cylinder & cone are equal.

Let "h" be height cand 'y' be radius of base then

Volume of cylinder

$(V_1)= \pi r^2h$

Volume of cone

$(V_2)= \frac{1}{3}\pi r^2h$

then

$\frac{V_2}{V_1}= \frac{1}{3} \frac {\pi r^2h}{\pi r^2h}=\frac{1}{3}$

$\therefore$ ratio of volume of cone to that of cylinder is

$1:3$

How many planks of size $2\ cm\times 25\ cm\times 8\ cm$ can be prepared from a wooden block $5\ m$ long. $70\ cm$ broad and $32\ cm$ thick, assuming that there is no wastage?

We know that volume of cuboid =length

$\times$ breadth

$\times$ height

So,

Volume of the block

$=500\times 70\times 21-112000\ cm^{3}$

Volume of each plank

$=200\times 25\times 8=40000\ cm^{3}$

Therefore total number of planks that can be made

$= \dfrac{Volume\ of\ the\ block}{Volume\ of\ each\ plank}$

$=112000/40000$

$=38$ planks

Fill in the blanks to make the correct statements true.
A cube of side 4 cm is painted on all its sides. If it is sliced in 1 cubic cm cubes, then number of such cubes that will have exactly two of their faces painted is ____________.

The correct fill up is 24.
Side a = 4
Vol. of cube =

$a^{3}$ =

$4^{3}$ =

$64 cm ^{3}$
On slicing it into

$1 cm^{3}$ , 64 small cubes will be obtained In each side of the larger cube, the smaller cubes in the edges will have more than one face painted.
The cubes which are situated at the corners of the big cube have three faces painted.
So, to each edge two small cubes are left which have two faces painted . As, the total number of edges in a cube are 12.
Hence, the number of small cubes with two faces painted=
=

$12 \times 2$ = 24.

Fill in the blanks the statements are true.
The surface area of a cylinder which exactly fits in a cube of side b is ________.

$\pi b^{2}$
Height = b and radius = b/2
Now, the curved surface area of the cylinder=

$2\pi rh=2\pi \times \frac{b}{2}\times b= \pi b^{2}$

Mr. Ahuja has to square plots of land which he utilises for two different purposes one for providing free education to the children below the age of $14$ years and the other to provide free medical services for the needy villagers. The sum of the areas of two square plots is $15425{m}^{2}$ . If the difference of their perimeter is $60m$ , find the sides of the two squares.
Which qualities of Mr. Ahuja are being depicted in the question?

Let

$x$ be the length of the side of first square and

$y$ be the length of side of the second square.
Then,

${x}^{2}+{y}^{2}=15425$ ...(i)
Let

$x$ be the length of the side of the bigger square

$4x-4y=60$

$\Rightarrow$

$x-y=15$ or

$x=y+15$ .....(ii)
Putting the value of

$x$ in terms of

$y$ from equation (ii) in equation (i) we get

${(y+15)}^{2}+{y}^{2}=15425\Rightarrow 2{y}^{2}+30y-15200=0$
or

${y}^{2}+15y-7600=0$ or

$(y+95)(y-80)=0$

$\Rightarrow$

$y=-95,80$
therefore,

$x=80+15=95$
Hence, sides of two squares are

$80m$ and

$95m$
Value: Caring, King, Social and generous

A wooden cylindrical pole is $7\ m$ high and its base radius is $10\ cm$ . Find its weight if the wood weights $225\ kg$ per cubic meter.

Given

$r===10\ cm=0.1\ m$ and

$h=7\ m$
But we know that volume of the cylinder

$=\pi r^2h$

$V=22/7\times 0.1\times 0.1 \times 7$

$V=0.22\ m^3$

Given the weight of the wood

$225\ kg$ per cubic meter
So, the weight of the pole

$=0.22 \times 225=49.5\ kg$

What figure is formed if only the height of a cube is increased or decreased?

If we only increase on decrease the height of a cube, the obtained figure is cuboid.
1653718_1793139_ans_22c1436e5ea44df496477393209dee75.png

Fill in the blanks to make the statement true.
The volume of a cylinder which exactly first in a cube of side a is ________.

Since, the cylinder that exactly first in cube of side a has its height equal to the edge of the cube and radius equal to half the edge of the cube.
Therefore, height= a and radius = a/2
Now, vol.of the cylinder =

$\pi r^{2}h = \pi \left ( \dfrac{a}{2} \right )^{2} a=\dfrac{1}{4}\pi a^{3}$

How many envelopes of size 25 cm $\times$ 15 cm can be made from a rectangular sheet of size 4 m $\times$ 1.2 m?

Size of envelope

$= 25 cm \times 15 cm$

Area of envelope

$= 25$

$\times$

$15 cm^2 = 375 cm^2$
Size of rectangular sheet

$= 4 m \times 1.2 m$
Area of rectangular sheet

$= 400 cm \times 120 cm = 48000 cm^2$

Total number of envolapes made from rectangular sheet

$=\dfrac{48000}{375}=128$

A solid sphere of radius $6cm$ is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is $5$ cm and its height is $32$ cm, then find the thickness of the cylinder.

Volume of sphere = Volume of hollow cylinder

$\left(\dfrac{4}{3}\right)\pi r^3=\pi h(R^2-r^2)$

$\left(\dfrac{4}{3}\right) 6^3=32(5^2-r^2)$

$25-r^2=[(\dfrac{4}{3})216]/32$

$25-r^2=[(4)72]/32$

$25-r^2=9$

$r^2=25-9=16$

$r=4$
Thickness

$R-r$

$=5-4$

$=1 cm$

A solid right circular cone of radius 4cm and height 7cm is melted to a sphere. Find the radius of sphere.

$\textbf{Step -1: Calculate the Volume of Cone.}$

$\text{Volume of Cone =}{\dfrac{1}{3}{\pi}r^2h}$

$\text{Volume of Cone =}{\dfrac{1}{3}{\pi}.4^2.7}$

$\text{Volume of Cone =}{\dfrac{112{\pi}}{3}}$

$cm^3$

$\textbf{Step -2: Calculate the Volume of Sphere.}$

$\text{Volume of Sphere =}{\dfrac{4}{3}{\pi}R^3}$

$\textbf{Step -3: Finding the radius.}$

$\text{Volume of Cone = Volume of Sphere}$

$\dfrac{112{\pi}}{3}={\dfrac{4}{3}{\pi}R^3}$

$R^3 = 28$

$R = 3.04$

$cm$

$\textbf{Hence, The radius of Sphere = 3.04}$

$cm$

State True(1) or False(0).

The curved surface area of a frustum of a cone is $\pi l (r_1+r_2)$ and $l=\sqrt{h^{2}+\left ( r_{1} -r_{2} \right )^{2}}$ where, $r_{1}\ and\ r_{2}$ are the radii of the two ends of the frustum and $h$ is the vertical height.

Curved Surface area of a frustum of a cone

$=\pi l({ r }_{ 1 }+{ r }_{ 2 })$ where

$l$ is the slant height

$= \sqrt { { h }^{ 2 }+{ ({ r }_{ 1 }-{ r }_{ 2 }) }^{ 2 } }$ ,

$h$ is the height,

${ r }_{ 1 }$ and

${ r }_{ 2 }$ are the radii of the lower and upper ends of a frustum of a cone.

A bucket is in the form of a frustum of a cone and holds $28.490$ litres of water. The radii of the top

and bottom are $28\ cm$ and $21\ cm$ respectively. Find the height of the bucket.

Bucket holds

$28.490$ litres of water

So, volume of the bucket

$=28.490 \times 1000cm^3$

$R=28$ cm &

$r=21$ cm

Volume of frustum

$=\dfrac{\pi h}{3} (R^2+Rr+r^2)$

$\Rightarrow V=\dfrac { \pi h }{ 3 } (R^{ 2 }+Rr+r^{ 2 })$

$\Rightarrow 28490=\dfrac { \pi h }{ 3 } (28^{ 2 }+28 \times21+21^{ 2 })$

$\Rightarrow 28490=\dfrac { \pi h }{ 3 } \times 1813$

$\Rightarrow h=15$ cm

State True(1) or False(0).

The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the Fig. 12.7 is
$\pi \cfrac{r^{2}}{3}(3h-2r)$

Volume of cylinder -volume of hemisphere

$=\pi r^2 h -\frac{2}{3} \pi r^3$

$=\pi \frac{r^2}{3} (3h-2r) cm^3$

Two identical solid hemispheres of equal base radius $r$ cm are stuck together along their bases. The total surface area of the combination is $k \pi r^2.$ Find $k$ .

By joining two hemispheres along their bases we have a sphere.
Total surface area of sphere =

$4 \times \pi { r }^{ 2 }$

So, k = 4

An ice cream cone full of ice cream having radius $5$ cm and height $10$ cm as shown. Calculate the volume of ice cream (to the nearest integer, in cm $^3$ ), provided that its $\dfrac{1}{6}$ th part is left unfilled with ice cream. Insert answer in nearest integer.

Volume of ice cream

$=$ Volume of hemisphere

$+$ Volume of cone

$=\dfrac{2}{3} \pi r^3 +\dfrac{1}{3} \pi r^2 h$

Radius of hemisphere

$=$ Radius of cone

Since height of hemisphere is 5 cm, then height of cone will be

$10-5=5 cm$

$\therefore$ Volume of ice cream

$=\dfrac{2}{3} \pi (5)^3+\dfrac{1}{3} \pi (5)^2\times 5$

$=125 \pi=392.85$

$\dfrac{1}{6}$ th of ice cream=

$\dfrac{392.85}{6}=65.475$

Volume of required portion of ice cream

$=392.85-65.47=327.375 cm^3$

A milk container of height $16\ cm$ is made of metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as $8\ cm$ and $20\ cm$ respectively. If the approximate cost of milk at the rate of $Rs.\ 22$ per litre which the container can hold is Rs___________.

$Given \Rightarrow$

$h =16 cm, R=20cm, r= 8cm$

Volume of frustum

$=\dfrac{\pi h}{3} (R^2+r^2+Rr) cm^3$

$\therefore$ volume of frustum

$=\dfrac{\pi (16)}{3} ((20)^2+8^2+(20)(8)) cm^3$

$=10459.428 cm^3=10.459l$

The cost of milk per litre

$=Rs.22$

$\therefore 10.459 l=22 \times 10.459= Rs.230.10$

The curved surface area of the frustum of a cone is $180$ sq. cm and the circumference of its circular bases are $18$ cm and $6$ cm respectively. Find the slant height of the frustum of a cone (in cm).

Curved Surface area of a frustum of a cone

$=\pi l({ r }_{ 1 }+{ r }_{ 2})$ where

$l$ is the slant height,

${ r }_{ 1 }$ and

${ r }_{ 1 }$ are the radii of the lower and upper ends of a frustum of a cone.
Circumference of the lower circular base

$= 2 \pi\ {r}_{1} = 18$

$\Rightarrow {r}_{1} = \dfrac {9}{\pi}$ cm
Circumference of the upper circular base

$= 2 \pi\ {r}_{2} = 6$

$\Rightarrow {r}_{2} = \dfrac {6}{\pi}$ cm
Given, curved Surface area of the frustum of a cone

$= \pi \times l \times \left(\dfrac {9}{\pi} + \dfrac {6}{\pi}\right) = 180$

$l = 12$ cm

A container open at the top, is in the form of a frustum of a cone of height 24 cm with radii of its lower and upper circular ends as 8 cm and 20 cm respectively. Find the cost of milk (correct upto two decimal places) which can completely fill the container at the rate of rs. 21 per litre. $[ Take \; \pi=\frac {22}{7}]$

1908862_261099_ans_e3e3f132fc6d4983b33f00eec7dc68b9.jpeg

A bucket is $40cm$ in diameter at the top and $28cm$ in diameter at the bottom. Find the capacity of the bucket in litres, if it is $21cm$ deep. Also find the cost of tin sheet used in making the bucket, if the cost of tin is $Rs.\space 1.50$ per sq dm.

Given:

$\Rightarrow r_1 = 20\space cm, r_2 = 14\space cm$ and

$h = 21\space cm$

Now, the required capacity (i.e. volume) of the bucket

$= \displaystyle\frac{\pi h}{3}(r_1^2 + r_1r_2 + r_2^2)$

$= \displaystyle\frac{22\times21}{7\times3}(20^2 + 20\times14 + 14^2)$

$= 22\times576\space cm^3$

$= 19272\space cm^3$

$= \displaystyle\frac{19272}{1000} litres = 19.272\space litres$

Now,

$l = \sqrt{(r_1 - r_2)^2 + h^2}$

$l= \sqrt{(20 - 14)^2 + 21^2}$

$l= \sqrt{6^2 + 21^2}$

$l= \sqrt{36+441} = \sqrt{477} = 21.84\space cm$

$\therefore$ Total surface area of the bucket (which is open at the top)

$= \pi l (r_1 + r_2) + \pi r_2^2$

$= \pi[(r_1 + r_2)l + r_2^2]$

$= \displaystyle\frac{22}{7}[(20+14)21.84 + 14^2] = 2949.76\space cm^2$

$\therefore$ Required cost of the tin sheet at the rate of

$Rs. 1.50$ per

$dm^2$ i.e.

$100\space cm^2$

$= \displaystyle\frac{1.50\times2949.76}{100} \approx Rs. 44.25$

From a cylinder whose height is 8 cm and radius is 6 cm, a conical cavity of height 8 cm and base radius 6 cm is hollowed out. Find the volume of the remaining solid. $\displaystyle \left ( \pi =\frac{22}{7} \right )$ .

Given that, the height

$(h)$ of the cylinder is

$8\ cm$ and radius

$(r)$ is

$6\ cm$ .

From this cylinder, a conical cavity of same height and radius is cut.

To find out: Volume of the remaining solid.

Volume of the remaining solid

$=$ Volume of cylinder

$-$ Volume of the conical cavity.

We know that,

Volume of a cylinder

$=\pi r^2h$

and volume of a cone

$= \dfrac { 1 }{ 3 } \times \pi { r }^{ 2 } h$

$\therefore \$ Volume of remaining solid

$\displaystyle =\pi r^{2}h-\frac{1}{3}\pi r^{2}h$

$=\dfrac{2}{3}\pi r^{2}h$

$\Rightarrow \dfrac{2}{3}\times \dfrac{22}{7}\times 6\times 6\times 8\quad \quad \left[\because \ \pi=\dfrac{22}{7}\right]$

$=\dfrac{12672}{21}$

$\displaystyle =603.43\ cm^{3}$

Hence, the volume of the remaining solid is

$603.43\ cm^3$ .

Three cubes whose edges are $3$ cm, $4$ cm and $5$ cm, respectively are melted to from a single cube. Find the surface area of the new cube.

Let

$x$ be the edge of the new cube

The volume of the new cube= Sum of the volumes of three cubes

$\implies\displaystyle x^{3}=3^{3}+4^{3}+5^{3}$

$\displaystyle =27+64+125$

$\displaystyle =216=6^{3}$

$\implies \displaystyle x=6$

Now, surface area of new cube

$\displaystyle =6\times \left ( edge \right )^{2}$

$\displaystyle =6\times 6\times 6$

$=\displaystyle 216cm^{2}$

Three solid spheres of gold whose radii are $1$ cm, $6$ cm and $8$ cm, respectively are melted into a single solid sphere. Find the radius of the sphere.

Given that,

Three solid spheres have radii

$1\ cm,\ 6\ cm$ and

$8\ cm$ , respectively.

A single solid sphere is made by melting the three spheres.

To find out,

The radius of the new sphere.

Let the radius of the new sphere be

$x\ cm$

We know that, when one solid is constructed by melting one or more objects, the volume remains same.

Hence, volume of new sphere

$=$ sum of volumes of the three sphere.

We know that, volume of a sphere

$=\dfrac43 \pi r^3$

Hence,

$\dfrac43 \pi x^3=\dfrac43 \pi (1)^3+\dfrac43 \pi (6)^3+\dfrac43 \pi (8)^3$

$\Rightarrow x^3=(1)^3+(6)^3+(8)^3$

$\Rightarrow x^3=1+216+512$

$\Rightarrow x^3=729$

$\Rightarrow x^3=9\times 9\times 9$

$\Rightarrow x^3=9^3$

$\therefore \ x=9\ cm$

Hence, the radius of the new sphere is

$9\ cm$ .

A spherical ball of radius $3\ cm$ is melted and recast into three spherical balls. The radii of two of the balls are $1.5\ cm$ and $2\ cm.$ Find the radius of the third ball.

Let the radius of the third ball be

$x.$

Given:

$R=3\ cm$

$r_1=1.5\ cm$

$r_2=2\ cm$

$r_3=x$

The volume of the sphere with radius

$3$ will be equal to the sum of the volume of the three spheres formed. Hence,

$\begin{aligned}{}\frac{4}{3}\pi {R^3} &= \frac{4}{3}\pi {r_1}^3 + \frac{4}{3}\pi {r_2}^3 + \frac{4}{3}\pi {r_3}^3\\{\left( 3 \right)^3}& = {\left( {1.5} \right)^3} + {\left( 2 \right)^3} + {x^3}\\27 &= 3.375 + 8 + {x^3}\\{x^3} + 11.375& = 27\\{x^3} &= 15.625\\x& = 2.5\ cm\end{aligned}$

Hence, the radius of the third ball is equal to

$2.5\ cm.$

If a sphere of radius $6 \ cm$ is melted and drawn into a wire of radius $0.02\ cm$ then what is the length of the wire ?

Given,

The radius of sphere

$= 6 \ cm$

Then volume of the sphere

$=\dfrac{4}{3}\pi r^{3}$

$=\dfrac{4}{3}\pi\times (6)^{3}\ cm^3$

$=\dfrac{4}{3}\pi\times 216\ cm^3$

$=288\pi\ cm^{3}$

The radius of wire

$=0.02\ cm$

Let, the length of wire be

$h$

Then volume of wire

$=\pi r^{2}h$

$=\pi (0.02)^{2}h\ cm^3$

As per question sphere is melted into wire

So, volume of wire

$=$ volume of sphere

$\therefore \pi (0.02)^{2}h=288\pi$

$\Rightarrow 0.0004h=288$

$\Rightarrow h=\dfrac{288}{0.0004}$

$\Rightarrow h=720000$

Hence, the length of the wire is

$720000\ cm$ i.e.

$7200\ m$

A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is $14\ cm$ and the total height of the vessel is $13\ cm$ . Find the inner surface area of the vessel.

The radius(r) of the cylindrical part and the hemispherical part is the same (i.e., 7 cm)

Height of hemispherical part = Radius

$= 7$ cm
Height of cylindrical part (h)

$= 13 - 7 = 6$ cm

Inner surface are of the vessel = CSA of cylindrical part + CSA of hemispherical part
=

$2\pi rh + 2\pi r^2$

Inner surface area of vessel =

$2\times \dfrac{22}{7}\times 7 \times 6+ 2\times \dfrac{22}{7}\times 7\times 7$
=

$44(6 + 7)$
=

$44 \times 13$
=

$572 cm^2$

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to $1\ cm$ and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$ .

Given:
Height (

$h$ ) of conical part = radius (

$r$ ) of conical part

$= 1\ cm$
Radius (

$r$ ) of hemispherical part = radius of conical part

$= 1\ cm$
Volume of solid = Volume of conical part + Volume of hemispherical part

$=\dfrac{1}{3}\pi r^2h+\dfrac{2}{3} \pi r^3$

$=\dfrac{1}{3}\pi \times 1^2\times 1+\dfrac{2}{3}\pi\times 1^3$

$=\dfrac{1}{3}\pi \times 1 + \dfrac{2}{3}\pi \times 1$

$=\dfrac{\pi}{3}+\dfrac{2\pi}{3}=\pi$

$cm^3$

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is $10\ cm$ , and its base is of radius $3.5\ cm$ , find the total surface of the article.

Given:
Height of Cylinder,

$h=10\ cm$
Radius of Cylinder

$=$ Radius of Hemispheres

$= r$

$=3.5\ cm$

Surface area of article

$= CSA$ of cylindrical part

$+ 2\times CSA$ of hemispherical part

$=2\pi rh+2\times 2\pi r^2$

$=2\pi \times 3.5\times 10 + 2\times 2\pi \times 3.5^2$

$= 70\pi + 49\pi$

$=119\pi = 119 \times \dfrac{22}{7}$

$= 17 \times 22$

$= 374\ cm^2$

A solid consisting of a right circular cone of height $120$ cm and radius $60$ cm standing on a hemisphere of radius $60$ cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is $60$ cm and its height is $180$ cm.

Given:
Radius (

$r$ ) of hemispherical part = Radius (

$r$ ) of conical part

$= 60$ cm
Height (

$h_1$ ) of conical part of solid

$= 120$ cm
Height (

$h_1$ ) of cylinder

$= 180$ cm
Radius (

$r$ ) of cylinder

$= 60$ cm

Volume of water left = Volume of cylinder - Volume of solid

= Volume of cylinder - (Volume of cone + Volume of hemisphere)

$=\pi r^2 h_1 -[\dfrac{1}{3}\pi r^2 h_2+\dfrac{2}{3}\pi r^3]$

$=\pi\times 60^2 \times (180)-[\dfrac{1}{3}\pi \times 60^2 \times 120 + \dfrac{2}{3}\pi \times 60^3]$

$= \pi (60)^2[180 - (40 + 40)]$

$= \pi (3600)[100]$

$= 3,60,000\times \dfrac{22}{7}=1131428.57 \space\$ cm

$^3$

A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Given:
Diameter of hemisphere = Edge of the cube

$= l$
Radius of hemisphere

$r=\dfrac{l}{2}$
Total surface area of solid = Surface area of cubical part + Closed Surface of hemisphere part - Area of base of hemispherical part

$= 6(Edge)^2 + 2\pi r^2 - \pi r^2= 6(Edge)^2+\pi r^2$

Total surface are of solid =

$6l^2 + \pi \times \left(\dfrac{l}{2}\right)^2$
=

$6l^2 +\dfrac{\pi l^2}{4}$

$= \dfrac{1}{4}[24+\pi] l^2$ sq. units

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are $15$ cm by $10$ cm by $3.5$ cm. The radius of each of the depressions is $0.5$ cm and the depth is $1.4$ cm. Find the volume of wood in the entire stand .

Depth (h) of each conical depression

$= 1.4$ cm
Radius (r) of each conical depression

$= 0.5$ cm

$\mbox{Volume of wood = Volume of cuboid - 4 (Volume of a conical depression)}$

$= l \times b\times h-4\times \dfrac{1}{3}\pi r^2 h$

$= 15 \times 10 \times 3.5 - 4 \times \dfrac{1}{3}\times \dfrac{22}{7}\times \left(\dfrac{1}{2}\right)^2\times 1.4$

$=525 - 1.47$

$=523.53 \space\ \text{cm}^3$

A spherical glass vessel has a cylindrical neck $8\ cm$ long, $2\ cm$ in diameter; the diameter of the spherical part is $8.5\ cm$ . By measuring the amount of water it holds, a child finds its volume to be $345$ $cm^3$ . Check whether she is correct, taking the above as the inside measurements, and $\pi$ $= 3.14$ .

Given:
Height (h) of cylindrial part

$= 8$ cm

Radius (

$r_2$ ) of cylindrical part =

$\dfrac{2}{2} = 1$ cm

Radius (

$r_1$ ) of spherical part =

$\dfrac{8.5}{2}=4.25$ cm

$\mbox{Volume of vessel = Volume of sphere + Volume of cylinder}$

$=\dfrac{4}{3}\pi r_{1}^3 + \pi r_{2}^2 h$

$=\dfrac{4}{3}\pi \left(\dfrac{8.5}{2}\right)^3 +\pi (1)^2 (8)$

$= \dfrac{4}{3}\times 3.14 \times 76.765625 + 8 \times 3.14$

$= 321.392 + 25.12$

$= 346.51$ cm

$^3$

So, the child measurement is wrong.

A cubical block of side $7\ cm$ is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

As per the question we can draw the diagram,
The diagram shows that the greatest diameter possible for such hemisphere is equal to the length of cube edge, i.e.,

$7$ cm.
Radius

$(r)$ of hemispherical part

$=\dfrac{7}{2}$

$= 3.5$ cm

Total surface area of solid

$=$ Surface area of remaining cubical part

$+$ CSA of hemispherical part

$+$ Area of base of hemispherical part

$=6(Edge)^2 + 2\pi r^2- \pi r^2 \\= 6(Edge)^2+ \pi r^2$

$=6(7)^2+\dfrac{22}{7}\times \dfrac{7}{2}\times \dfrac{7}{2}$

$=294 + 38.5$

$=332.5$

$cm^2$

The frustum cone curved surface area is $210 m^2$ . The upper and lower circle area of a cone is $24$ and $15 m^2$ . Find the total surface area.

Total surface area

$=\pi (R + r)s + A_{1} + A_{2}$
Curved surface area

$=\pi (R + r)s = 210$

$A_{1} = 24 m^2$ ,

$A_{2} = 15 m^2$
Therefore, total surface area

$= 210 + 24 + 15 = 249 m^2$

Find the volume of a frustum cone, whose base and upper area of a circle is $40cm^2$ and $90cm^2$ . (Insert answer in nearest integer).

Volume of a frustum cone =

$\cfrac{1}{3}(A_1+A_2+\sqrt{A_{1}A_{2}})h$
=

$\cfrac{1}{3}(40 + 90 + \sqrt{40\times 90})10$
=

$\cfrac{1}{3}(130 + 60)10$
=

$633.33 cm^3$

The total surface area of frustum cone is $1200 ft^2$ . The radius of a top cone is $3$ and the bottom cone is $6$ ft. What is the curved surface area of a frustum of cone?

Total surface area

$=\pi (R + r)s + \pi R^2 + \pi r^2$
Curved surface area

$=\pi (R + r)s$
Therefore,

$1200 =$ Curved surface area

$+\pi (6^2 + 3^2)$

$1200 =$ Curved surface area

$+ 3.14 \times (36 + 9)$

$1200 -141.3 =$ Curved surface area
Curved surface area

$= 1058.7$ ft.

The radii of the circular ends of a frustum of height 4 cm are 7 cm and 4 cm respectively. Find the total surface area. (Use $\pi = 3$ ).

Slant height

$=\sqrt{(R-r)^2+h^2}$

$=\sqrt{(7-4)^2+4^2}$

$=\sqrt{9+16}$

$l = 5$ cm
Total surface area

$=\pi (R+r)l+\pi R^2+\pi r^2$

$=\pi[(7+4)5+49+16]$

$=3[120]$

$=360 cm^2$

Find the volume of the frustum cone whose base and top radius is $8$ cm and $4$ cm respectively. The height of the cone is $5$ cm. (Use $\pi = \dfrac{22}{7}$ ).
(Insert answer in nearest integer).

Volume of the frustum cone =

$\dfrac{1}{3}\pi h(R^2+r^2+Rr)$

$=\cfrac{1}{3}\pi \times 5(8^2+4^2+32)$

$=\cfrac{1}{3}\pi \times 560$

$=\cfrac{1}{3}\times \cfrac{22}{7}\times 280$

$=586.66 cm^3$

The base and top radius of a cone is $2.1$ m and $0.8$ m respectively. The height of the cone is $24$ m. What is the volume of frustum of a cone? (Use $\pi = 3.14$ ). Insert answer in nearest integer,

Volume of the frustum cone =

$\cfrac{1}{3}\pi h(R^2+r^2+Rr)$

$=\cfrac{1}{3}\pi \times 24(2.1^2+0.8^2+1.68)$

$=\cfrac{1}{3}\pi \times 161.52$

$=\cfrac{1}{3}\times 3.14 \times 161.52$

$=169.05 m^3$

A cone has base radius cm is cut from the top. Calculate the volume of the frustum. Insert answer in terms of $\pi$ .

Volume of the cone missing from the top =

$\cfrac{1}{3}\pi r^2h = \cfrac{1}{3}\pi \times 2^2\times 6$
=

$8\pi$

The larger cone is similar to the smaller cone.

So, the volume of the whole cone including the missing top which shouldn't really be there is

$\cfrac{1}{3}\pi \times 5^2\times 6 =50\pi$

The volume of the frustum =

$50\pi - 8 \pi = 42 \pi$

The figure shows a 3-D view of a metal nut. The metal nut has two regular hexagonal faces. The side of a hexagonal face is $2\ cm$ and has a thickness of $1\ cm.$ A hole of a diameter of $2\ cm$ is present in the metal nut. Given the density of the metal is $7.9$ grams per cubic cm., calculate the mass of this nut.

A regular hexagon can be divided into

$6$ equilateral triangles as shown

The area of an equilateral triangle

$=\dfrac { 1 }{ 2 } (b)\left( \dfrac { b\sqrt { 3 } }{ 2 } \right)$

$=\dfrac { 1 }{ 2 } (2)\left( \dfrac { 2\sqrt { 3 } }{ 2 } \right)$

$=\sqrt { 3 }$

There are

$6$ of those, so the area of the hexagon is

$6\sqrt { 3 }$ square cms.

The diameter of the hole is

$2\ cm$ , so it has a radius of

$1\ cm$

Therefore the area of the hole

$=π$ square cms.

$\therefore$ Area of the hexagonal face of the nut

$=6\sqrt { 3 } -π$ square cms.

Hence volume of the nut

$=(6\sqrt { 3 } -π)\times 1$

$=6\sqrt { 3 } -π$ cubic cms.

We know that

$\mathrm{Density=\dfrac{Mass}{Volume}}$

Given, the density of nut

$=7.9$ grams per cubic cm

$\Rightarrow 7.9=\dfrac { m }{ 6\sqrt { 3 } -π }$

$\Rightarrow 7.9(6\sqrt { 3 } -π)=m$

$\Rightarrow 57.2706=m$

$\Rightarrow m=57$

Hence, the mass of the nut is

$57$ grams (approx).

A container open at the top, is in the form of a frustum of a cone of height $24$ cm with radii of its lower and upper circular ends, as $8$ cm and $20$ cm respectively. Find the cost of milk which can completely fill the container at the rate of $21$ per litre. [Use $\displaystyle \pi =\frac { 22 }{ 7 }$ ].

Given: $h = 24$ cm

Upper radius = $r_1=20$

Lower radius = $r_2=8$

Volume of frustum of a cone = $\cfrac{1}{3}\pi h [r_1^2+r_2^2+r_1r_2]$

= $\cfrac{1}{3}\times \cfrac{22}{7}\times 24[20^2+8^2+20\times 8]$

= $15689.14$ $cm^3$

= $15.69$ litre

The cost of milk which can completely fill the container at the rate of $Rs. 21$ per litre = $21 \times 15.69 = Rs. 329.49$

The slant height of the frustum of a cone is $4$ cm, and the perimeter of its circular bases are $18$ cm and $6$ cm respectively. Find the curved surface area and total surface area of the frustum.

Given,

$2\pi r_1=18cm,$

$2\pi r_2=6cm, l=4cm$

$\therefore r_1=\displaystyle\frac{9}{\pi}cm$ ,

$r_2=\displaystyle\frac{3}{\pi}$ cm
Curved surface area of the frustum

$=\pi(r_1+r_2)l$

$=\pi\left(\displaystyle\frac{9}{\pi}+\frac{3}{\pi}\right)4=\pi\times \displaystyle\frac{12}{\pi}\times 4$

$\therefore$ CSA of the frustum

$=48cm^2$
Total surface area of the frustum

$=\pi(r_1+r_2)l+\pi r^2_1+\pi r^2_2$

$=\displaystyle 48+\pi\times \frac{9}{\pi}\times \frac{9}{\pi}+\pi\times \frac{3}{\pi}\times \frac{3}{\pi}=48+\frac{81}{\pi}+\frac{9}{\pi}$

$=\displaystyle 48+\frac{81\times 7}{22}+\frac{9\times 7}{22}=48+25.8+2.9$

$\therefore$ TSA of the frustum

$=76.7cm^2$ .

The radii of two circular ends of a frustum shaped dust bin are $15$ cm and $8$ cm. If its depth is $63$ cm, find volume of the dust bin.

Given

$r_1=15$ cm,

$r_2=8$ cm,

$h=63$ cm
Volume of the dustbin (frustum)

$=\displaystyle\frac{1}{3}\pi h(r^2_1+r^2_2+r_1r_2)$

$=\displaystyle\frac{1}{3}\times \frac{22}{7}\times 63(15^2+8^2+15\times 8)$

$=\displaystyle\frac{1}{3}\times \frac{22}{7}\times 63(225+64+120)$

$=66\times 409=26, 994$

$\therefore$ Volume of the dustbin

$=26, 994cm^3$ .

From the top of a cone of base radius $12$ cm and height $20$ cm, a small cone of base radius $3$ cm is to be cut off. How far down the vertex is the cut to be made? Find the volume of the frustum so obtained.

Given,

$r_1=12cm, h_1=20cm, r_2=3cm, h_2=?$
We know,

$\displaystyle\frac{r_1}{r_2}=\frac{h_1}{h_2}$

$\therefore \displaystyle\frac{12}{3}=\frac{20}{h_2}$

$\therefore h_2=\displaystyle\frac{3\times 20}{12}=5cm$

$\therefore$ The cut to be made at

$5$ cm from the vertex.
Volume of the frustum

$=\displaystyle\frac{1}{3}\pi h(r^2_1+r^2_2+r_1r_2)$

$=\displaystyle\frac{1}{3}\times \frac{22}{7}\times 15(12^2+3^2+12\times 3)=\frac{110}{7}\times (144+9+36)$

$=\displaystyle\frac{110}{7}\times 189=2,970$

$\therefore$ Volume of the frustum

$=2,970cm^3$ .

A flower vase is in the form of a frustum of a cone. The perimeter of the ends are $44$ cm and $8.4\pi$ cm. If the depth is $14$ cm, find how much water it can hold?

A flower vase is in the form of frustum of a cone.

The perimeter of its bases are

$44\;\text{cm}$ and

$8.4\pi\text{ cm}$ .

${ C }_{ 1 }=44cm\Rightarrow 2\pi { r }_{ 1 }=44cm\Rightarrow { r }_{ 1 }=\dfrac { 44\times 7 }{ 44 } =7cm$

${ C }_{ 2 }=8.4\pi \Rightarrow 2\pi { r }_{ 2 }=8.4\pi \Rightarrow { r }_{ 2 }=\dfrac { 8.4 }{ 2 } =4.2cm$

Volume of frustum of flower vase

$=\dfrac { 1 }{ 3 } \pi \left( { r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+{ r }_{ 1 }{ r }_{ 2 } \right) \times h$

$=\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times 14\left( { 7 }^{ 2 }+{ 4.2 }^{ 2 }+7\times 4.2 \right) =\dfrac { 44 }{ 3 } \times 96.04$

$= 1408.58$

${ cm }^{ 3 }$

A vessel is in the form of a frustum of a cone. Its radius at one end is $8\ cm$ and the height is $14\ cm$ . If its volume is $\displaystyle\dfrac{5676}{3}\ cm^3$ , find the radius of the other end.

$R = 8\ cm, h = 14\ cm$ ,
Volume of frustum

$=\dfrac { 5676 }{ 3 }\ { cm }^{ 3 }$

$\Rightarrow \quad \dfrac { 1 }{ 3 } \pi h\left( { R }^{ 2 }+{ r }^{ 2 }+R\times r \right) =\dfrac { 5676 }{ 3 }$

$\Rightarrow \dfrac { 22 }{ 7 } \times 14\left( { 8 }^{ 2 }+{ r }^{ 2 }+8r \right) =5676$

$\Rightarrow \quad 64+{ r }^{ 2 }+8r=\dfrac { 5676 }{ 44 } =129\quad$

$\Rightarrow { r }^{ 2 }+8r+64-129=0$

$\Rightarrow \quad { r }^{ 2 }+8r-65=0\quad$

$\Rightarrow { r }^{ 2 }+13r-5r-65=0\quad$

$\Rightarrow \left( r+13 \right) \left( r-5 \right) =0$

$\therefore$

$r = -13 \ and\ 5$ , neglect negative value.

$\therefore$

$r = 5$

Hence radius of other end is

$5$ cm.

A petrol tank is in the shape of a cylinder with hemispheres of same radius attached to both ends. If the total length of the tank is $6$ m and the radius is $1$ m, what is the capacity of the tank in litres.

Tank is in the shape of a cylinder with hemispheres of same radius attached to both ends.

Given : Length of the tank

$=6\ m$

Radius

$r=1\ m$ is same for all shapes

Height of hemispheres

$=$ Radius

$(r)=1\ m$

$\therefore$ Height of the cylinder

$(h)=6-2=4\ m$

And radius of hemisphere

$=1\ m$

Now, volume of the tank

$=$ volume of two hemipheres

$+$ volume of a cylinder

Volume of a cylinder

$=\pi r^2 h$

$=\pi (1)^2\times 4$

$=4 \pi$

Volume of a hemisphere

$=\dfrac{2}{3}\pi r^3$

$=\dfrac{2}{3}\pi (1)^3$

$=\dfrac{2 \pi}{3}$

$\therefore$ Volume of the tank

$=2\times \dfrac{2\pi}{3}+4\pi$

$=\dfrac{16\pi}{3}$

$=16.75516\ m^3$

$=16577.16$ litres

A storage tank consists of a circular cylinder with a hemisphere adjoined on either ends. The external diameter of the cylinder is $1.4$ m and length is $8$ m, find the cost of painting it on the outside at the rate of Rs. $10$ per $m^2$ .

External diameter of cylinder

$= 1.4m$

$\therefore$ External radius =

$\dfrac { 1.4 }{ 2 } =0.7m$

or Radius of hemisphere = 0.7 m

Height of cylinder, h = 8 m

Total outer area to be painted =

$2\pi rh+\pi { r }^{ 2 }+2\pi { r }^{ 2 }$

$\pi r\left( 2h+3r \right)$

$\dfrac { 22 }{ 7 } \times 0.7\left( 2\times 8+3\times 0.7 \right)$

$2.2\times 18.1=38.82{ m }^{ 2 }$

Cost of painting = Rs.

$\left( 38.82\times 10 \right) =Rs.388.2$

A solid is composed of a cylinder with hemispherical ends. If the whole lengthe of the solid is $104$ cm and the radius of each hemispherical end is $7$ cm, find the cost of polishing its surface at the rate of Rs. $4$ per $100cm^2$ .

Hemispherical part:

$r=7cm$
Cylindrical part:

$r=7cm, h=104-(7+7)=90cm$
Total surface area of the solid

$=$ CSA of cylindricall part

$+2\times$ CSA of hemispherical part

$=2\pi rh+2\times 2\pi r^2=2\pi r(h+2r)$

$=2\times \displaystyle\frac{22}{7}\times 7[90+2(7)]cm^2$

$=44\times 104cm^2=4576cm^2$

$\therefore$ TSA of the solid

$=4.576cm^2$
Cost of polishing at the rate of Rs.

$4$ per

$100cm^2$

$=4576\times\displaystyle\frac{Rs.4}{100}=Rs.183.04$

$\therefore$ Cost of polishing the solid

$=Rs. 183.04$ .

A cup is in the form of a hemisphere surmounted by a cylinder. The height of the cylindrical portion is $8\ cm$ and the total height of teh cup is $11.5\ cm.$ Find the total surface area of the cup.

Given:

Height of the cylinder

$=8\ cm$

Total height of the cup

$=11.5\ cm$

Height of hemisphere

$=11.5-8$

$=3.5\ cm$

The radius of the hemisphere will be equal to the height of the hemisphere that is

$3.5\ cm.$

Radius,

$r=3.5\ cm$

Now, total surface area of the cup

$=$ Curved surface area of cylinder

$+$ Curved surface area of hemisphere

$=2\pi rh+2\pi r^2$

$=2\pi r(h+r)$

$=2\pi\times 3.5(8 + 3.5)$

$=7\pi(11.5)$

$=253\ cm^2$

Hence, the total surface area of the cup is

$253\ cm^2.$

From the top of a cone of base radius $24$ cm and height $45$ cm, a cone of slant height $17$ cm is cut off. What is the volume of the remaining frustum of the cone?

${ L }^{ 2 }={ R }^{ 2 }+{ H }^{ 2 }\Rightarrow L=\sqrt { { 24 }^{ 2 }+{ 45 }^{ 2 } } =\sqrt { 2601 } =51cm$

we know that,

$\triangle OAB\sim \triangle OCD$ , therefore

$\dfrac { r }{ 24 } =\dfrac { h }{ 45 } =\dfrac { 17 }{ 51 }$

$\Rightarrow \quad \dfrac { r }{ 24 } =\dfrac { 17 }{ 51 } \Rightarrow r=\dfrac { 24 }{ 3 } =8cm$

$\Rightarrow \quad \dfrac { h }{ 45 } =\dfrac { 17 }{ 51 } \Rightarrow h=\dfrac { 45 }{ 3 } =15cm$

$\therefore$ Height of frustum

$= 45 - 15 = 30cm$

$\therefore$ Volume of frustum =

$\dfrac { 1 }{ 3 } \pi \left( { R }^{ 2 }+{ r }^{ 2 }+Rr \right) h=\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times 30\left( { 24 }^{ 2 }+{ 8 }^{ 2 }+24\times 8 \right)$

$\dfrac { 220 }{ 7 } \times 832=26148.57{ cm }^{ 3 }$

A toy is in the form of a cone mounted on a hemisphere with the same radius. The diameter of the conical portion is $6$ cm and its height is $4$ cm. Determine the surface area and volume of the solid.

Conical portion

$r=3cm, h=4cm$

$l^2=r^2+h^2=3^2+4^2=9+16$

$l^2=25$

$l=\sqrt{25}=5$ cm
Spherical portion

$r=3cm$
Surface area of the toy

$=$ CSA of hemisphere

$+$ CSA of cone

$=2\pi r^2+\pi rl=2\times\displaystyle\frac{22}{7}\times 3\times 3cm^2+\frac{22}{7}\times 3\times 5cm^2$

$=\displaystyle\frac{396}{7}cm^2+\frac{330}{7}cm^2=\frac{726}{7}cm^2=103.71cm^2$
Surface area of the toy

$=103.71cm^2$
Volume of the toy

$=$ Volume of hemisphere

$+$ Volume of cone

$=\displaystyle\frac{2}{3}\pi r^3+\frac{1}{3}\pi r^2h=\displaystyle\frac{2}{3}\times \frac{22}{7}\times 3\times 3\times 3cm^3+\frac{1}{3}\times \frac{22}{7}\times 3\times 3\times 4cm^3$

$=\displaystyle\frac{396}{7}cm^3+\frac{264}{7}cm^3=\frac{660}{7}cm^3$

$\therefore$ Volume of the toy

$=94.28cm^3$ .

A wooden toy is in the form of a cone surmounted on a hemisphere. The diameter of the base of the cone is $16$ cm and its height is $15$ cm. Find the cost of painting the toy at $7$ per Rs. $100cm^2$ .

Total area to be painted =

$\pi rl+2\pi { r }^{ 2 }$

$\pi r\left( l+2r \right)$

$\dfrac { 22 }{ 7 } \times 8\left( \sqrt { { r }^{ 2 }+{ h }^{ 2 } } +2\times 8 \right)$

$\dfrac { 176 }{ 7 } \left( \sqrt { { 8 }^{ 2 }+{ 15 }^{ 2 } } +16 \right)$

$\dfrac { 176 }{ 7 } \times \left( 17+16 \right) =\dfrac { 176\times 33 }{ 7 } =829.71{ cm }^{ 2 }$

Cost of painting = Rs.

$\left( \dfrac { 7 }{ 100 } \times 829.71 \right) =Rs.58.08$

A rocket is in the shape of a cylinder with a cone attached to one end and a hemisphere attached to the other. All of them are of the same radius of $1.5$ m. The total length of the rocket is $7$ m and height of the cone is $2$ m. Calculate the volume of the rocket.

Given :

Radius

$r=1.5\ m$ , Total length =

$7\ m$

Height of the cone

$(h_1)=2\ m$

Height of hemisphere

$=$ Radius of hemisphere

$=1.5\ m$

Height of cylinder

$(h_2)=$ Total length

$-h_1-r=7-2-1.5=3.5\ m$

Now, volume of the rocket

$=$ Volume of the cylinder

$+$ volume of a cone

$+$ volume of a hemisphere ........

$(i)$

We know that, Volume of the cylinder

$=\pi r^2 h_2$

$=\pi(1.5)^2\times 3.5$

$=7.875\pi\ m^3$

Volume of a cone

$=\pi r^2 \dfrac{h_1}{3}$

$=\pi(1.5)^2\times \dfrac{2}{3}$

$=1.5\pi\ m^3$

Volume of a hemisphere

$=\dfrac{2}{3}\pi r^3$

$=\dfrac{2}{3}\pi\times (1.5)^3$

$=2.25\pi\ m^3$

From

$(i)$

Volume of the rocket

$=7.875\pi+1.5\pi+2.25\pi$

$=11.625\pi\ m^3$

$=36.521\ m^3$

A bucket is in the shape of a frustum with the top and bottom circles of radii $15$ cm and $10$ cm. Its depth is $12$ cm. Find its curved surface area and total surface area. (Express the answer in terms of $\pi$ ).

$R = 15cm, r = 10cm, h = 12cm$ $

$l=\sqrt { { \left( R-r \right) }^{ 2 }+{ h }^{ 2 } } =\sqrt { { \left( 15-10 \right) }^{ 2 }+{ 12 }^{ 2 } } =\sqrt { 169 } =13cm$

CSA =

$\pi \left( R+r \right) l=\pi \left( 15+10 \right) \times 13=325\pi { cm }^{ 2 }$

TSA =

$\pi l\left( R+r \right) +\pi \left( { R }^{ 2 }+{ r }^{ 2 } \right) =325\pi +\pi \left( { 15 }^{ 2 }+{ 10 }^{ 2 } \right)$

$325\pi +325\pi =650\pi { cm }^{ 2 }$

Two similiar cones have volumes 12 $\pi$ cu. units and 96 $\pi$ cu. units. If the curved surface area of the smaller cone is 15 $\pi$ sq. units, what is the curved surface area of the larger one

Let the radius of cone one is

$r$ and cone two is

$R$ and height of cone one is

$h$ and cone two is

$H$

We know that volume of cone =

$\dfrac{1}{3}\pi r^{2}h$

Taking ratios of volume of two cones

$\therefore \dfrac{12\pi }{96\pi }=\dfrac{\pi r^{3}}{\pi R^{3}}$

$\Rightarrow \dfrac{1}{8}=\dfrac{r^{3}}{R^{3}}$

$\Rightarrow \dfrac{1}{4}=\dfrac{r^{2}}{R^{2}}$

$\dfrac{CAS_{1}}{CAS_{2}}=\dfrac{\pi rl}{\pi RL}=\dfrac{r^{2}}{R^{2}}$ (the cone are similar )

$\therefore \dfrac{15\pi }{CAS_{2}}=\dfrac{1}{4}$

$\Rightarrow CAS_{2}=4\times 15\pi =60 \pi$ sq unit

A bucket is in the form of frustum of cone and holds $28.490$ litres of water. The radii of the top and bottom are $28$ cm and $21$ cm respectively. Find the height of the bucket.

Given : Volume

$=28.490$ litres

$=28490{ cm }^{ 3 }$
Radius of the top of cone

$={ r }_{ 1 }=28$ cm.
Radius of the bottom of cone

$={ r }_{ 2 }=21$ cm.
Find : The height of the bucket.
Volume of frustrum

$=\dfrac { 1 }{ 3 } \pi \left( { r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+{ r }_{ 1 }\times { r }_{ 2 } \right) h$

$\therefore 28490=\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times \left( { 28 }^{ 2 }+{ 21 }^{ 2 }+28\times 21 \right) \times h$

$\therefore 28490=\dfrac { 22 }{ 21 } \times \left( 784+441+588 \right) \times h$

$\therefore h=\dfrac { 28490\times 21 }{ 22\times 1813 } =15$ cm.

$\therefore$ height of the bucket

$=15$ cm.

A test tube has diameter $20\ mm$ and height is $15\ cm$ . The lower portion is a hemisphere. Find the capacity of the test tube. $(\pi = 3.14)$

Given that, a test tube has a diameter of

$20\ mm$ and height of

$15\ cm$ . The lower portion of the test tube is a hemisphere.

To find out: The capacity of the test tube.

We can see that the test tube combines two shapes, a hemispherical bottom and cylindrical body.

Radius of hemisphere

$= \dfrac {20}{2}mm = 1\ cm\quad \quad [\because \ 1\ cm=10\ mm]$
We know that, the volume of a hemisphere

$= \dfrac {2}{3}\pi r^{3}$

$\therefore \$ Volume of the hemispherical bottom of the test tube

$= \dfrac {2}{3} \times 3.14\times (1)^{3}$

$\Rightarrow 2.09\ cm^{3}$

Now, height of the cylindrical portion

$=$ height of the test tube

$-$ radius of the hemispherical bottom

$\Rightarrow 15 - 1 = 14\ cm$

We know that volume of a cylinder

$= \pi r^{2}h$

$\therefore \$ Volume of the cylindrical portion

$= 3.14\times (1)^{2} \times 14$

$\Rightarrow 43.96 \ cm^3$

Capacity of the test tube

$=$ Volume of cylindrical portion

$+$ Volume of hemispherical portion

$\therefore \$ Capacity

$= 43.96 + 2.09$

$\Rightarrow 46.05\ cm^{3}$

Hence, the capacity of the given test tube is

$46.05\ cm^{3}$ .

The volume of a cone is the same as that of volume of a cylinder whose height is 9 cm and diameter 40 cm. Find the radius of the base of the cone if its height is 108 cm.

Let

$r_1, h_1$ and

$r_2, h_2$ be the radius and height of of cone and cylinder respectiely.

Given,

$h_2 = 9 cm, d_2 = 40 cm$

$r_2 = \dfrac{40}{2} = 20 cm, h_1 = 108 cm$ .
Volume of the cone

$=$ Volume of the cylinder.

$\dfrac{1}{3} \pi r_1^2 h_1 = \pi r_2^2 h_2$

$\dfrac{1}{3} \pi r_1^2 (108) = \pi (20)^2 (9)$

$r_1^2 (108) = 3 \times 9 \times (20)^2$

$r_1^2 = \dfrac{3 \times 9 \times 400}{108}$

$r_1^2 =100 cm$ .

$r_1 = 10 cm$
Thus, the radius of the base of the cone is

$10 cm$ .

Two cubes each of volume $27\ cm^{3}$ are joined end to end to form a solid. Find the surface area (cm $^2$ ) of the resulting cuboid.

Volume of one cube,

$v=27\ {cm}^3$

Therefore, side of a cube

$s = v^{1/3} = {27}^{1/3} = 3\ cm$

Two cubes of the same volume joined together to form a cuboid.

Therefore, the length of cuboid,

$L = 2s = 6\ cm$

Width of cuboid (B) = Height of cuboid (H)

$=s = 3 \ cm$

Surface Area of cuboid,

$S.A = 2(LB+BH+LH) = 2(6\times3 + 3\times3 + 6\times3) = 2(18+9+18) = 90\ {cm}^2$

The perimeter of the ends of a frustum of a cone are $44$ cm and $8.4$ $\pi$ cm. If the depth is $14$ cm, then find its volume.

Height

$h=14$ cm
Given,

$2\pi R=44$ cm and

$2\pi r=8.4\pi$

$\Rightarrow 2\times \dfrac { 22 }{ 7 } \times R=44$ and

$2\times \dfrac { 22 }{ 7 } \times r=8.4$

$\Rightarrow R=7$ cm and

$r=4.2$ cm
Volume of the frustum

$=\dfrac { 1 }{ 3 } \pi h\left( { R }^{ 2 }+{ r }^{ 2 }+Rr \right)$

$=\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times 14=\left( { 7 }^{ 2 }+{ \left( 4.2 \right) }^{ 2 }+7\left( 4.2 \right) \right)$

$=\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times 14=\left( 49+17.64+29.4 \right)$

$=\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times 14\times 96.04=1408.58{ cm }^{ 3 }$

$=1408.6{ cm }^{ 3 }$

The radii of two circular ends of a frustum of a cone shaped dust bin are $15\ cm$ and $8\ cm$ . If its depth is $63\ cm$ , find the volume of the dust bin.

Solution:

Let

$r_1=15\ cm, r_2=8\ cm$ and

$h=63\ cm$

Volume of the dustbin (frustrum)

$=\cfrac13\times\pi\times h\times (r_1^2+r_2^2+r_1r_2)$

$=\cfrac13\times \cfrac{22}{7}\times63\times(15^2+8^2+15\times8)$

$=66(225+64+120)\ cm^3$

$=66\times409\ cm^3$

$=26994\ cm^3$

The radii of two circular ends of a frustum shaped bucket are $15$ cm and $8$ cm. If its depth is $63$ cm, find the capacity of the bucket in litres. (Take $\pi = \dfrac{22}{7}$ )

Here,

$h = 63$ cm,

$R = 15$ cm,

$r = 8$ cm
The volume of frustum

$= \dfrac{1}{3} \pi h (R^2 + r^2 + Rr)$ Square cm

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 63 (15^2 + 8^2 + 15(8))$

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 63 (225 + 64 + 120)$

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 63 \times 409 = 26994$ cubic cm

$= \dfrac{26994}{1000}$ litre

$= 26.994$ litre

Therefore, capacity of the bucket in litres is

$26.994$ .

The curved surface area of a cone is $1159\dfrac{5}{7}\ cm^2$ .Area of its base is $254\dfrac{4}{7}\ cm^2$ .Find its volume.

Given curved surface area of the cone is

$1159\cfrac{5}{7}cm^{2}$ and the area of its base is

$254\cfrac{4}{7}m^{2}$

Then area of base =

$\pi r^{2}=254\cfrac{4}{7}=\cfrac{1782}{7}$

$\Rightarrow \cfrac{22}{7}r^{2}=\cfrac{1782}{7}$

$\Rightarrow r^{2}=\cfrac{1782}{22}=81$

$\Rightarrow r=9$ cm

The curved surface area

$= \pi rl =1159\cfrac{5}{7}$

$\Rightarrow \cfrac{22}{7}\times 9\times l=\cfrac{8118}{7}$

$\Rightarrow l=\cfrac{8118}{22\times 9}=41$ cm

Then height of cone =

$\sqrt{(41)^{2}-(9)^{2}}=\sqrt{1681-81}=\sqrt{1600}=40$ cm

So, volume of cone =

$\cfrac{1}{3}\pi r^{2}h=\cfrac{1}{3}\times \cfrac{22}{7}\times (9)^{2}\times 40$

$=\cfrac{22\times 81\times 40}{3\times 7}=\cfrac{71280}{21}=3394\cfrac{2}{4} cm^{3}$

A solid is in the form of one mounted on a right circular cylinder both having same radii of their bases. The base of the cone is placed on the top base of the cylinder. If the radius of the base and height of the cone be $7\ cm$ and $10\ cm$ respectively and the total height of the solid be $30\ cm$ , find the volume of the solid.

Given : Radius of a cone

$(r_{1})=7\ cm$ , height of the cone

$(h_1)=10\ cm$

Height of the cylinder

$(h_{2})=20\ cm$ .

Area of base of the cone

$(b)=\pi r_{1}^2=3.14\times (7)^2=153.86\ cm^2$

Volume of the cone

$=\dfrac{1}{3}\times b\times h_{1}$

$=\dfrac{1}{3}\times 153.86\times 10=512.8667\ cm^3$

Volume of the cylinder

$=\pi r_{1}^2 h_{2}$

$=3.14\times 49\times 20=3077.2\ cm^3$

Volume of the solid

$=$ volume of the cone

$+$ volume of a cylinder

$=512.8667+3077.2=3590.06\ cm^3$

$\therefore$ Volume of the solid

$=3590.06\ cm^3$

A solid wooden toy is in the form of a cone surmounted on a hemisphere. If the radii of the hemisphere and the base of the cone are $3.5\text{ cm}$ each and the total height of the toy is $17.5\text{ cm}$ , then find the volume of wood used in the toy. Use $\pi =\cfrac { 22 }{ 7 }$

Hemispherical portion:

Radius

$r=3.5\text{ cm}$

Conical portion:

Radius

$r=3.5\text{ cm}$

Heignt,

$h=17.5-3.5=14\text{ cm}$

Volume of the wood

$=$ Volume of the hemisphere

$+$ Volume of the cone

$=\cfrac { 2 }{ 3 } \pi { r }^{ 3 }+\cfrac { 1 }{ 3 } \pi { r }^{ 2 }h$

$=\cfrac { \pi { r }^{ 3 } }{ 3 } (2r+h)$

$=\cfrac { 22 }{ 7 } \times \cfrac { 3.5\times 3.5 }{ 3 } \times (2\times 3.5+14)$

$=269.5\text{ cm}^3$

Hence, the volume of the wood used in the toy is equal to

$269.5\text{ cm}^3$

1073939_622542_ans_95f70e0ea4c9451cbb15d87aad305a2d.PNG

On a Hemisphere, frustrum of a cone shaped shuttle-cock is used for playing Badminton. The outer radius of the frustrum of cone is $5cm$ and its inner radius is $2cm$ . The height of the entire shuttle-cock is $7cm$ . Find the outer surface area of the shuttle-cock.

Radius of hemisphere

$r=$ smaller radius of frustrum of cone

$=2cm$
For the frustrum of cone

${r}_{1}=$ bigger radius

$=5cm$

${r}_{2}=$ smaller radius

$=2cm$
Height

$(h)=$ total height

$-$ radius of hemisphere

$=7-2=5\ cm$
Now,

$l=\sqrt { { h }^{ 2 }+{ \left( { r }_{ 1 }-{ r }_{ 2 } \right) }^{ 2 } } =\sqrt { { 5 }^{ 2 }+{ (5-2) }^{ 2 } } =\sqrt { 25+9 } =\sqrt { 34 }$

$l=5.83\ cm=$ slant height of the frustrum of a cone.
External surface area of the shuttle cock

$=$ Curved Surface Area of frustum of a cone

$+$ Curved Surface Area of hemisphere

$=\pi l\left( { r }_{ 1 }+{ r }_{ 2 } \right) +2\pi { r }^{ 2 }$

$=\pi \left[ l\left( { r }_{ 1 }+{ r }_{ 2 } \right) +2{ r }^{ 2 } \right]$

$=\cfrac { 22 }{ 7 } \left[ 5.83(5+2)+2(4) \right]$

$=\cfrac { 22 }{ 7 } \left[ 5.83\times 7+8 \right]$

$=\cfrac { 22 }{ 7 } \times 48.81=153.40\ { cm }^{ 2 }$

Hence, outer surface area is

$153.40\ cm^2$

A cuboid shaped slab of iron whose dimensions are $55cm\times 40cm\times 15cm$ is melted and reacast into a pipe. The outer diameter and thickness of the pipe are $8cm$ and $1cm$ respectively. Find the length of the pipe. (Take $\pi =\cfrac { 22 }{ 7 }$ )

Let

${h}_{1}$ be the length of the pipe.

Let

$R$ and

$r$ be the outer and inner radii of the pipe respectively.

Iron slab:

Length

$=55cm$

Breadth

$=40cm$

Height

$=15cm$

Volume

$=lbh=55\times 40\times 15\ cm^3$

Iron pipe:

Outer diameter,

$2R=8cm$

$\therefore$ Outer radius,

$R=4cm$

Thichness,

$w=1cm$

$\therefore$ Inner radius

$r=R-w=(4-1)cm=3cm$

Now,

The volume of the iron pipe

$=$ Volume of iron slab

$\Rightarrow \pi { h }_{ 1 }(R^2-r^2)=lbh$

$\Rightarrow \cfrac { 22 }{ 7 } \times { h }_{ 1 }(4^2-3^2)=55\times 40\times 15$

$\Rightarrow \cfrac { 22 }{ 7 } \times { h }_{ 1 }\times 7=55\times 40\times 15$

$\Rightarrow 22 \times { h }_{ 1 }=55\times 40\times 15$

$\Rightarrow { h }_{ 1 }=\dfrac{55\times 40\times 15}{22}$

$\Rightarrow { h }_{ 1 }=1500$

Thus, the length of the pipe,

${ h }_{ 1 }=1500cm=15m$

1035306_622552_ans_b15a38cf183d48369035af8554133229.png

A vessel is in the form of a frustum of a cone. Its radius at one end and the height are $8cm$ and $14cm$ respectively. If its volume is $\cfrac{5676}{3}{cm}^{3}$ , then find the radius(in cm) at the other end.

We have,

$R=8\ cm$

$h=14\ cm$

Let the other end radius be

$r$ .

Volume of the frustum cone

$=\dfrac{5676}{3}\ cm^3$

We know that the volume of the frustum cone

$=\dfrac{1}{3}\pi (R^2+r^2+R r)h$

So,

$\dfrac{1}{3}\times \dfrac{22}{7} (8^2+r^2+8\times r)\times 14=\dfrac{5676}{3}$

$22(64+r^2+8\times r)\times 2=5676$

$44\times(64+r^2+8\times r)=5676$

$64+r^2+8\times r=129$

$r^2+8r-65=0$

$(r+13)(r-5)=0$

$r=5, -13$

Since,

$r$ can not be negative.

Hence, the radius of other end is

$5\ cm$ .

The radii of two circular ends of a frustum shaped bucket are $15\text{ cm}$ and $8\text{ cm}$ . If its depth is $63\text{ cm}$ , find the capacity of the bucket in litres. (Take $\pi =\cfrac { 22 }{ 7 }$ )

Let

$R$ and

$r$ are radii of the circular ends at the top and bottom and

$h$ be the depth of the bucket respectively.
Given that

$R=15\text{ cm},r=8\text{ cm}$ and

$h=63\text{ cm}$
The volume of the bucket (frustum)

$=\cfrac { 1 }{ 3 } \pi ({ R }^{ 2 }+{ r }^{ 2 }+Rr)\times h$

$=\cfrac { 1 }{ 3 } \times \cfrac { 22 }{ 7 } \times 63\times \left( { 15 }^{ 2 }+{ 8 }^{ 2 }+15\times 8 \right) =29664\text{ cm}^3$

$=\cfrac { 26994 }{ 1000 }$ litres

$[\because 1000\text{ cm}^3=1$ litre

$]$
Thus, the capacity of the bucket

$=26.994$ litres.
1035325_622529_ans_0f33eb66d9c94e85aec40678bc3de5e2.png

A hollow sphere of external and internal diameters of $8cm$ and $4cm$ respectively is melted and made into another solid in the shape of a right circular cone of base diameter $8cm$ . Find the height of the cone.

Let

$R$ and

$r$ be the external and internal radii of the hollow sphere.
Let

$h$ and

${r}_{1}$ be the height and the radius of the cone to be made.
Hollow sphere
External:

$2R=8cm$

$\Rightarrow$

$R=4cm$
Internal

$2r=4cm$

$\Rightarrow$

$r=2cm$
Cone

$2{r}_{1}=8$

$\Rightarrow$

${r}_{1}=4$
When the hollow sphere is melted and made into a solid cone, we have
Volume of the cone

$=$ Volume of the hollow sphere

$\cfrac { 1 }{ 3 } \pi { r }_{ 1 }^{ 2 }h=\cfrac { 4 }{ 3 } \pi \left[ { R }^{ 3 }-{ r }^{ 3 } \right]$

$\cfrac { 1 }{ 3 } \times \pi \times { 4 }^{ 2 }\times h=\cfrac { 4 }{ 3 } \times \pi \times \left( { 4 }^{ 3 }-{ 2 }^{ 3 } \right)$

$h=\cfrac { 64-18 }{ 4 } =14$
Hence, the height of the cone

$h=14cm$

1029083_622545_ans_394c672058364dc7b089230049dd7faf.bmp

A circus tent is to be erected in the form of a cone surmounted on a cylinder. The total height of the tent is $49m$ . Diameter of the base is $42m$ and height of the cylinder is $21m$ . Find the cost of canvas needed to make the tent, if the cost of canvas is Rs. $12.50/{m}^{2}$ . (Take $\pi =\cfrac { 22 }{ 7 }$ )

Cylindrical Part
Diameter

$2r=42m$
Radius,

$r=21m$
Height

$h=21m$
Conical Plant
Radius,

$r=21m$
Height

${h}_{1}=49-21=28m$
Slant height,

$l=\sqrt { { h }_{ 1 }^{ 2 }+{ r }^{ 2 } }$

$=\sqrt { { 28 }^{ 2 }+{ 21 }^{ 2 } } =7\sqrt { { 4 }^{ 2 }+{ 3 }^{ 2 } } =35m$
Total area of the canvas needed

$=$ CSA of the cylindrical part

$+$ CSA of the conical part

$=2\pi rl+\pi rl=\pi r(2h+l)$

$=\cfrac { 22 }{ 7 } \times 21(2\times 21+35)=5082$
Therefore, area of the canvas

$=5082{ m }^{ 2 }$
Now, the cost of the canvas per sq.m

$=$ Rs.

$12.50$
Thus, the total cost of the canvas

$=5082\times 12.5=$ Rs.

$63525$

1035310_622544_ans_3ee9ff3528a446d69b72c9a9f49c4ea6.PNG

The radius of a cone is $7cm$ and its height is $9cm$ . The volume of this cone is equal to lateral surface area of another cone which has same radius. Find the slant height of the cone.

We know that the radius of cone

$r=7cm$

and its height

$h=9cm$

and let the slant height of the second cone be

$=l$

$cm$

Then, according to the question.

Numerically: The lateral surface of the second cone

$=$ volume of the first cone

$\Rightarrow \pi rl=\cfrac { 1 }{ 3 } \pi { r }^{ 2 }h$

$\Rightarrow l=\cfrac { 1 }{ 3 } r h$

On substituting the values we get,

$l=\cfrac { 1 }{ 3 }\times 7 \times 9$

$\therefore \,\,l = 21 \,\,cm$

Thus the required slant height of the second cone

$=21cm$

A cylinder has hemispherical ends having radius 7 cm and total height of solid is 104 cm. If its outer surface is to polished & cost of polish is Rs. 100 per sq. mtr. find the total cost of polish.

Radius of each hemisphere = r = 7 cm
Height of total solid = 104 c
Height of cylinder,

$h = 104 - 2 \times 7 = 90 cm$
Then,
T.S.A. of the solid = C.S.A. of the cylinder + 2

$\times$ C.S.A. of the hemisphere

$= 2 \pi r h + 2 \times 2 \pi r^2$

$= 2 \pi r (h + 2r)$

$= 2 \times \dfrac{22}{7} \times 7 (90 + 2 \times 7)$

$= 44 \times 104$

$= 4576 cm^2$

$= 0.4576 m^2$

$\therefore$ Total surface of the given solid is 0.4576 m

$^2$ .
Cost of polish is Rs. 100 per sq. mtr.

$\therefore$ Cost of polishing the solid of area

$0.4576 m^2 = 0.4576 \times 100 = 45.76$
Total cost of polish is Rs. 45.76.

A tank is in the shape of a cylinder with two hemispheres attached to both ends as shown in the figure. Its common diameter is $2\text{ m}$ and total length is $8\text{ m}$ .
Find the total cost of painting the outer surface of this tank at the rate of $\text{Rs. }60$ per square metre. (use $\pi = 3.14$ as an approximation)

Given, Common diameter

$= 2\text{ m}$
Radius

$= 1\text{ m}$
Total length of tank

$= 8\text{ m}$
Length of cylinder

$= h = 8 - 2(1) = 6 \text{ m}$

Surface area of the tank
= Surface area of two hemispheres + lateral surface area of the cylinder

$=2\times 2\pi r^2+2\pi rh$

$=2\pi r(2r+h)$

$=2\times 3.14\times 1(2\times 1+6)$

$=50.26 \text{ m}^2$

Cost of painting

$=50.26\times 60 = \text{Rs. } 3015.92$

$\therefore$ Cost of painting the tank is

$\text{Rs. }$

$3015.92$

A cylindrical container having a diameter of $16$ cm and a height of $40$ cm is full of ice cream. The ice cream is to be filled into cones of height $12$ cm and diameter $4$ cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

Radius of cylinder

$= r = 8$ cm
Height of cylinder

$= h = 40$ cm
Volume of cylinder

$= \pi r^2 h = \pi \times (8)^2 \times 40$
Ice-cream inside the cylinder

$= \pi \times (8)^2 \times 40$ ........ (i)
Radius of cone

$= r = 2$ cm
Height of cone

$= h = 12$ cm
Volume of cone

$= \dfrac{\pi}{3} r^2 h = \dfrac{\pi}{3} \times (2)^2 \times 12 = 16 \pi$
Radius of hemisphere

$= r = 2$ cm
Volume of hemisphere

$= \dfrac{2}{3} \pi r^3 = \dfrac{2}{3} \times \pi \times (2)^3 = \dfrac{16}{3} \pi$
Ice-cream in one cone

$=$ Volume of cone + Volume of hemisphere

$= 16 \pi + \dfrac{16}{3} \pi = \dfrac{64}{3} \pi$
Number of cones which can be filled with ice-cream

$= \dfrac{\text{Ice-cream inside the cylinder}}{\text{Ice-cream in one cone}}$

$= \dfrac{\pi \times (8)^2 \times 40}{\dfrac{64}{3} \pi}$

$= 120$

A wooden toy is in the shape of a cone mounted on a cylinder as shown in the figure. If the height of the cone is $24\ \mathrm{cm}$ , the total height of the toy is $60\ \mathrm{cm}$ and the radius of the base of the cone is equal to $10\ \mathrm{cm}$ and is twice the radius of the base of the cylinder. Find the total surface area of the toy. [Take $\pi=3.14$ ]

Given:

Height of cylinder,

$h$ = Total height of toy - Height of cone

$=60-24$

$=36\ \mathrm{cm}$

Radius of cone,

$r=10\ \mathrm{cm}$

Slant height,

$l=\sqrt { { 10 }^{ 2 }+{ 24 }^{ 2 } }$

$=26\ \mathrm{cm}$

Surface Area of cone

$=\pi { r }^{ 2 }+\pi rl$

$=\pi \left( { 10 }^{ 2 }+10\times 26 \right)\ \mathrm{cm}^2$

$=360\pi \ \mathrm{cm}^2$

According to given condition in the question:

Radius of cylinder

$R=5\ \mathrm{cm}\dots (\dfrac {10}2=R)$ or

$(r=2R)$

Surface Area of cylinder

$=\pi { r }^{ 2 }+2\pi rh$

$=\pi \left( { 5 }^{ 2 }+2\times 5\times 36 \right)\ \mathrm{cm}^2$

$=385\pi \mathrm{cm}^2$

Total Surface area of the toy

$=\left( 360+385 \right) \pi \ \mathrm{cm}^2$

$=2340.5\ \mathrm{cm}^2$

In an experiment of Maths, frustum of cone is made up of seven rings.The radius of uppermost ring is 4 cm and the radius of each ring is increased by 1 cm, so that the radius of the last ring is 10 cm. If the width of each ring is 3 cm. Then find the volume of entire frustum. (avoid the spaces between rings).

The radius of lower ring is

$r_1 = 10 cm$
The radius of upper ring is

$r_2 = 4 cm$
Width of each ring is

$3\ cm$ .
So, width of

$7$ rings

$= 21 cm$ .

$\therefore$ Height of frustum

$= 21\ cm$ .
Volume of frustum

$= \dfrac{1}{3} \pi h [r_1^2 + r_2^2 + r_1 r_2]$

$= \dfrac{1}{3} \times \dfrac{22}{7} \times (21) [(10)^2 + (4)^2 + 40]$

$= 22[156]$

$= 3432 cm^3$

From a solid right circular cylinder of a height $2.4$ cm and radius $0.7$ cm, a right circular cone of same height and same radius is cut. Find the total surface area of the remaining solid.

For both Cylinder and Cone :

${ h } = 2.4$ cm,

$r = 0.7$ cm

Total surface area of remaining solid

$=$ Total surface area of cylinder

$-$ Total surface area of cone ....

$(a)$

Total surface area of cylinder

$= 2\pi r (r+h)$

$= 2\pi\times \dfrac {7}{10} \times \left(\dfrac {7} {10}+\dfrac {24}{10}\right)$

$=\dfrac {14 \pi \times 31}{100}$ ....(1)

Total surface area of cone

$= \pi r ( r +l )$

$= \pi \times \dfrac{7}{10} \left(\dfrac{7}{10} + \sqrt{h^2 + r^2}\right)$

$= \pi \times \dfrac{7}{10} \left[\dfrac{7}{10} +\sqrt{\left(\dfrac{24}{10}\right)^2+\left(\dfrac{7}{10}\right)^2}\right]$

$=\dfrac{7\pi}{10}\left(\dfrac{7}{10}+\sqrt{\dfrac{576+46}{100}}\right)$

$=\dfrac{7\pi}{10} \left(\dfrac{7}{10}+\dfrac{25}{10}\right)$

$=\dfrac{7\pi}{10}\times \dfrac{32}{10} \ \ \ \ \ ...(2)$

Substituting

$(1)$ and

$(2)$ in

$(a)$ ,

We get,

Total surface area of remaining solid

$=\pi \left(\dfrac{434}{100}-\dfrac{224}{100}\right)$

Total surface area of remaining solid

$= 6.6 cm^2$ .

A solid consisting of a right circular cone of height $120\ cm$ and radius $60\ cm$ standing on a hemisphere of radius $60\ cm$ is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, (in $m^{3}$ ) if the radius of the cylinder is $60\ cm$ and its height is $180\ cm$ ?

In the given question,

Volume of water in the cylinder

$=$ Volume of cylinder

$-$ Volume of Solid

Given:

Radius of the base of the cone

$r_1=60\ cm$

Height of the cone

$h_1=120\ cm=2r1$

Radius of the hemisphere

$r_2=60\ cm=r_1$

Height of the cylinder

$h_2=180\ cm=3r_1$

Radius of the base of the cylinder

$r_3=60\ cm=r_1$

Now,

Volume of the solid

$=\dfrac{2}{3}\pi (r_1)^3+\dfrac{1}{3}\pi (r_1)^2(2r_1)$

$=\dfrac{2}{3}\pi (60)^3+\dfrac{1}{3}\pi (60)^2(2\times 60)\\$

$=\dfrac{1}{3}\pi (60)^3[2+2]\\$

$=\dfrac{4}{3}\pi (60)^3$

Again,

Volume of the cylinder

$=\pi (r_1)^2(3r_1)$

$=3\pi (60)^3$

Therefore,

Required volume of water

$=\pi (60)^3\left [3-\dfrac{4}{3}\right]$

$=\dfrac{5}{3}\pi (60)^3$

$=1130973.35\ cm^3$

$=1.131m^3$ (approx.)

1017352_805676_ans_e37e2393e18d4cf0a22b2793d5ff6bac.png

A gulab jamun, contains sugar syrup up to about $30\%$ of its volume. Find approximately how much syrup would be found in $45$ gulab jamuns, each shaped like a cylinder with two, hemispherical ends with length $5\ cm$ and diameter $2.8\ cm$ .

${\textbf{Step 1: Find out the volume}}{\textbf{.}}$

${\text{Given }}h = 5cm$

${\text{Length of cyclinder }}h = 5 - 2.8 = 2.2$

${\text{Volume of gulab jamun = volume of cyclinder + volume of 2 hemisphere}}$

$= 3\pi {r^2}h + \dfrac{4}{3}\pi {r^3}$

$= \dfrac{{22}}{7} \times 1.4 \times 1.4 \times \dfrac{{\left( {3 \times 2.2 + 4 \times 1.4} \right)}}{3}$

$= \dfrac{{6.16\left( {12.2} \right)}}{3}$

$= \dfrac{{75.152}}{3}$

$= 25.05cm$

${\text{There are 45 gulab jamuns}}{\text{.}}$

${\textbf{Step 2: Find the volume of 45 gulab jamuns}}{\textbf{.}}$

${\text{Volume = 45}} \times {\text{25}}{\text{.05}}$

$= 1127.2830{\text{ percentage of syrup}}$

$= 1127.28 \times \dfrac{{30}}{{100}}$

$= 338.184cm$

${\textbf{Hence, there are }}\mathbf{338.184}{\textbf{ cm of syrup be found in }}\mathbf{45}{\textbf{ gulab jamuns}}{\textbf{.}}$

A solid is composed of a cylinder with a hemisphere at one end and a cone at other end as shown in the figure. If the radius of each of these solids is 7 cm and height of the cylinder is equal to slant height of the cone, find the total surface area of the solid if slant height is 4 cm.

For all figures,

$r=7 \ cm$

$l\ (slant\ height)= \ h\ (height\ of\ cylinder)=4 \ cm$

T.S.A of the figure= C.S.A of Cone+C.S.A of cylinder +C.S.A of hemisphere.

CSA of cone:

$\pi rl=(3.14)\times (7cm)\times (4cm)=87.92 \ cm^2$

CSA of cylinder:

$2\pi r h=(2)\times (3.14)\times (7cm)\times (4cm)=175.84 \ cm^2$

CSA of hemisphere:

$2\pi r^2=(2)\times (3.14)\times (7cm)^2=307.72 \ cm^2$

Adding all: TSA of figure=

$571.48 \ cm^2$

Two dairy owners $A$ and $B$ sell flavoured milk filled to capacity in mugs of negligible thickness, which are cylindrical in shape with a raised hemispherical bottom. The mugs are $14\ cm$ high and have diameter of $7\ cm$ as shown in given figure. Both $A$ and $B$ sell flavoured milk at the rate of $Rs. 80$ per litre. The dairy owner $A$ uses the formula $\pi {r}^{2}h$ to find the volume of milk in the mug and charges $Rs. 43.12$ for it. The dairy owner $B$ is of the view that the price of actual quantity of milk, should be charged. What according to him should be the price of one mug of milk? Which value is exhibited by the diary owner $B$ ? (Use $\pi =\dfrac { 22 }{ 7 }$ )

Let us first find the actual quantity of milk in mug.

Actual capacity of the glass

$=$ Volume of cylinder

$-$ Volume of the hemisphere.

Volume of the cylinder:

Given:diameter

$=7$ cm, radius

$r=\dfrac{d}{2}=\dfrac{7}{2}$ cm and height

$=14$ cm.

Volume of cylindrical glass

$=\pi{r}^{2}h=\dfrac{22}{7}\times\dfrac{7}{2}\times\dfrac{7}{2}\times 14=539{cm}^{3}$

Volume of hemisphere

$=\dfrac{2}{3}\pi{r}^{3}=\dfrac{2}{3}\times \dfrac{22}{7}\times\dfrac{7}{2}\times\dfrac{7}{2}\times \dfrac{7}{2}=\dfrac{539}{6}{cm}^{3}$

Actual capacity of the glass

$=$ Total volume of cylinder

$-$ Volume of hemisphere.

$=539{cm}^{3}-\dfrac{539}{6}{cm}^{3}=\dfrac{2695}{6}{cm}^{3}$

The owner charges Rs.

$80$ per liter.

Amount for

$1$ liter

$=$ Rs.

$80$

Amount for

$1000{cm}^{3}=$ Rs

$\dfrac{80}{100}$ since

$1\ l=1000{cm}^{3}$

Amount for

$1{cm}^{3}=\dfrac{80}{100}$

Now,

$1$ mug has

$\dfrac{2695}{6}{cm}^{3}$ capacity.

Amount for

$\dfrac{2695}{6}{cm}^{3}= \dfrac{80}{100}\times \dfrac{2695}{6}=$ Rs.

$35.93$

$VALUE$ :Since owner

$B$ is honest, the value exhibited by him is

$HONESTY$

A toy is in the form of a cone mounted on a hemisphere as shown in the figure. If the radius of each of these solids is $\displaystyle \frac{7}{2}$ cm and height of the cone is 5 cm, find the volume of the toy.

Volume of the toy=Volume of cone + Volume of hemisphere.

Given:

$r=3.5cm$

$h=5cm$

Volume of cone

$=\dfrac { 1 }{ 3 } \pi r^{ 2 }h$

$=\dfrac{1}{3} \times (3.14)\times (3.5cm)^2 \times(5cm)\\=64.11cm^3$

Volume of hemisphere

$=\dfrac { 2 }{ 3 } \pi r^{ 3 }$

$=\dfrac{2}{3}\times (3.14) \times (3.5cm)^3 \\= 89.75cm^3$

Adding both,

Volume of toy

$= 153.86cm^3$

A bucket is in the form of a frustum of a cone and it can hold $28.49$ litres of water. If radii of its circular ends are $28cm\ and\ 21cm$ , find the height of the bucket. Hence find its slant height.

$r_1=28cm,r_1=21cm$

Volume of frustum

$=28.49cm^3\times1000$

$\cfrac{1}{3}\pi(R_2^2-R_1^2)h=28490$

$\implies \cfrac{1}{3}\pi(28^2-21^2)h=28490$

$\implies h=\cfrac{4.07}{22\times 7}\times 300cm=872.14cm$

$\triangle ABC$

$l=\sqrt {h^2+r^2}=\sqrt{28^2+872.14^2}=872.59cm$

1168851_878969_ans_f4ca2d7f7eb7402e8ddf66be92ea76c2.PNG

A bucket is in the form of a frustum of a cone and holds 28.490 litres of water.The radii of the top and bottom are 28 cm and 21 cm respectively.Find the height of the buckrt.

Volume of frusturn of a cone

$=\cfrac { h }{ 3 } \pi \left[ { R }^{ 2 }+{ r }^{ 2 }+Rr \right] =28.49ℓ$

$\Rightarrow \cfrac { 1 }{ 3 } \times \cfrac { 22 }{ 7 } \times h\left[ { 28 }^{ 2 }+{ 21 }^{ 2 }+28\times 21 \right] =28490㎤$

$\Rightarrow \cfrac { 1 }{ 3 } \times \cfrac { 22 }{ 7 } \times h\left[ 784+441+588 \right] =28490\Rightarrow h=\cfrac { 28490\times 21 }{ 22\times 1813 }$

$\Rightarrow h=15㎝$

The following figure represents a solid consisting of a right circular cylinder with a hemisphere at one end and a cone at the other. Their common radius is $7$ cm. The height of the cylinder and cone are each of $4$ cm. Find the volume of the solid.

The solid can be divided into

$3$ parts.

The first part is the cone with height

$4\;cm$ and radius

$7\;cm$ .

Volume of cone

$\displaystyle V_1=\dfrac{1}{3}\pi r^2 h\\ = \dfrac{1}{3}\cdot \dfrac{22}{7}\cdot 7^2 \cdot 4\\= 205.34\; cm^3$

The second part is the cylinder with height

$4\;cm$ and radius

$7\;cm$ .

Volume of cylinder

$\displaystyle V_2=\pi r^2 h\\=\dfrac{22}{7} \cdot 7^2 \cdot 4 \\= 616\; cm^3$

The third part is the hemisphere with radius

$7\;cm$ .

Volume of hemisphere

$\displaystyle V_3=\dfrac{2}{3}\pi r^3\\ = \dfrac{2}{3}\cdot \dfrac{22}{7}\cdot 7^3\\= 718.67\;cm^3$

Volume of total solid

$V_1+V_2+V_3=1540\; cm^3$

The diameters of the lower and upper ends of a bucket in the form of a frustum of a cone are $10\ cm$ and $30\ cm$ respectively. If its height is $24\ cm$ , find
(i) The area of the metal sheet used to make the bucket ?
(ii) Why we should avoid the bucket made by ordinary plastic (Use $\pi=3.14$ )

Given: Height of the frustrum

$(h)=24$ cm

radius of lower end of bucket

$(r_1)=\dfrac{10}{2}=5 cm$

Radius of upper end of bucket

$(r_2)=\dfrac{30}{2}=15cm$

$\therefore$ Slant height

$(l)=\sqrt{h^2+(r_2-r_1)^2}$

$=\sqrt{(24)^2+(15-5)^2}$

$=\sqrt{(24)^2+(10)^2}$

$=\sqrt{576+100}=\sqrt{676}=26 cm$

Now, Area of the metal sheet used to make the bucket = T.S.A of the

bucket (excluding the upper end)

Total Surface Area(TSA)

$=\pi (r_1+r_2)l+\pi r_1^2$

$=(3.14)(15+5)(26)+3.14(5)^2$

$=1632.8+78.5=1711.3 cm^2$

(b) We should avoid the bucket made by ordinary plastic as plastic is non biodegradable and hazardous to environment.

A cylindrical bucket, $32cm$ high and $18cm$ of radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is $24cm$ , find the radius and slant height of the heap.

Given dimensions for cylinder :

$h=32 cm \ r=18cm$

Dimensions of cone :

$R=?\ H=24cm$

The volume of sand should remain the same.

Hence

$\pi r^2h=\dfrac13 \pi R^2H$

$\Rightarrow R^2= \dfrac{3r^2h}{H}=\dfrac{3(18)^232}{24}$

$R=36cm$

By Pythagoras theorem

$L^2=R^2+H^2$

Hence

$L=43.266cm$

The length of a road roller is $2.1$ m and its diameter is $1.4$ m. For levelling a ground $500$ rotations of the road roller were required. How much area of ground was levelled by the road roller? Find the cost of levelling at the rate of Rs $7$ per sq.m

$l=2.1m,d=1.4m\implies r=0.7m$

$n=500$

Area covered

$=2n\pi\times r\times l\\=2\times 500\times 0.7\times 2.1\pi=4620m^2$

Cost of travelling

$=4620\times 7=Rs.32,340$

A cylinder with the greatest lateral surface is inscribed in a cone. Find the ratio of the altitude of the cone to that of the cylinder.

Let radius and height of cone be R and H respectively and r and h be the radius and height of cylinder

Then,

$S=2\pi rh$

Now,

$h=-\cfrac { H }{ R } r+H$

$\Rightarrow S=2\pi r\left[ -\cfrac { H }{ R } r+H \right] =2\pi rH-\cfrac { 2\pi H }{ R } { r }^{ 2 }$

$\cfrac { dS }{ dr } =0\Rightarrow 2\pi H-\cfrac { 4\pi H }{ R } r=0\Rightarrow R-2r=0\Rightarrow r=\cfrac { R }{ 2 }$

$\therefore h=\cfrac { -H }{ R } \times \cfrac { R }{ 2 } +H\Rightarrow h=\cfrac { H }{ 2 } \Rightarrow \cfrac { H }{ h } =2$

Set up an equation of a tangent to the graph of the following function.
Find the altitude of the cylinder with the greatest lateral area which can be inscribed in a sphere of radius B.

Radius of sphere

$=B$ , let radius of base of cylinder be 'r' and height be 'h'. Then,

$S=2\pi rh$

Now,

$r=B\sin { \alpha } ,h=2B\cos { \alpha }$ where

$\alpha$ be the angle between radius and axis of the cylinder. Then,

$S=2\pi { r }^{ 2 }\sin { \alpha } \cos { \alpha }$

For maximum, differentiate S w.r.t.

$\alpha$ and put equals to

$0$

$\cfrac { dS }{ d\alpha } =2\pi { B }^{ 2 }\left[ \cos ^{ 2 }{ \alpha } -\sin ^{ 2 }{ \alpha } \right]$

Now,

$\cfrac { dS }{ d\alpha } =0\Rightarrow 2\pi { B }^{ 2 }\left[ \cos ^{ 2 }{ \alpha } -\sin ^{ 2 }{ \alpha } \right] =0$

$2\pi { B }^{ 2 }\neq 0\quad \therefore \cos ^{ 2 }{ \alpha } -\sin ^{ 2 }{ \alpha } =0\Rightarrow \tan ^{ 2 }{ \alpha } =1\Rightarrow \alpha =45°$

Now,

$h=2B\cos { 45° } =\sqrt { 2 } B$

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is $2\ cm$ and the diameter of the base is $4\ cm$ . Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volume of the cylinder and toy. (Take $\pi = 3.14)$ .

Volume of the toy

$=\cfrac{1}{3}\pi r^2h+\cfrac{2}{3}\pi r^3$

$=\cfrac{1}{3}\pi\times 2^2\times 2+\cfrac{2}{3}\pi\times 2^3$

$=25.12cm^3$

Volume of the cylinder circumscribing

$h=4cm$

$r=2cm$

$v=\pi r^2h=\pi\times 4\times 4=50.24cm^3$

Difference in volume

$=(50.24-25.12)cm^3=25.12cm^3$

1169126_879059_ans_2fb3c8de2609463383c1cad7721e6f15.png

Three solid cubes of sides 1cm,6cm and 8 cm are melted to form a new cube.Find the surface area of the cube so formed.

Volume of new cube

$=\left( { 1 }^{ 3 }+{ 6 }^{ 3 }+{ 8 }^{ 3 } \right) { cm }^{ 3 }=729{ cm }^{ 3 }$

side length of new cube

$=\sqrt { 729 } cm=9cm$

Surface area of new cube

$=\left( 6\times 9\times 9 \right) { cm }^{ 2 }=486{ cm }^{ 2 }$

A tent of height $77dm$ is in the form of a right circular cylinder of diameter $36m$ and height $44dm$ surmounted by a right circular cone. Find the cost of the canvas at Rs. $3.50$ per ${m}^{2}$ (Use $\pi =22/7$ )

Given total height of the tent as

$7.7m$ and length of cylindrical part as

$4.4m$

Hence , the height of the cone part

$=7.7-4.4=3.3m$

$r=18m$ for both Cylinder and cone

The area of canvas required=Surface area of Cylindrical and cone parts of tent

$A=2\pi rh+\pi r(\sqrt{h^2+r^2})$

$A=2\pi \times 18\times 4.4+\pi \times 18(\sqrt{18^2+3.3^2})$

$A=1352.5m^2$

Cost of canvas=

$1352.5\times 3.5=5363.6 \ rupees$

A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The radius and height of the cylindrical part are $5cm$ and $13cm$ respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Find the surface area of the toy if the total height of the toy is $30cm$ .

Height of the cylindrical part

$=13$ cm

Radius of cone , cylinder and hemi sphere

$=5$ cm

r=5cm for hemisphere cylinder and cone.

Height of cone

$h=$

$30-5-13=12$

The area of canvas required=Surface area of hemishphere,cylinder and cone parts of tent

$A=2\pi r^2+2\pi r H+\pi r(\sqrt{h^2+r^2})$

$A=2\pi \times 5\times 5+2\pi \times 5\times 13+\pi \times 5(\sqrt{5^2+12^2})$

$A=770cm^2$

A toy is in the form of a cone surmounted on a hemisphere. The diameter of the base and the height of the cone are $6\ \text{cm}$ and $4\ \text{cm}$ respectively. Determine the surface area of the toy. (Use $\pi=3.14$ )

Height of the cone part,

$h=4\ \text{cm}$

Radius of cone and hemi sphere,

$r=3\ \text{cm}$

The area of canvas required=Surface area of hemi-sphere and conical parts of tent:

$A=2\pi r^2+\pi r(\sqrt{h^2+r^2})$

$A=2\pi \times 3\times 3+\pi \times 3(\sqrt{3^2+4^2})$

$A=103.67\ \text{cm}^2$

A cylindrical tub of radius $5cm$ and length $9.8cm$ is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed in the tub. If the radius of the hemisphere is immersed in the tub. If the radius of the hemisphere is $3.5cm$ and height of the cone outside the hemisphere is $5cm$ , find the volume of the water left in the tub (Take $\pi=22/7$ )

Volume of water in tub

$=\pi r^2 h=\pi \times 5^2 \times 9.8cm^3$

Volume of Hemishpere

$=\dfrac{2}{3}\times \pi \times 3.5^3cm^3$

Volume of cone

$=\dfrac{1}{3}\pi \times 3.5^2\times 5cm^3$

Volume of water left in the tub = Volume of water in tub-Volume of toy

Volume left =

$\pi \times 25\times 9.8-\left(\dfrac{2}{3}\times \pi\times 3.5^3+\dfrac{1}{3}\pi\times 3.5^2\times 5\right)$

$V=\pi(245-49)=\dfrac{22}{7}\times 196=22\times28=616cm^3$

A rocket is in the form of a circular cylinder closed at the lower end with a cone of the same radius attached to the top. The cylinder is of radius $2.5m$ and height $21m$ and the cone has the slant height $8m$ . Calculate the total surface area and the volume of the rocket.

Total surface area = Area of base of cylinder+ Area of curved surface of cylinder+Area of curved surface of cone

$A=\pi\times r^2+2\pi rh+\pi r l$

h=height of cylindrical part

l=slant height of conical part

r=radius of cylinder=radius of cone

$A=\pi \times 2.5^2+2\pi \times 2.5\times 21+\pi \times 2.5\times 8$

$A=412.334m^2$

Volume of tent

$=\pi r^2 h+\dfrac{1}{3}\pi r^2 h$

$V=\pi \times 2.5^2\times 21+\dfrac{1}{3}\pi \times 2.5^2\times 7.6$

$V=462.076m^3$

A solid is in the form of a right circular cylinder, with a hemisphere at one end and a cone at the other end. The radius of the common base is $8\ cm$ and the heights of the cylindrical and conical portions are $10cm$ and $6cm$ respectively. Find the total surface area of the solid. (Use $\pi=22/7$ )

Height of the cylindrical part and cone are

$10cm \ and \ 6cm$ respectively

Radius of cone , cylinder and hemi sphere

$=3.5cm$

$r=3.5$ cm for both Cylinder and cone

The area of canvas required=Surface area of hemishphere,cylinder and cone parts of tent

$A=2\pi r^2+2\pi r h+\pi r(\sqrt{h^2+r^2})$

$A=2\pi \times 3.5\times 3.5+2\pi \times 3.5\times 10+\pi \times 3.5(\sqrt{3.5^2+6^2})$

$A=373.3cm^2$

A petrol tank is a cylinder of base diameter $21cm$ and length $18cm$ fitted with conical ends each of axis length $9cm$ . Determine the capacity of the tank.

Given

$r=11.5cm$ , height of cylinder

$=18cm$ , height of cone

$=9cm=$

$h_1$

$V=\pi r^2 h+2\times \dfrac{1}{3}\pi r^2 h_1$

$V=\pi \times 20.5^2\times 18+\dfrac{2}{3}\pi \times 20.5^2\times 9$

$V=8316cm^3$

A tent of height $77\ dm$ is in the form of a right circular cylinder of diameter $36\ m$ and height $44\ dm$ surmounted by a right circular cone. Find the cost of the canvas at $Rs.\ 3.50$ per square meter.

The radius of cylinder and radius of cone both will be equal to

$18\ m.$

As we know that,

$1\ m=10\ dm$

So,

$1\ dm=0.1\ m$

Therefore,

Height of the tent

$=77\ dm$

$=77\times 0.1\ m$

$=7.7\ m$

Height of cylinder

$= 44\ dm$

$=44\times 0.1\ m$

$=4.4\ m$

Height of cone

$=7.7-4.4$

$=3.3m$

The slant height of the cone can be calculated by using height and the radius of the cone,

$\begin{aligned}{}l& = \sqrt {{R^2} + {H^2}} \\ &= \sqrt {{{\left( {18} \right)}^2} + {{\left( {3.3} \right)}^2}} \\ &= \sqrt {324 + 10.89} \\& = \sqrt {334.89} \\& = 18.3\ m\end{aligned}$

Total surface area of tent

$= \pi Rl+2\pi RH_{cylinder}$

$=\pi R(l+2H_{cylinder})$

$=\pi \times18(18.3+2\times4.4)$

$=18\pi \times27.1$

$=1532.46\ m^2$

The total cost of canvas

$=$ Cost of canvas per square meter

$\times$ Surface area of the tent in square meters

$=1532.46 \times 3.5$

$=Rs.\ 5363.64$

Sushant has a vessel, of the form of an inverted cone, open at the top, of height $11cm$ and radius of top as $2.5cm$ and is full of water. Metallic spherical balls each of diameter $0.5cm$ are put in the vessel due to which $(\cfrac{2}{5})$ th of the water in the vessel flows out. Find how many balls were put in the vessel. Sushant made the arrangement so that the water that flows out irrigates the flower beds. What value has been shown by Sushant?

Given

$h=11cm,r_1=2.5cm,r_2=0.25cm$

$25\times$ volume of cone

$=n\times$ Volume of each spherical ball

$25\times \dfrac{1}{3}\pi r_1^2 h=n\times \dfrac{4}{3} \pi r_2^3$

$r_1^2h=n\times 10r_2^3$

$2.5^2 \times 11=n\times 10(0.25^3)$

$n=440$

A circus tent has cylindrical shape surmounted by a conical rood. The radius of the cylindrical base is $20m$ . The heights of the cylindrical and conical portions are $4.2m$ and $2.1m$ respectively. Find the volume of the tent.

2152729_975153_ans_03bf751a7b254c3e93fb02590cc87f4f.jpg

A conical hole is drilled in a circular cylinder of height $12cm$ and base radius $5cm$ . The height and the base radius of the cone are the same as the Cylinder. Find the Curved Surface Area and volume of the remaining solid.

Given,

Height of cylindrical part

$=12cm$

Height of the concical part

$=12cm$

$r=5cm$ for both Cylinder and cone

The Curved Surface Area required

$=$ Curved Surface area of Cylindrical

$+$ Curved Surface area of Cone.

$\therefore A=2\pi rh+\pi r(\sqrt{h^2+r^2})$ {

$\because l=\sqrt{h^2+r^2}$ }

$\Rightarrow A=2\pi \times 12\times 5+\pi \times 5(\sqrt{5^2+12^2})$

$\Rightarrow A=581.2cm^2$

Now to Calculate Volume,

Volume of the solid

$=$ Volume of Cylinder

$-$ Volume of Cone

$\therefore Volume=\pi r^2 h-\dfrac{1}{3}\pi r^2 h$

$\Rightarrow V=\pi \times 5^2\times 12-\dfrac{1}{3}\pi \times 5^2\times 12$

$\Rightarrow V=628.319cm^3$

An wooden toy is made by scooping out a hemisphere of same radius from each end of a solid cylinder. If the height of the cylinder is $10cm$ , and its base is of radius $3.5cm$ , find the volume of wood in the toy. (Use $\pi=22/7$ )

Height of cylinder

$=10cm$

Volume of the solid

$=\pi r^2 h+2\times2\pi r^2$

$V=\pi \times 3.5^2\times 10+4\pi \times 3.5^2$

$V=539cm^3$

A boiler is in the form of a cylinder $2m$ long with hemispherical ends each of $2$ metre diameter. Find the volume of the boiler.

$r=h=1m$

$Total \ volume= 2\times \dfrac{2}{3}\pi r^3+\pi r^2 h$

$V=\dfrac{4\pi}{3}+\pi \times 1^2\times 2=\dfrac{10\pi}{3}=\dfrac{220}{21}m^3$

A toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of the hemisphere. If the radius of the base of the cone is $21cm$ and its volume is $\dfrac{2}{3}$ of the volume of the hemisphere, calculate the height of the cone and the surface area of the toy. (Use $\pi=22/7$ )

Volume of cone =

$\dfrac23$ Volume of hemisphere

$\dfrac13 \pi r^2h=\dfrac23\left(\dfrac23\pi r^3\right)$ ; r =Common radius h = Cone height

$h=\dfrac43r=\dfrac43(21)=28 cm$

Slant height,

$l =\sqrt{r^2+h^2}=\sqrt{21^2+28^2}=35cm$

Total surface area =

$CSA_{cone}+CSA_{hemisphere}$

$= \pi r l +2\pi r^2=\pi r(l+2r)=5082 cm^2$

A solid is composed of a cylinder with hemispherical ends. If the length of the cylindrical part is $104cm$ and the radius of each of the hemispherical ends is $7cm$ , find the cost of polishing its surface at the rate of Rs. $10$ per ${dm}^{2}$ .

Total area of polishing = CSA of cylinder + 2

$\times$ CSA of hemisphere

Given:

$r=7cm, h=104cm$

Therefore,

$Surface\ Area= 2\pi rh+2\times 2\pi r^2$

$A=2\pi\times 7\times 104+4\pi \times 7^2$

$A=5190cm^2=51.9dm^2$ {

$1dm = 10cm$ }

Cost of polishing

$=51.9\times 10=519Rs.$

A solid toy is in the form of a hemisphere surmounted by a right circular cone. Height of the cone is $2cm$ and the diameter of the base is $4cm$ . If a right circular cylinder circumscribes the toy, find how much more space it will cover.

Volume of toy

$=$ Vol. of cone + Vol.of hemisphere

$=\dfrac { 1 }{ 3 } \pi { r }^{ 2 }H+\dfrac { 2 }{ 3 } \pi { r }^{ 3 }=\dfrac { 1 }{ 3 } \pi { r }^{ 2 }\left[ H+2r \right] h$

$=\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times 2\times 2\left[ 2+2\times 2 \right]$

$=\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times 4\times { 6 }^{ 2 }=25.12{ cm }^{ 3 }$

Volume of cylinder

$=\pi { r }^{ 2 }h=\dfrac { 22 }{ 7 } \times 2\times 2\left[ 2+2 \right] =\dfrac { 22 }{ 7 } \times 16=50.24{ cm }^{ 3 }$

Difference in volume

$=50.24-25.12=25.12{ cm }^{ 3 }$

Hence, cylinder cover

$25.12{ cm }^{ 3 }$ more space.

A vessel in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is $14cm$ and the height of cylindrical part is $13cm$ . Find the inner surface area of the vessel.

Inner surface area of vessel

$=$ CSA of hemisphere + CSA of cylinder

$=2\pi { r }^{ 2 }+2\pi rh=2\pi r\left( r+h \right)$

$=2\times \dfrac { 22 }{ 7 } \times 7\left( 7+13 \right) =44\times 20=880{ cm }^{ 2 }$

A solid iron pole having a cylindrical portion $110\text{ cm}$ high and of base diameter $12\text{ cm}$ is surmounted by a cone $9\text{ cm}$ high. Find the mass of the pole, given that the mass of $1\text{ cm}^{3}$ of iron is $8\text{ gm}$ .

Given: Diameter of cone

$=12\text{ cm}$

So, Radius of cone

$=6\text{ cm}$

and height of cone

$=9\text{ cm}$

Volume of cone

$=\dfrac 13 \pi r^2h$

$=\dfrac{1}{3}\times \dfrac{22}{7}\times 6\times 6\times 9$

$=339.428\text{ cm}^3$

Radius of the cylinder,

$R=$ Radius of cone

$=6\text{ cm}$

Height of the cylinder,

$H=110\text{ cm}$

Volume of cylinder

$=\pi R^2H$

$=\dfrac {22}{7}\times 6\times 6\times 110$

$=12445.714\text{ cm}^3$

Hence,

Volume of pole

$=$ Volume of the conical part

$+$ Volume of the cylindrical part

$=339.43+12445.714$

$=12785.142\text{ cm}^3$

Required mass of pole

$=8\times 12785.142$

$=102281.13\text{ gm}$

$=102.281\text{ kg}$

A tent is in the form of a cylinder of diameter $20m$ and height $2.5m$ , surmounted by a cone of equal base and height $7.5m$ . Find the capacity of the tent and the cost of the canvas at RS. $100$ per square metre.

Given, length of the cylindrical part as

$2.5\ m$

Hence, the height of the cone part

$=7.5\ m$

For both Cylinder and cone,

$r=20\ m$

The area of canvas required

$=$ Surface area of Cylindrical and cone parts of the tent

$A=2\pi rh+\pi r(\sqrt{h^2+r^2})$

$A=2\pi \times 20\times 2.5+\pi \times 20(\sqrt{20^2+7.5^2})$

$A=2\times 3.14\times 20 \times 2.5+3.14\times 20\times 21.36$

$A=1656.25\ m^2$

Cost of canvassing

$=Area\times Rs.100$

$=1656.25\times 100$

$=Rs.165625$

A solid is in the shape of a cone surmounted on a hemisphere, the radius of each of them being $3.5cm$ and the total height of solid is $9.5cm$ . Find the volume of the solid. (Use $\pi=22/7$ )

Height of cone part

$=$ total height

$-$ radius of the hemisphere

Height of cone

$= 9.5-3.5=6cm$

Volume of the solid

$= \dfrac{1}{3}\pi r^2 h+\dfrac{2}{3}\pi r^3$

$V=\dfrac{1}{3}\pi \times 3.5^2\times 6+\dfrac{2}{3}\pi \times 3.5^3$

$V=166.7cm^3$

A solid is composed of a cylinder with hemispherical ends. If the length of the whole solid is $108cm$ , and the diameter of the cylinder is $36cm$ , find the cost of polishing the surface at the rate of $7$ paise per ${cm}^{2}$ .

CSA of solid=CSA of cylinder

$+2\times$ CSA of hemisphere

$=2\pi rh+2\times 2\pi { r }^{ 2 }$

$=2\pi r\left( h+2r \right)$

$=2\times \cfrac { 22 }{ 7 } \times 18\left( \left( 108-36 \right) +2\times 18 \right)$

$=12219.43㎠$

Cost of polishing=Rs.

$\left( 12219.43\times 0.07 \right)$ =Rs.

$855$

The interior of a building is in the form of a cylinder of base radius $12m$ and height $3.5m$ surmounted by a cone of equal base and slant height $12.5m$ . Find the internal curved surface are and the capacity of the building.

Internal CSA

$=2\pi rH+\pi rl$

$=\pi r\left[ 2h+l \right]$

$=\cfrac { 22 }{ 7 } \times 12\left[ 2\times 3.5+12.5 \right]$

$=\cfrac { 22 }{ 7 } \times 12\times 19.5=735.43m^{3}$

Capacity of building

$=\cfrac { 1 }{ 3 } \pi { r }^{ 2 }h+\pi { r }^{ 2 }H$

$=\pi { r }^{ 2 }\left[ \cfrac { 1 }{ 3 } \sqrt { { l }^{ 2 }-{ r }^{ 2 } } +H \right] =\cfrac { 22 }{ 7 } \times { 12 }^{ 2 }\left[ \cfrac { 1 }{ 3 } \sqrt { { 12.5 }^{ 2 }-{ 12 }^{ 2 } } +3.5 \right]$

$=\cfrac { 22 }{ 7 } \times 144\left[ \cfrac { 3.5 }{ 3 } +3.5 \right] =2112m^{3}$

A toy is in the form of a cone mounted on a hemisphere with the same radius. The diameter of the base of the conical portion is $6\ cm$ and its height is $4\ cm$ . Determine the surface area of the toy. (Use $\pi=3.14$ )

Surface area of the toy

$=\pi rl+2\pi { r }^{ 2 }$

$=\pi r\left( l+2r \right) =3.14\times 3\left[ \sqrt { { r }^{ 2 }+{ h }^{ 2 } } +2\times 3 \right]$

$=9.42\left( \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } +6 \right) =9.42\left( 5+6 \right)$

$=103.62\ {cm}^2$

Find the volume of a solid in the form of a right circular cylinder with hemispherical ends whose length is $2.7\ m$ and the diameter of each hemispherical end is $0.7\ m$

Volume of solid

$=\pi { r }^{ 2 }h+2\times \cfrac { 2 }{ 3 } \pi { r }^{ 3 }$

$\Rightarrow \pi { r }^{ 2 }\left[ h+\cfrac { 4 }{ 3 } r \right] =\cfrac { 22 }{ 7 } \times { \left( \cfrac { 0.7 }{ 2 } \right) }^{ 2 }\left[ \left( 2.7-2\times \cfrac { 0.7 }{ 2 } \right) +\cfrac { 4 }{ 3 } \times \cfrac { 0.7 }{ 2 } \right]$

$=\cfrac { 22 }{ 7 } \times \cfrac { 0.7 }{ 2 } \times \cfrac { 0.7 }{ 2 } \times 2.47=0.95\ m^{3}$

An iron pillar consists of a cylindrical portion $2.8m$ high and $20cm$ in diameter and a cone $42cm$ high is surrounding it. Find the weight of the pillar, given that $1$ cubic $cm$ of iron weights $7.5gm$ .

Given,

An iron pillar consists of a cylindrical portion and a cone mounted on it.

The height of the cylindrical portion of the pillar,

$H=2.8\ m=280\ cm$

The height of the conical portion of the pillar,

$h=42\ cm$

The diameter of the cylindrical portion of the pillar = diameter of the circular base of cone =

$D=20\ cm$

The radius of the circular base of cylinder/ cone

$r=\dfrac{D}{2}=10\ cm$

Now,

Volume of the pillar,

$(V)$ = Volume of the cylindrical portion of pillar + volume of the conical portion of the pillar.

$\rightarrow V=\pi r^{2}H + \dfrac{1}{3}\pi r^{2}h$

$\rightarrow V=\bigg(\dfrac{22}{7} \times 10^{2}\times280 + \dfrac{1}{3}\times\dfrac{22}{7} \times10^{2}\times42\bigg)\ cm^{3}$

$\rightarrow V=\bigg(22 \times 100\times40 + 22 \times100\times2\bigg)\ cm^{3}$

$\rightarrow V=\bigg(88000 + 4400\bigg)\ cm^{3}$

$\rightarrow V=92400\ cm^{3}$

Hence, volume of iron pillar is

$92400\ cm^{3}$

Given,

Weight of

$1\ cm^{3}$ iron

$=7.5\ gm.$

Hence, weight of

$92400\ cm^{3}$ iron

$=7.5\times92400\ gm.$

$=693000\ gm.$

$=693\ Kg.$

$\because 1Kg=1000\ gm.$

Hence, the weight of iron piller is

$693\ Kg.$

A toy is in the form of a cone mounted on a hemisphere of radius $3.5cm$ . The total height of the toy is $15.5cm$ , find the total surface area and volume of the toy.

We know radius of cone

$=$ radius of hemisphere

$=r=3.5$ cm

Slant height of the cone

$l=\sqrt {r^2+h^2}$

$=\sqrt {(3.5)^2+(12)^2}$

$=12.5$

$cm^2$

TSA of toy

$=\pi rl+2\pi { r }^{ 2 }$

$=\pi r\left( l+2r \right)$

$=\cfrac { 22 }{ 7 } \times 3.5\left( 12.5 +2\times 3.5 \right)$

$=214.5\ cm^{2}$

Volume of toy

$=\cfrac { 1 }{ 3 } \pi { r }^{ 2 }h+\cfrac { 2 }{ 3 } \pi { r }^{ 3 }$

$=\cfrac { 1 }{ 3 } \pi { r }^{ 2 }\left[ h+2r \right]$

$=\cfrac { 1 }{ 3 } \times \cfrac { 22 }{ 7 } \times 3.5\times 3.5\left( 12+2\times 3.5 \right)$

$=243.83\ cm^{3}$

A circus tent is cylindrical to a height of $3$ metres and conical above it. If its diameter is $105m$ and the slant height of the conical portion is $53m$ , calculate the length of the canvas $5m$ wide to make the required tent.

Canvas required

$=\pi rl+2\pi rh=\pi r\left( l+2h \right)$

$=\cfrac { 22 }{ 7 } \times \cfrac { 105 }{ 2 } \left( 53+3\times 2 \right)$

$=165\times 59$

$=9735㎡$

Let length of canvas be l then

$l\times 5=9735㎡\Rightarrow l=\cfrac { 9735 }{ 5 } =1947m$

Therefore, length of canvas

$=1947m$

A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of the base of the cylinder or the cone is $24m$ . The height of the cylinder is $11m$ . If the vertex of the cone is $16m$ above the ground, find the area of the canvas required for making the tent. (Use $\pi=22/7$ )

Area of canvas required

$=\pi rl+2\pi rH$

$\Rightarrow \pi r\sqrt { { r }^{ 2 }+{ h }^{ 2 } } +2\pi rH=\pi r\left( \sqrt { { r }^{ 2 }+{ h }^{ 2 } } +2H \right)$

$=\cfrac { 22 }{ 7 } \times 12\left( \sqrt { { 12 }^{ 2 }+{ 5 }^{ 2 } } +2\times 11 \right) =1320m^{2}$

A solid is composed of a cylinder with hemispherical ends. If the length of the whole solid is $108\ cm$ and the diameter of the cylinder is $36\ cm$ , find the cost of polishing the surface at the rate of $7$ paise per ${cm}^{2}$ . (Use $\pi=3.1416$ )

$C.S.A$ of solid

$=2\pi rh+2\times 2\pi { r }^{ 2 }=2\pi r\left( h+2r \right)$

$=2\times 3.1416\times 18\left( \left( 108-2\times 18 \right) +2\times 18 \right)$

$=2\times 3.1416\times 18\times 108=12214.5408㎠$

Cost of polishing =Rs.

$\left( 12214.5408\times 0.07 \right)$ =Rs.

$855.02$

In the figure, from the top of a solid cone of height $12cm$ and base radius $6cm$ , a cone of height $4cm$ is removed by a plane parallel to the base. Find the total surface area of the remaining solid. (Use $\pi=22/7$ and $\sqrt{5}=2.236$ )

$\triangle OAB\sim \triangle OCD$ [By AA similarity]

$\dfrac { OA }{ OC } =\dfrac { AB }{ CD }$

$AB=\dfrac { 4\times 6 }{ 12 }$

$=2cm$

TSA of frustum

$=\pi \left( { R }_{ 1 }+{ R }_{ 2 } \right) l+\pi \left( { R }_{ 1 }^{ 2 }+{ R }_{ 2 }^{ 2 } \right)$

$=\pi \left[ \left( { R }_{ 1 }+{ R }_{ 2 } \right) \sqrt { { \left( { R }_{ 1 }-{ R }_{ 2 } \right) }^{ 2 }+{ h }^{ 2 } } +{ R }_{ 1 }^{ 2 }+{ R }_{ 2 }^{ 2 } \right]$

$=\dfrac { 22 }{ 7 } \left[ \left( 6+2 \right) \sqrt { { \left( 6-2 \right) }^{ 2 }+{ 8 }^{ 2 } } +{ 6 }^{ 2 }+{ 2 }^{ 2 } \right]$

$=\dfrac { 22 }{ 7 } \left[ 8\times 4\sqrt { 5 } +36+4 \right]$

$=\dfrac { 22 }{ 7 } \left[ 32\times 2.236+40 \right]$

$=350.59{ cm }^{ 2 }$

The radii of the ends of a frustum of a right circular cone are $5$ metres and $8$ metres and its lateral height is $5$ metres. Find the lateral surface area and volume of the frustum.

Given, radii of the ends of a frustum are

$r=5\ m$ and

$R=8\ m$

Lateral height

$l=5\ m$

LSA of frusturn

$=\pi \left( R+r \right) l$

$=\cfrac { 22 }{ 7 } \left( 8+5 \right) \times 5\ m^2$

$=204.29\ m^2$

Height

$h=\sqrt { { l }^{ 2 }-{ \left( R-r \right) }^{ 2 } }$

$=\sqrt { { 5 }^{ 2 }-{ \left( 8-5 \right) }^{ 2 } } \ m$

$=\sqrt { 16 } \ m$

$=4\ m$

Volume of the frustum

$=\cfrac { 1 }{ 3 } \pi h\left( { R }^{ 2 }+{ r }^{ 2 }+Rr \right)$

$=\cfrac { 1 }{ 3 } \times \cfrac { 22 }{ 7 } \times 4\left( { 8 }^{ 2 }+{ 5 }^{ 2 }+8\times 5 \right)$

$=\cfrac { 1 }{ 3 } \times \cfrac { 22 }{ 7 } \times 4\times 129$

$=540.57\ m^{2}$

Hence, the lateral surface area is

$204.29\ m^2$ and volume is

$540.57\ m^{2}$ .

A solid is in the form of a cylinder with hemispherical ends. Total height of the solid is $19cm$ and the diameter of the cylinder is $7cm$ . Find the volume and total surface area of the solid.

Volume of solid

$=\pi { r }^{ 2 }h+2\times \cfrac { 2 }{ 3 } \pi { r }^{ 3 }$

$=\pi { r }^{ 2 }\left[ h+\cfrac { 4r }{ 3 } \right] =\cfrac { 22 }{ 7 } \times \cfrac { 7 }{ 2 } \times \cfrac { 7 }{ 2 } \left[ \left( 19-7 \right) +\cfrac { 4\times 7 }{ 3\times 2 } \right]$

$=38.5\left( 12+\cfrac { 14 }{ 3 } \right) =38.5\times \cfrac { 50 }{ 3 }$

$=641.67cm^3$

TSA of solid

$=2\pi rh+2\times 2\pi { r }^{ 2 }=2\pi r\left( h+2r \right)$

$=2\times \cfrac { 22 }{ 7 } \times \cfrac { 7 }{ 2 } \left[ \left( 19-7 \right) +2\times \cfrac { 7 }{ 2 } \right]$

$=22\left( 12+7 \right) =22\times 19$

$=418cm^2$

The radii of the ends of a bucket of height $24cm$ are $15cm$ and $5cm$ . Find its capacity. (Take $\pi=22/7$ )

Capacity of bucket

$=\cfrac { 1 }{ 3 } \pi h\left( { R }^{ 2 }+{ r }^{ 2 }+Rr \right) =\cfrac { 1 }{ 3 } \times \cfrac { 22 }{ 7 } \times 24\left( { 15 }^{ 2 }+{ 5 }^{ 2 }+15\times 5 \right)$

$=\cfrac { 176 }{ 7 } \times 325=8171.43cm^{2}$

A golf ball has a diameter equal to $4.2\;cm$ . Its surface has $200$ dimples each of radius $2\;mm$ . Calculate the total surface area which is exposed to the surroundings assuming that the dimples are hemispherical.

The total surface area of golf ball exposed to surroundings will be equal to the surface area of a golf ball without dimples plus area of

$200$ dimples and then we have to subtract the area of flat surface hidden by dimples

$\begin{aligned}{}A& = 4\pi {r^2} + 200 \times 2\pi r_1^2 - 200 \times \pi r_1^2\\& = 4\pi \left[ {{{\left( {2.1} \right)}^2} + 100 \times {{\left( {0.2} \right)}^2} - 50 \times {{\left( {0.2} \right)}^2}} \right]\\ &= 80.58\;c{m^2}\end{aligned}$

An oil funnel of tin sheet consists of a cylindrical portion $10cm$ long attached to a frustum of a cone. If the total height be $22cm$ , the diameter of the cylindrical portion $8cm$ and the diameter of the top of the funnel $18cm$ , find the area of the tin required.

Area of tin required

$=\pi \left( R+r \right) l+2\pi r{ h }_{ 1 }$

$=\pi \left[ \left( R+r \right) \sqrt { { \left( R-r \right) }^{ 2 }+{ h }_{ 2 }^{ 2 } } \right] +2\pi \times 4\times 10$

$=\cfrac { 22 }{ 7 } \left[ \left( 9+4 \right) \sqrt { { \left( 9-4 \right) }^{ 2 }+{ 12 }^{ 2 } } +2\times 4\times 10 \right]$

$=\cfrac { 22 }{ 7 } \left[ 13\times \sqrt { 169 } +80 \right] =\cfrac { 22 }{ 7 } \times \left( \left( 13\times 13 \right) +80 \right)$

$=\cfrac { 22 }{ 7 } \times 249=249\pi cm^{2}$

The largest sphere is carved out of a cube of side $10.5cm$ . Find the volume of the sphere.

Let

$r$ be the radius of sphere.
Diameter of sphere

$=$ side of cube =

$2r=10.5 \ cm$

$\therefore$ Radius

$=\cfrac { 10.5 }{ 2 } =5.25 \ cm$

Volume

$=\cfrac { 4 }{ 3 } \pi { r }^{ 3 }=\cfrac { 4 }{ 3 } \times \cfrac { 22 }{ 7 } \times { \left( 5.25 \right) }^{ 3 }$

$=606.375cm^{3}$

If a cone of radius $10cm$ is divided into two parts by drawing a plane through the mid-point of its axis, parallel to its base. Compare the volumes of the two parts.

Volume of cone

$=\cfrac { 1 }{ 3 } \pi { r }_{ 1 }^{ 2 }{ h }_{ 1 }=\cfrac { 1 }{ 3 } \pi { \left( \cfrac { r }{ 2 } \right) }^{ 2 }\times \cfrac { h }{ 2 }$

$=\cfrac { 1 }{ 3 } \pi { \left( \cfrac { 10 }{ 2 } \right) }^{ 2 }\times \cfrac { h }{ 2 }$

$=\cfrac { 25\pi h }{ 6 } cm^3$

Volume of frustum ABCD

$=\cfrac { 1 }{ 3 } \pi { h }_{ 2 }\left( { R }^{ 2 }+{ r }^{ 2 }+Rr \right)$

$=\cfrac { 1 }{ 3 } \pi \times \cfrac { h }{ 2 } \left( { 10 }^{ 2 }+{ 5 }^{ 2 }+10\times 5 \right)$

$=\cfrac { 175\pi h }{ 6 }$

Required ratio

$=\cfrac { \cfrac { 25\pi h }{ 6 } }{ \cfrac { 175\pi h }{ 6 } } =\cfrac { 25 }{ 175 } =\cfrac { 1 }{ 7 } =1:7$

A frustum of a cone is $9cm$ thick and the diameters of its circular ends are $28cm$ and $4cm$ . Find the volume and lateral surface area of the frustum. (Take $\pi=22/7$ )

Volume of frusturn

$=\cfrac { 1 }{ 3 } \pi h\left( { R }^{ 2 }+{ r }^{ 2 }+Rr \right)$

$=\cfrac { 1 }{ 3 } \times \pi \times 9\left( { 14 }^{ 2 }+{ 2 }^{ 2 }+14\times 2 \right)$

$=684\pi cm^{2}$

LSA of frusturn

$=\pi \left( R+r \right) l=\pi \left( R+r \right) \sqrt { { \left( R-r \right) }^{ 2 }+{ h }^{ 2 } }$

$=\pi \left( 14+2 \right) \sqrt { { \left( 14-2 \right) }^{ 2 }+{ 9 }^{ 2 } } =16\pi \times \sqrt { 144+81 } =16\pi \sqrt { 225 }$

$=16\pi \times 15=240\pi cm^{2}$

A container open at the top, is in the form of a frustum of a cone of height $24cm$ with radii of its lower and upper circular ends as $8cm$ and $20cm$ respectively. Find the cost of milk which can completely fill the container at the rate of Rs. $21$ per litre. (Use $\pi=22/7$ )

Capacity of container

$=\cfrac { 1 }{ 3 } \pi h\left( { R }^{ 2 }+{ r }^{ 2 }+Rr \right)$

$=\cfrac { 1 }{ 3 } \times \cfrac { 22 }{ 7 } \times 24\left( { 20 }^{ 2 }+{ 8 }^{ 2 }+20\times 8 \right)$

$=\cfrac { 176 }{ 7 } \times 624=15689.14cm^{3}$

$=15.68914ℓ$

Cost of milk=Rs.

$\left( 15.68914\times 21 \right) \approx$

$Rs.$

$329.47$

A conical vessel of radius $6cm$ and height $8cm$ is completely filled with water. A sphere is lowered into water and its size is such that when it touches the sides, it is just immersed as shown in figure. What fraction of water over flows

Let the radius of the sphere be

$r$ cm
In

$\triangle VO'A$

we have

$\tan { \theta } =\cfrac { 6 }{ 8 } =\cfrac { 3 }{ 4 } \Rightarrow \sin { \theta } =\cfrac { 3 }{ 5 }$

$\triangle VPO$ , we have

$\sin { \theta } =\cfrac { r }{ VO } \Rightarrow \cfrac { 3 }{ 5 } =\cfrac { r }{ 8-r } \Rightarrow 24-3r=5r\Rightarrow r=3cm$

Volume of the sphere

${ V }_{ 1 }=\cfrac { 4 }{ 3 } \pi { \times 3 }^{ 3 }{ cm }^{ 3 }=36\pi { cm }^{ 3 }\quad$

Volume of the water

${ V }_{ 2 }=\cfrac { 1 }{ 3 } \pi \times { 6 }^{ 2 }\times 8{ cm }^{ 3 }=96\pi { cm }^{ 3 }\quad$

Clearly the volume of the water that flows out of the cone is same as the volume of the sphere

${V}_{1}$

Fraction of the water that flows out

${ V }_{ 1 }:{ V }_{ 2 }=36\pi :96\pi \quad=36:96$

A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is $19cm$ and the diameter of the cylinder is $7cm$ . find the volume and total surface area of the solid (Use $\pi =22/7$ )

Let

$r$ cm be the radius and

$h$ cm the height of the cylinder. Then,

$r=\cfrac { 7 }{ 2 } cm ; h=\left( 19-2\times \cfrac { 7 }{ 2 } \right) cm=12cm$
Also radius of hemisphere

$=\cfrac { 7 }{ 2 } cm=rcm\quad$

Now,
volume of the solid = volume of the cylinder + volume of two hemisphere

$\left\{ \pi { r }^{ 2 }h+2\left( \cfrac { 2 }{ 3 } \pi { r }^{ 3 } \right) \right\} { cm }^{ 3 }=\pi { r }^{ 2 }\left( h+\cfrac { 4r }{ 3 } \right) { cm }^{ 3 }$

$=\left\{ \cfrac { 22 }{ 7 } \times { \left( \cfrac { 7 }{ 2 } \right) }^{ 2 }\times \left( 12+\cfrac { 4 }{ 3 } \times \cfrac { 7 }{ 2 } \right) \right\} { cm }^{ 3 }=\cfrac { 22 }{ 7 } \times \cfrac { 7 }{ 2 } \times \cfrac { 7 }{ 2 } \times \cfrac { 50 }{ 3 } { cm }^{ 3 }=641.66{ cm }^{ 3 }\quad$

Surface area of the solid

$=$ curved surface are of cylinder + surface area of two hemisphere

$=\left( 2\pi rh+2\times 2\pi { r }^{ 2 } \right) { cm }^{ 2 }$

$=2\pi r\left( h+2r \right) { cm }^{ 2 }$

$=2\times \cfrac { 22 }{ 7 } \times \cfrac { 7 }{ 2 } \left( 12+2\times \cfrac { 22 }{ 7 } \times \cfrac { 7 }{ 2 } \right) { cm }^{ 2 }$

$=\left( 2\times \cfrac { 22 }{ 7 } \times \cfrac { 7 }{ 2 } \times 19 \right) { cm }^{ 2 }=418{ cm }^{ 2 }$

A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The height and radius of the cylindrical part are $13cm$ and $5cm$ respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Calculate the surface area of the toy if the height of the conical part is $12cm$ .

Let

$r$ cm be the radius and

$h$ cm the height of the cylindrical part. Then,

$r=5cm,h=13cm$

Clearly radii of the spherical part and base of the conical part are also

$r$ cm. Let

${h}_{1}$ cm be the height,

$l$ cm be the slant height of the conical part.

Then

${ l }^{ 2 }={ r }^{ 2 }+{ h }_{ 1 }^{ 2 }\Rightarrow l=\sqrt { { r }^{ 2 }+{ h }_{ 1 }^{ 2 } } \Rightarrow l=\sqrt { { 5 }^{ 2 }+{ 12 }^{ 2 } } =13cm\quad$

Now
surface area of the toy

$=$
curved surface area of the cylindrical part + curved surface area of hemispherical part + curved surface area of conical part

$=\left( 2\pi rh+2\pi { r }^{ 2 }+\pi rl \right) { cm }^{ 2 }$

$=\pi r\left( 2h+2r+l \right) { cm }^{ 2 }$

$=\left\{ \cfrac { 22 }{ 7 } \times 5\times \left( 2\times 13+2\times 5+13 \right) \right\} { cm }^{ 2 }$

$=770{ cm }^{ 2 }\quad$

1029924_1010654_ans_d1e31faafe20412d8e6179dbc6002f29.png

A vessel is in the form of a hemispherical bowl mounted by a hollow cylinder. The diameter of the sphere is $14cm$ and the total height of the vessel is $13cm$ . Find its capacity. (Take $\pi =22/7$ )

Let

$r$ be the radius of the hemispherical bowl and

$h$ be the height of the cylinder.

Then,

$r=7cm$ and

$h=6cm$

Total capacity of the bowl

$=$ volume of the cylinder + volume of the hemisphere

$=\left( \pi { r }^{ 2 }h+\cfrac { 2 }{ 3 } \pi { r }^{ 3 } \right) { cm }^{ 3 }$

$=\pi { r }^{ 2 }\left( h+\cfrac { 2 }{ 3 } r \right) { cm }^{ 3 }$

$=\cfrac { 22 }{ 7 } \times { 7 }^{ 2 }\times \left( 6+\cfrac { 2 }{ 3 } \times 7 \right) { cm }^{ 3 }$

$=1642.66{ cm }^{ 3 }$

1041262_1010658_ans_31be84169ecc47379714db9728ba970d.png

In figure two circular flower beads have been shown on two sides of a square lawn ABCD of side 56 m. If the center of each circular flower bed is the point of intersection of the diagonals of the square lawn, find the sum of the areas of the lawns and the flower beds.

We have,

$\Delta ABC$ by Pythagoras theorem

$AC = BD =$

$\sqrt{AB^2 \, + \, BC^2} \,$

$AC = BD =$

$\sqrt{56^2 \, + \, 56^2} \, = \, 56\sqrt{2} \, m$

$\therefore$

$OA = OB =$

$\dfrac{1}{2}$

$AC = 28$

$\sqrt{2}\ m$

So, the radius of the circle having a center at the point of intersection of diagonals is

$28\sqrt{2}\ m$

Now,

Area of one circular end

$=$ Area of a segment of angle

$90^{\circ}$ in a circle of radius

$28\sqrt{2}$

Area

$= \, \left(\dfrac{\pi \theta}{360} -\sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} \right )r^2$

Substituting

$r=28\sqrt{2}$ and

$\theta = 90^\circ$

$= \left\{\dfrac{22}{7} \times \dfrac{90}{360} -\sin 45^{\circ} \cos 45^{\circ} \right\} \times (28\sqrt{2})^2 \, m^2$

$= 28 \times 28 \times 2 \times \dfrac{4}{14}\ cm^2$

$= 448 \, m^2$

$\therefore$ Area of two flower beds

$=2$

$\times$

$448$

$m^2 = 896 \, m^2$

Area of the square lawn

$=56 \times 56 m^2 = 4032 \, m^2$

Total area

$= 896 + 4032 = 4928 \ m^2$

A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their respective volume is

Volume of cone

$=\dfrac { 1 }{ 3 } \pi { r }^{ 2 }h$

Volume of cylinder

$=\pi { r }^{ 2 }h$

Volume of hemisphere

$=\dfrac { 2 }{ 3 } \pi { r }^{ 3 }$

These stand on common base & of same height, so that

$r=h$

Now,

$\dfrac { 1 }{ 3 } \pi { r }^{ 3 }:\dfrac { 2 }{ 3 } \pi { r }^{ 3 }:\pi { r }^{ 3 }=1:2:3$

The radius of a solid iron sphere is $8cm$ . Eight rings of iron plate of external radius $6\cfrac { 2 }{ 3 } cm$ and thickness $3cm$ are made by melting this sphere. Find the internal diameter of each ring.

We have
Volume of solid iron sphere

$\cfrac { 4 }{ 3 } \pi \times { 8 }^{ 3 }{ cm }^{ 3 }=\cfrac { 2048 }{ 2 } \pi { cm }^{ 3 }\quad$

External radius of each iron ring

$=6\cfrac { 2 }{ 3 } cm=\cfrac { 20 }{ 3 } cm\quad$

Let the internal radius of each ring be

$r$ cm

Since each ring forms a hollow cylindrical shell of external and internal radii

$\cfrac { 20 }{ 3 } cm\quad$ and

$r$ cm respectively and height

$3cm$

Volume of each ring

$=\pi \left\{ { \left( \cfrac { 20 }{ 3 } \right) }^{ 2 }-{ r }^{ 2 } \right\} \times 3{ cm }^{ 3 }$

volume of 8 such rings

$=8\pi \left( \cfrac { 400 }{ 9 } -{ r }^{ 2 } \right) \times 3{ cm }^{ 3 }=24\pi \left( \cfrac { 400 }{ 9 } -{ r }^{ 2 } \right) { cm }^{ 3 }$

Clearly volume of 8 rings= volume of the sphere

$\Rightarrow 24\pi \left( \cfrac { 400 }{ 9 } -{ r }^{ 2 } \right) =\cfrac { 2048 }{ 2 } \pi$

$\Rightarrow \cfrac { 400 }{ 9 } -{ r }^{ 2 }=\cfrac { 2048 }{ 2 } \pi \times \cfrac { 1 }{ 24\pi } \quad$

$\Rightarrow { r }^{ 2 }=16$

$\Rightarrow r=4cm$

The interior of a building is in the form of cylinder of diameter $4.3m$ and height $3.8m$ surrounded by a cone whose vertical angle is a right angle. Find the area of the surface and the volume of the building (Take $\pi =3.14$ )

We have
Radius of the base of the cylinder

${r}_{1}=\cfrac { 4.3 }{ 2 } m=2.15m$

Radius of base of the cone

$={r}_{1}=2.15m$

Height of the cylinder

${h}_{1}=3.8m$

$\triangle {VOA}$ we have

$\sin { { 45 }^{ o } } =\cfrac { OA }{ VA }$

$\Rightarrow \cfrac { 1 }{ \sqrt { 2 } } =\cfrac { 2.15 }{ VA }$

$\Rightarrow VA=\left( \sqrt { 2 } \times 2.15 \right) m=3.04m$

Clearly

$\triangle {VOA}$ is an isosceles triangle

Therefore,

$VO=OA=2.15m$

Thus, we have

height of the cone

$={h}_{2}=VO=2.15m$

Slant height of the ocne

${l}_{2}=VA=3.04m$

Surface area of the building = Surface area of the cylinder + Surface area of cone

$=\left( 2\pi { r }_{ 1 }^{ }{ h }_{ 1 }+\pi { r }_{ 2 }^{ }{ l }_{ 2 } \right) { m }^{ 2 }=\left( 2\pi { r }_{ 1 }^{ }{ h }_{ 1 }+\pi { r }_{ 1 }^{ }{ l }_{ 2 } \right) { m }^{ 2 }$

$=\pi { r }_{ 1 }^{ }\left( 2{ h }_{ 1 }+{ l }_{ 2 } \right) { m }^{ 2 }=3.14\times 2.15\times \left( 2\times 3.8+3.04 \right) { m }^{ 2 }=3.14\times 2.15\times 10.64{ m }^{ 2 }=71.83{ m }^{ 2 }\quad$

Volume of the building = volume of the cylinder + volume of the cone

$=\left( \pi { r }_{ 1 }^{ 2 }{ h }_{ 1 }+\cfrac { 1 }{ 3 } \pi { r }_{ 2 }^{ 2 }{ h }_{ 2 } \right) { m }^{ 3 }=\left( \pi { r }_{ 1 }^{ 2 }{ h }_{ 1 }+\cfrac { 1 }{ 3 } \pi { r }_{ 1 }^{ 2 }{ h }_{ 2 } \right) { m }^{ 3 }\left[ \because { r }_{ 2 }={ r }_{ 1 } \right]$

$=\pi { r }_{ 1 }^{ 2 }\left( { h }_{ 1 }+\cfrac { 1 }{ 3 } { h }_{ 2 } \right) { m }^{ 3 }=3.14\times 2.15\times 2.15\times \left( 3.8+\cfrac { 2.15 }{ 3 } \right) { m }^{ 3 }=65.55{ m }^{ 3 }\quad$

A solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is $3.5cm$ and the height of the cone is $4cm$ . The solid is placed in a cylindrical tub, full of water, in such a way that the whole solid is submerged in water. If the radius of the cylinder is $5cm$ and its height is $10.5cm$ , find the volume of water left in the cylindrical tub $\left (Use \ \pi=\dfrac{22}7 \right )$

We have,

$VO=4cm,OA=OB=OO'=3.5cm$

Volume of solid = volume of its conical part + volume of its semi-spherical part

Volume of the solid

$\left\{ \cfrac { 1 }{ 3 } \times \cfrac { 22 }{ 7 } \times { (3.5) }^{ 2 }\times 4+\cfrac { 2 }{ 3 } \times \cfrac { 22 }{ 7 } \times { (3.5) }^{ 3 } \right\} { cm }^{ 3 }\quad$

$=\cfrac { 1 }{ 3 } \times \cfrac { 22 }{ 7 } \times { (3.5) }^{ 2 }\left( 4+2\times 3.5 \right) { cm }^{ 3 }=\left\{ \cfrac { 1 }{ 3 } \times \cfrac { 22 }{ 7 } \times { (\cfrac { 7 }{ 2 } ) }^{ 2 }\times 11 \right\} { cm }^{ 3 }$

Clearly when the solid is submerged in the cylindrical tub the volume of water that flows out of the cylinder is equal to the volume of the solid

hence,
Volume of water left in the cylinder = volume of cylinder - volume of the solid

$=\left\{ \cfrac { 22 }{ 7 } \times { (5) }^{ 2 }\times 10.5-\cfrac { 1 }{ 3 } \times \cfrac { 22 }{ 7 } \times { (\cfrac { 7 }{ 2 } ) }^{ 2 }\times 11 \right\} { cm }^{ 3 }$

$=683.83{ cm }^{ 3 }\quad$

A circus tent is cylindrical up to a height of $3m$ and conical above. If the diameter of the base is $105m$ and the slant height of the conical part is $53m$ , find the total canvas used in making the tent.

We have

Total canvas used = curved surface area of cylinder + curved surface area of cone

Total canvas used

$=\left( 2\times \cfrac { 22 }{ 7 } \times 52.3\times 3+\cfrac { 22 }{ 7 } \times 52.5\times 53 \right) { m }^{ 3 }\left[ \because S=2\pi rh+\pi rl \right]$

$=\cfrac { 22 }{ 7 } \times 52.5\left( 6+53 \right) { m }^{ 2 }=9735{ m }^{ 2 }$

1041268_1010608_ans_803c2b5076224a43b2cf200c24b16f5f.png

Mayank made a bird-bath for his garden in the shape of cylinder with a hemispherical depression at one end as shown in figure. The height of the hollow cylinder is $1.45\ m$ and its radius is $30\ cm$ . Find the total surface area of the bird-bath. (Take $\pi=22/7$ )

Let

$r$ be the common radius of the cylinder and hemisphere and

$h$ be the height of the hollow cylinder.

Then

$r=30cm ; h=1.45m=145cm$

Total surface are of the bird-bath = curved surface area of the cylinder + curved surface area of the hemisphere

$=2\pi rh+2\pi { r }^{ 2 }$

$=2\pi (h+r)$

$=2\times \cfrac { 22 }{ 7 } \times 30(145+30){ cm }^{ 2 }$

$=3.3{ m }^{ 2 }\quad$

A solid toy is in the form of a right circular cylinder with a hemispherical shape at one end and a cone at the other end. Their common diameter is $4.2cm$ and the height of the cylindrical and conical portions are $12cm$ and $7cm$ respectively. Find the volume of the solid toy. (Use $\pi =22/7$ )

We have
volume of the solid toy

$=$ volume of the conical portion

$+$ volume of the cylindrical portion

$+$ volume of the hemispherical portion

Let

$V$ be the volume of toy

$V=\dfrac13 \pi r^2 h+\pi r^2h+\dfrac23 \pi r^3$

$=\cfrac { 1 }{ 3 } \pi \times { (2.1) }^{ 2 }\times 7+\pi \times { (2.1) }^{ 2 }\times 12+\cfrac { 2 }{ 3 } \times \pi \times { (2.1) }^{ 3 }$

$=(32.31+166.16+19.38) cm^3$

$=217.85cm^3$

1041251_1010664_ans_dbc0d3d3b7684154bcf054d69045339a.png

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to $1\ cm$ and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$ .

We have
Radius of the cone

$={r}_{1}=1cm$
Height of the cone

$={h}_{1}=1cm$
Radius of the hemisphere

$={r}_{2}=1cm$

Volume of the solid = volume of the cone +volume of the hemisphere

$=\cfrac { 1 }{ 3 } \pi { r }_{ 1 }^{ 2 }{ h }_{ 1 }^{ 2 }+\cfrac { 2 }{ 3 } \pi { r }_{ 2 }^{ 3 }$

$=\left( \cfrac { 1 }{ 3 } \pi \times { 1 }^{ 2 }\times 1+\cfrac { 2 }{ 3 } \pi \times { 1 }^{ 3 } \right) { cm }^{ 3 }$

$=\pi \ { cm }^{ 3 }$

A Gulab Jamun when completely ready for eating contains sugar syrup up to about $30\%$ of its volume. Find approximately how much syrup would be found in $45$ Gulab Jamun shaped like a cylinder with two hemispherical ends, if the complete length of each of the Gulab Jamun is $5 \ cm$ and its diameter is $2.8 \ cm$ .

Given:

Diameter of the hemispherical end of Gulab Jamun is

$2.8\ cm$

Therefore its radius will be

$\dfrac{2.8}{2}=1.4\ cm$

Here, each Gulab Jamun is in the shape of a cylinder with 2 hemispherical ends

And the total height of the Gulab Jamun

$=5\ cm$ [given]

So the height of the cylindrical part of the Gulab Jamun

$=5-2\times (1.4)=2.2\ cm$

$\therefore$ The volume of 1 Gulab Jamon

$=\dfrac{2}{3}\pi r^3+ \pi r^2h + \dfrac{2}{3}\pi r^3$

$=\pi r^2\left ( h+\dfrac{4}{3}r \right )$

$=\dfrac{22}{7}\times \left(1.4\right)^2 \left ( 2.2+\dfrac{4}{3} \times 1.4\right )$

$=25.05\ cm^3$

Therefore, the volume of

$45$ Gulab jamuns

$= 45 \times 25.05 \ cm^3$

Given: Volume of syrup in each Gulab Jamun

$=30\%$ of its volume.

$\therefore$ Volume of syrup in

$45$ gulabjamoons

$=\dfrac{30}{100} \times 45 \times 25.05$

$=338.176 \ cm^3$

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is $2cm$ and the diameter of the base is $4cm$ . If a right circular cylinder circumscribes the solid. Find how much more space it will cover.

Let

$BPC$ be the hemisphere and

$ABC$ be the cone mounted on the base of the hemisphere.

Let

$EFGH$ be the right circular cylinder circumscribing.
We have
Height of the cone

$=OA+2cm$
Diameter of the base of the cone

$=BC=4cm$

Radius of the hemisphere

$=BO=\cfrac { 1 }{ 2 } BC=\cfrac { 4 }{ 2 } cm=2cm$

$OP=2cm$

$AP=OP+OA=2+2=4cm$

Volume of the right circular cylinder

$\Rightarrow\pi r^2h=\pi \times { 2 }^{ 2 }\times 4{ cm }^{ 3 }=16\pi { cm }^{ 3 }$

volume of the solid toy

$=\left\{ \cfrac { 2 }{ 3 } \pi \times { 2 }^{ 3 }+\cfrac { 1 }{ 3 } \pi \times { 2 }^{ 2 }\times 2 \right\} { cm }^{ 3 }=8\pi { cm }^{ 3 }$

Required space

$=$ volume of the right circular cylinder - volume of the toy

$=16\pi { cm }^{ 3 }-8\pi { cm }^{ 3 }$

$=8\pi \ { cm }^{ 3 }$

1041248_1010677_ans_66c79d11b414417aa924b427fac39d97.png

A wooden toy rocket is in the shape of a cone mounted on a cylinder as shown in figure. The height of the entire rocket is $26cm$ , while the height of the conical part is $6cm$ . The base of the conical portion has a diameter of $5cm$ , while the base diameter of the cylindrical portion is $3cm$ . If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these coloures . (Take $\pi=22/7$ )

Let

$r$ be the radius of the base of the cone and its slant height be

$l$ . Further, let

${r}_{1}$ be the radius of the cylinder and

${h}_{1}$ be its height

We have

$r=2.5cm,h=6cm$

${ r }_{ 1 }=1.5cm;{ h }_{ 1 }=20cm$

Now

$\quad l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \Rightarrow l=\sqrt { { (2.5) }^{ 2 }+{ 6 }^{ 2 } } =6.5cm=\sqrt { 4.25+36 } =\sqrt { 40.25 }$

Now
area to be painted orange = curved surface area of the cone + base area of the cone - base area of the cylinder

$\Rightarrow \pi rl+\pi { r }^{ 2 }-\pi { r }_{ 1 }^{ 2 }$

$=\pi \left\{ rl+{ r }^{ 2 }-{ r }_{ 1 }^{ 2 } \right\}$

$=\pi \left\{ 2.5\times 6.6+{ (2.5) }^{ 2 }-{ (1.5) }^{ 2 } \right\} { cm }^{ 2 }$

$=3.14\left( 16.25+6.25-2.25 \right) { cm }^{ 2 }=63.585{ cm }^{ 2 }$

Area to be painted yellow

$=$ curved surface area of the cylinder + area of the base of the cylinder

$=2\pi { r }_{ 1 }{ h }_{ 1 }+\pi { r }_{ 1 }^{ 2 }$

$=\pi { r }_{ 1 }\left( 2{ h }_{ 1 }+{ r }_{ 1 } \right)$

$=3.14\times 1.5(2\times 20+1.5){ cm }^{ 2 }$

$=195.465{ cm }^{ 2 }\quad$

1035650_1010712_ans_c8cc3a04c9d3427d9d84c0f8927e9fe1.png

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figures. If the height of the cylinder is $10cm$ and its base is of radius $3.5cm$ , find the total surface area of the article.

We have

Height of cylinder= 10cm

radius of base= 3.5cm

Total surface area of the article =Curved surface area of the cylinder +2 surface area of a hemisphere

$=2\pi rh+2\left( 2\pi { r }^{ 2 } \right)$

$=2\pi r\left( h+2r \right)$

$=2\times \cfrac { 22 }{ 7 } \times 3.5(10+2\times 3.5){ cm }^{ 2 }$

$=374{ cm }^{ 2 }$

From a solid cylinder whose height is $2.4cm$ and diameter $1.4cm$ , a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest ${cm}^{2}$ .

We have,

Height of solid cylinder

$=$ 2.4cm

diameter of solid cylinder

$=$ 1.4cm

clearly,
Total surface area of the remaining solid

$=$ Curved surface area of the cylinder + Area of the base of the cylinder + Curved surface area of the cone.

$=2 \pi rh+\pi r^2+\pi r \sqrt{{r^2} +{h^2}}$

$=2\times \cfrac { 22 }{ 7 } \times 0.7\times 2.4+\cfrac { 22 }{ 7 } \times { (0.7) }^{ 2 }+\cfrac { 22 }{ 7 } \times 0.7\times \sqrt { { (2.4) }^{ 2 }+{ (0.7) }^{ 2 } } \left[ \because l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \right]$

$=10.56+1.54+5.5c{ m }^{ 2 }$

$=17.6c{ m }^{ 2 }\quad \quad$

1040749_1010784_ans_28d55c38872a488a926670852e9e3e9e.png

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are $15\ cm \times$ $10\ cm \times$ $3.5\ cm.$ The diameter of each of the depressions $1\ cm$ and the depth is $1.4\ cm.$ Find the volume of the wood in the entire stand.

It is given that the dimensions of the cuboidal part are

$15\ cm \times$

$10\ cm\times$

$\ 3.5cm$ .

volume of the cuboid

$=(15\times 10\times 3.5){cm}^{3}=525{cm}^{3}$

It is given that there are four conical depressions such that the radius of each depression is

$0.5\ cm$ and the depth is

$1.4\ cm$

Volume of wood taken out to make four cavities

$=4\times$ volume of a cone of base radius

$0.5\ cm$ and height

$1.4\ cm$

$=4 \times \dfrac13 \pi r^2 h$

$=4\times \cfrac { 1 }{ 3 } \times \cfrac { 22 }{ 7 } \times 0.5\times 0.5\times 1.4{ cm }^{ 3 }$

$=\cfrac { 22 }{ 15 } { cm }^{ 3 }=1.47{ cm }^{ 3 }$ (approximately)

Hence volume of the wood in the entire stand

$=\left( 525-1.47 \right) { cm }^{ 3 }=523.53\ { cm }^{ 3 }$

1040762_1010794_ans_6a6a97375dac44f0943bd042dc7029a5.png

A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are $2.1\ \text{m}$ and $4\ \text{m}$ , and slant height of the top is $2.8\ \text{m}$ , find the area of the canvas used for making the tent. Also, find the cost of canvas of the tent at the rate of $\text{Rs.}\ 500/\text {m}^{2}$

For conical portion we have

$r=2\text{ m} ~; l=2.8\text{ m}$

${S}_{1}=$ curved surface area of conical portion

$\Rightarrow { S }_{ 1 }=\pi rl\quad \quad$

$\Rightarrow { S }_{ 1 }=\pi \times 2\times 2.8\text{ m }^{ 2 }\quad \quad$

$\Rightarrow { S }_{ 1 }=5.6\pi \text{ m }^{ 2 }$

for the cylindrical portion, we have

$r=2\text{ m}~ ; l=2.1\text{ m}$

${S}_{2}=$ curved surface area of the cylindrical portion

$\Rightarrow { S }_{ 2 }=2\pi rh=2\pi \times 2\times 2\times 2.1\text{ m}=8.4\pi \text{ m }^{ 2 }\quad \quad$

Area of the canvas used for makine the tent

$={ S }_{ 1 }+{ S }_{ 2 }=\left( 5.6\pi +8.4\pi \right)\text { m }^{ 2 }=14\times \cfrac { 22 }{ 7 } \text{ m }^{ 2 }=44\text{ m }^{ 2 }\quad$

Total cost of the canvas at the rate of

$\text{Rs. }500/\text{m }^{ 2 }=\text{Rs. }(500\times 44)=\text{ Rs. } 22000$

1049086_1010726_ans_52d10283e4f246c4b968214f7190aa25.png

A bucket is in the form of a frustum of a cone and holds $28.490$ litres of water. The radii of the top and bottom are $28cm$ and $21cm$ respectively. Find the height of the bucket.

Let the height of the bucket be

$h$ cm

We have

${r}_{1}=28cm,{r}_{2}=21cm$

$V=$ volume of the bucket

$=28.490litres=28.490\times 1000{ cm }^{ 3 }=28490{ cm }^{ 3 }$

$V=28490{ cm }^{ 3 }\quad \Rightarrow V=\cfrac { 1 }{ 3 } \pi \left( { r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+{ r }_{ 1 }{ r }_{ 2 } \right) h=284490$

$\Rightarrow \cfrac { 1 }{ 3 } \times \cfrac { 22 }{ 7 } \times h\times \left( { 28 }^{ 2 }+28\times 21+{ 21 }^{ 2 } \right) =28490$

$\Rightarrow \cfrac { 22 }{ 21 } \times h\times (784+588+441)=28490\quad \quad$

$\Rightarrow h=\cfrac { 28490\times 21 }{ 22\times 1813 } cm$

$\Rightarrow h=15cm$

If the radii of the circular ends of a conical bucket which is $45cm$ high are $28cm$ and $7cm$ , find the capacity of the bucket (Use $\pi =22/7$ )

Clearly, bucket forms a frustum of a cone such that the radii of its circular ends are

${r}_{1}=28cm,{r}_{2}=7cm$ and height

$h=45m$ .

Therefore the capacity of the bucket = volume of the frustum

Capacity of the bucket

$=\cfrac { 1 }{ 3 } \pi h\left( { r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+{ r }_{ 1 }{ r }_{ 2 } \right)$

$=\cfrac { 1 }{ 3 } \times \cfrac { 22 }{ 7 } \times 45\left( { 28 }^{ 2 }+{ 7 }^{ 2 }+28\times 7 \right)$

$=22\times 15\times \left( 28\times 4+7+28 \right)$

$=330\times 147{ cm }^{ 3 }$

$=48510{ cm }^{ 3 }$

A decorative block shown in figure, is made of two solids- a cube and a hemisphere. The base of the block is a cube with edge $5cm$ , and the hemisphere fixed on the top has a diameter $4.2cm$ . Find the total surface area of the block (Take $\pi =22/7$ )

We have

Edge of cube= 5 cm

Diameter of hemisphere= 4.2 cm

Clearly

Surface area of decorative block

$=$ total surface area of the cube - base area of hemisphere + curved surface area of hemisphere

$=\left( 6\times (edge)^2-\pi { r }^{ 2 }+2\pi { r }^{ 2 } \right) { cm }^{ 2 }$

$=\left( 25 \times 6+\pi { r }^{ 2 } \right) { cm }^{ 2 }$

$=\left\{ 150+\cfrac { 22 }{ 7 } \times { \left( 2.1 \right) }^{ 2 } \right\} { cm }^{ 2 }$

$=163.86{ cm }^{ 2 }$

The slant height of the frustum of a cone is $4cm$ , and the perimeter of its circular bases are $18cm$ and $6cm$ . Find the curved surface area of the frustum of a cone

Let

${r}_{1}$ and

${r}_{2}$ be the radii of the circular bases of the frust,

$l$ be the slant height and

$h$ be the height

We have

$\Rightarrow l=4cm,2\pi { r }_{ 1 }=18,2\pi {r}_{2}=6$

$\Rightarrow l=4cm,{ r }_{ 1 }=\cfrac { 9 }{ \pi } ,{ r }_{ 2 }=\cfrac { 3 }{ \pi }$
Curved surface area

$=\pi \left( { r }_{ 1 }+{ r }_{ 2 } \right) l=\pi \left( \cfrac { 9 }{ \pi } +\cfrac { 3 }{ \pi } \right) \times 4{ cm }^{ 2 }=48{ cm }^{ 2 }\quad$

A container, open from the top, made up of a metal sheet is in the form of a frustum of a cone of height $8\ cm$ with radii of its lower and upper ends as $4\ cm$ and $10\ cm$ respectively. Find the cost of milk which can completely fill the container at the rate of Rs. $50$ per litre and the cost of metal sheet used, if the costs Rs. $50$ per $100\ cm^{2}$ . (Use $\pi =3.14$ )

Let

$l$ be the slant height of the frustum

We have

$\quad { r }_{ 1 }=20cm,{ r }_{ 2 }=8cm;h=16cm$

$\therefore l=\sqrt { { \left( { r }_{ 1 }-{ r }_{ 2 } \right) }^{ 2 }+{ h }^{ 2 } } \Rightarrow l=\sqrt { { (20-8) }^{ 2 }+{ 16 }^{ 2 } } =20cm\quad$

Now, volume of the container

$=\cfrac { 1 }{ 3 } \pi \left( { r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+{ r }_{ 1 }{ r }_{ 2 } \right) h$

$=\cfrac { 1 }{ 3 } \pi \left( { 20 }^{ 2 }+{ 8 }^{ 2 }+20\times 8 \right) \times 16{ cm }^{ 3 }=\cfrac { 3.14\times 624\times 16 }{ 3 } { cm }^{ 3 }$

$=10449.92{ cm }^{ 3 }=10.45\quad litres$

Cost of milk at the rate of

$15$ Rs. per litre

$=Rs.(10.45\times 15)=Rs. 156.75$

Total surface area of the frustum

$=\pi \left( { r }_{ 1 }+{ r }_{ 2 } \right) l+\pi { r }_{ 2 }^{ 2 }\quad$

$=\left\{ 3.14(20+8)\times 20+3.14\times { 8 }^{ 2 } \right\} { cm }^{ 2 }\quad$

$=3.15\times \left( 560+64 \right) { cm }^{ 2 }=3.14\times 624{ cm }^{ 2 }=1959.36{ cm }^{ 2 }$

Cost of metal used

$=Rs.\left( \cfrac { 1959.36\times 5 }{ 100 } \right) =Rs.97.96$

Shanta runs an industry in a shed that was in the shape of a cuboid surmounted by half-cylinder. If the base of the shed is $7\ m \times 15\ m$ and height of the cuboidal portion is $8\ m$ , find the volume of the air that the shed can hold. If the industry requires machinery which would occupy a total space of $300\ {m}^{3}$ , and there are $20$ workers each of whom would occupy $0.08\ {m}^{3}$ space on an average, how much air would be in the shed when it is working. $\left (Take \ \pi=\dfrac{22}{7} \right )$

Clearly, the volume of air inside the shed (when there is no people or machinary) is equal to the volume of air inside the cuboid and inside the half-cylinder taken together

For cuboidal part, we have

Length

$=15m$ ; breadth

$=7m$ and height

$=8m$
Volume of cuboidal part

$=15\times 7\times 8{ m }^{ 2 }=840{ m }^{ 3 }$

Clearly
Radius

$=r=\cfrac{7}{2}m$

Height (length) of half-cylinder

$=h=$ Length of cuboid

$=15m$

Volume of half cylinder

$=\cfrac { 1 }{ 2 } \pi { r }^{ 2 }h=\cfrac { 1 }{ 2 } \times \cfrac { 22 }{ 7 } \times { \left( \cfrac { 7 }{ 2 } \right) }^{ 2 }\times 15{ m }^{ 3 }=288.75{ m }^{ 3 }$

Volume of air inside the shed when there is no people or machinary

$=\left( 840+288.75 \right) { m }^{ 3 }=1128.75{ m }^{ 3 }$

Total space occupied by 20 workers

$=20\times 0.08{ m }^{ 3 }=1.6{ m }^{ 3 }\quad$

Total space occupied by the machinery

$=300{ m }^{ 3 }$

Volume of the air inside the shed when there are machine and workers inside it

$=(1128.75-1.6-300){ m }^{ 3 }=827.15{ m }^{ 3 }$

A friction clutch is in the form of a frustum of a cone, the diameter of the ends being $32cm$ and $20cm$ and length $8cm$ . Find its bearing surface and volume.

Let ABB'A be the friction clutch of slant height

$l$ cm

We have

${r}_{1}=16cm,{r}_{2}=10cm$ and

$h=8cm$

${ l }^{ 2 }={ h }^{ 2 }+{ \left( { r }_{ 1 }-{ r }_{ 2 } \right) }^{ 2 }\quad \Rightarrow { l }^{ 2 }=64+36\Rightarrow l=10cm$

Now Bearing surface of the clutch

$=$ lateral surface of the furstum

$=\pi \left( { r }_{ 1 }+{ r }_{ 2 } \right) l$

$=\cfrac { 22 }{ 7 } \times \left( 16+10 \right) \times 10{ cm }^{ 2 }$ $

$=817.14{ cm }^{ 2 }\quad$

Volume

$=\cfrac { 1 }{ 3 } \pi \left( { r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+{ r }_{ 1 }{ r }_{ 2 } \right) h\quad$

$=\cfrac { 1 }{ 3 } \times \cfrac { 22 }{ 7 } \times 8\times \left( { 16 }^{ 2 }+16\times 10+{ 10 }^{ 2 } \right) { cm }^{ 3 }\quad$

$=\cfrac { 176 }{ 21 } \times 516{ cm }^{ 3 }$

$=4324.57{ cm }^{ 3 }\quad$

1040796_1010918_ans_1315196d399e4811b69d6bcc8b3b8949.png

The radius of the base of a right circular cone is $r$ . It is cut by a plane parallel to the base at a height $h$ from the base. The distance of the boundary of the upper surface from the centre of the base of the frustum is $\sqrt { { h }^{ 2 }+\cfrac { { r }^{ 2 } }{ 9 } }$ . Show that the volume of the frustum is $\cfrac{13}{27}\pi { r }^{ 2 }h$

We have

$OA=r,OO'=h,OB'\sqrt { { h }^{ 2 }+\cfrac { { r }^{ 2 } }{ 9 } }$

$\triangle {OO'B}$ we have

${ OB' }^{ 2 }={ OO' }^{ 2 }+{ O'B }^{ 2 }$

$\Rightarrow { h }^{ 2 }+\cfrac { { r }^{ 2 } }{ 9 } ={ h }^{ 2 }+{ O'B' }^{ 2 }$

$\Rightarrow O'B'=\cfrac { r }{ 3 }$

Volume of the frustum

$=\cfrac { 1 }{ 3 } \pi \left( { r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+{ r }_{ 1 }{ r }_{ 2 } \right) h$

volume of the frustum

$=\cfrac { 1 }{ 3 } \pi \left\{ { r }^{ 2 }+{ \left( \cfrac { r }{ 3 } \right) }^{ 2 }+r\times \cfrac { r }{ 3 } \right\} h\quad \quad$

$=\cfrac { 1 }{ 3 } \pi \left\{ { r }^{ 2 }+\cfrac { { r }^{ 2 } }{ 9 } +\cfrac { { r }^{ 2 } }{ 3 } \right\} h$

$=\cfrac { 13 }{ 27 } \pi { r }^{ 2 }h$

$[Hence \ proved]$

1032444_1010966_ans_ac4a08f7795f425386fdd181071808fb.png

A solid consisting of a right cone standing on a hemisphere is placed upright in a right circular cylinder full of water and touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is $60cm$ and its height is $180cm$ , the radius of the hemisphere is $60cm$ and height of the cone is $120cm$ , assuming that the hemisphere and the cone have common base.

For the cylinder we have
Radius of the base

$=r=60cm$
Height

$=h=180cm$

Volume of water that the cylinder contains

$=V=\pi { r }^{ 2 }h=\left\{ \pi \times { 60 }^{ 2 }\times 180 \right\} {cm}^{3}$

For conical part, we have
Radius of the base

$=r=60cm$
Height

$={h}_{1}=120cm$

Volume of conical part

${V}_{1}=\cfrac { 1 }{ 3 } \pi { r }^{ 2 }{ h }_{ 1 }=\cfrac { 1 }{ 3 } \pi \times { 60 }^{ 2 }\times 120{ cm }^{ 3 }=\left\{\pi \times { 60 }^{ 2 }\times 40 \right\} { cm }^{ 3 }$
For hemispherical part, we have
Radius

$=r=60cm$

Volume of the hemisphere

${ V }_{ 2 }=\left\{ \cfrac { 2 }{ 3 } \pi \times { 60 }^{ 3 } \right\} { cm }^{ 3 }\quad$

$\Rightarrow { V }_{ 2 }=\left\{ 2\pi \times { 60 }^{ 2 }\times 20 \right\} { cm }^{ 3 }=\left( 40\pi \times { 60 }^{ 2 } \right) { cm }^{ 3 }\quad \quad$

Hence volume of the water left out in the cylinder

$\Rightarrow V={ V }_{ 1 }-{ V }_{ 2 }\quad =\left\{ \pi \times { 60 }^{ 2 }\times 180-\pi \times { 60 }^{ 2 }\times 40-40\pi \times { 60 }^{ 2 } \right\} { cm }^{ 3 }=\pi \times { 60 }^{ 2 }\times \left\{ 180-40-40 \right\} { cm }^{ 3 }$

$=\cfrac { 22 }{ 7 } \times 3600\times 100{ cm }^{ 3 }$

$=\cfrac { 22\times 36 }{ 700 } { m }^{ 3 }$

$=1.1314{ m }^{ 3 }$

1035678_1010814_ans_fb121d8ccd464e198db8b6bc8bc41337.png

The radii of the circular ends of a frustum of height $6\ cm$ are $14\ cm$ and $6\ cm$ respectively. Find the lateral surface area and total surface area of the frustum.

We have

${r}_{1}=14\ cm ; {r}_{2}=6\ cm;h=6\ cm$

Let

$l$ be the slant height of the frustum.

Then

$l=\sqrt { { h }^{ 2 }+{ \left( { r }_{ 1 }{ -r }_{ 2 } \right) }^{ 2 } } \ cm \Rightarrow l=\sqrt { 36+64 } \ cm=10\ cm$

Lateral surface area

$=\pi \left( { r }_{ 1 }{ +r }_{ 2 } \right)l$

$=\cfrac { 22 }{ 7 } \times (14+6)\times10\ { cm }^{ 2 }=628.57{ cm }^{ 2 }$

total surface area

$=\pi \left\{ { r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+({ r }_{ 1 }+{ r }_{ 2 })l \right\}$

$=\cfrac { 22 }{ 7 } \times \left( 196+36+20\times 10 \right) { cm }^{ 2 }=1357.71{ cm }^{ 2 }$

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see figure). The length of the entire capsule is $14mm$ and the diameter of the capsule is $5mm$ . Find its surface area.

From the figure, it can be seen that, the capsule is a combination of cylinder and two hemispheres.

Total surface area of capsule will be sum of surface area of cylinders and twice the surface area of hemisphere.

From the figure, length of cylinder

$=9\ cm$

radius of cylinder

$=$ radius of hemisphere

$= \dfrac{5}{2}$

Surface area of cylinder :

$=2\pi r \ell$

$=2\pi (\dfrac{5}{2})(9)$

$=45\pi$

Surface area of two hemispheres :

$=2 \times 2\pi r^2$

$=2(2\pi (\dfrac{5}{2})^{2})$

$=4\pi \dfrac{25}{4}$

$= 25\pi$

Total surface area :

$=25\pi +45\pi$

$=70\pi cm^{2}$

$=\dfrac{22}{7}\times 70$

$=220 cm^{2}$

2105047_1090856_ans_f334f2c438b84aeab854a29fe1407d38.png

A toy is in the form of a cone of radius $3.5cm$ mounted on a hemisphere of same radius. The total height of the toy is $15.5cm$ . Find the total surface area of the toy.

Given that Radius of cone

$=3.5cm=$ Radius of hemisphere

Height of toy

$=15.5cm$

Now,

$T.S.A$ of toy

$=C.S.A$ of hemispere+

$C.S.A$ of cone

$\Rightarrow slant\space height =\sqrt{h^2+r^2}$

$\Rightarrow slant\space height =\sqrt{(3.5)^2+(15.5-3.5)^2}$

$\Rightarrow slant\space height =\sqrt{12.25+144}=\sqrt {156.25}=12.5cm$

Now,

$\Rightarrow T.S.A$ of toy

$=2\pi r^2+\pi rl$

$\Rightarrow T.S.A$ of toy

$=2\pi (3.5)^2+\pi (3.5)(12.5)$

$\Rightarrow T.S.A$ of toy

$=68.25\pi\space cm^2$

A cylinder and $a$ cone have equal bases. The height of the cylinder is $3\ cm$ and the area of its base is $100\ cm^{2}$ . The cone is placed upon the cylinder. Volume of the solid figure so formed is $500\ cm^{3}$ . Find the total height (in $cm$ ) of the figure.

Consider the following question.

The height of the cylinder $=3cm$

the area of its base $=100c{{m}^{2}}$

Volume of the solid figure $=500c{{m}^{3}}$

Find the total height of the figure-

Total volume of solid figure $=$ volume of cylinder $+$ volume of cone

$A=\pi {{r}^{2}}$

$\dfrac{100}{\pi }={{r}^{2}}\,\,\,\,\,\,\,\,....\left( 1 \right)$

Put the value of eq.1 in eq 2

$500=A\times {{h}_{1}}+\dfrac{1}{3}\pi {{r}^{2}}{{h}_{2}}$

$500=100\times 3+\dfrac{1}{3}\pi \times \dfrac{100}{\pi }\times {{h}_{2}}$

$500-300=\dfrac{100{{h}_{2}}}{3}$

$200=\dfrac{100{{h}_{2}}}{3}$

${{h}_{2}}=6cm$

$H={{h}_{1}}+{{h}_{2}}$

$=3+6$

$=9cm$

Hence, this is the required answer.

The shape of a garden is rectangular in the middle and semi-circular at the ends as shown in the diagram. Find the area and the perimeter of this garden. (Length of a rectangle is $20-(3.5+3.5)$ metres)

Area = Area of rectangle

$+ 2\times$ area of semicircle

Area

$=l \times b+2\times \dfrac{1}{2} (\pi \times r^2 )$

$=(20-3.5-3.5)(7)+2\times \dfrac{1}{2}\left(\pi \times \dfrac{7}{2}\times \dfrac{7}{2}\right)$

$=13\times 7 +\dfrac{22\times 7}{2\times 2}$

$=91+38.5$

Area

$=129.5\ m^2$

Perimeter =

$2(20-(3.5+3.5))+2\times \dfrac{22}{7}\times \dfrac{7}{2}$

$=26+22$

$=48\ m$

A vessel is in the form of a hollow hemisphere mounted on a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

$\textbf{Step 1: Find Surface Area of Hemisphere}$

$\text{Given, a hollow hemisphere of has diameter 14 cm.}$

$\text{The height of the vessel is}$

$13 \ cm$ .

$\text{Height of the hemisphere=Radius of hemisphere(r)}$

$=\dfrac{14}{2}$

$=7\ cm$

$\therefore \text{C.S.A. of hemisphere}$

$=2\pi r^{2}$

$=2\times \dfrac{22}{7}\times 7^2$

$\textbf{Step 2 : Find Surface Area of Cylinder}$

$\text{Height of cylinder (h)= Height of vessel $$-$$ Height of hemisphere}$

$\Rightarrow h=13-7$

$=6 \ cm$

$\therefore \text{C.S.A. of Cylinder}$

$=2\pi rh$

$= 2\times \dfrac{22}{7}\times 6$

$\textbf{Step 3: Find Inner Surface Area of Vessel}$

$\text{Inner surface area of vessel = C.S.A. of Hemisphere+ C.S.A. of cylinder }$

$\therefore$

$\text{Inner surface area of vessel }$

$=[2\pi r^2+2\pi rh]$

$= 2\pi r(r+h)$

$= 2\times \dfrac{22}{7}\times 7(7+6)$

$\boldsymbol{\left[\because \ \pi=\dfrac{22}{7}\right]}$

$=2\times 22\times 13$

$= 572\ cm^{2}$

$\textbf{Hence, the inner surface area of the vessel is}$

$\boldsymbol{572\ cm^2}$ .

A cubical block of side $7cm$ is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

The hemisphere is mounted on the cubical block

The diameter of the hemisphere $=$ the side of the cube $= 7cm$

$\Rightarrow TSA$ of the solid $= TSA$ of cube $-$ Area of base of hemisphere $+ CSA$ of hemisphere

$\Rightarrow TSA = (6\times 7^2) –(\pi \times (3.5)^2) +( 2\times \pi \times (3.5)^2)$

$\Rightarrow TSA=(6\times 7^2) + (\pi \times (3.5)^2)$

$\Rightarrow TSA = 332. 465\ cm^2$

992852_1086214_ans_ab6f9a84a4004ddb99bf080ab152ef94.PNG

A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are $3.5\ m$ and $2\ m$ respectively, and the slant height of the top is $3.5\ m$ , find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs. $1000$ per ${m}^{2}$ . (Note that the base of the tent will not be covered with canvas)

$Area \ of\ Canvas=Curved \ Area \ of \ Cone+Curved \ Surface \ Area \ of \ the \ Cylinder$

$Lets\ calculate\ the\ Curved\ Surface\ area\ of\ Cone:$

$Diameter \ of \ cone=Diameter \ of \ Cylinder=4 \ m$

$\implies radius=\dfrac{4}{2}=2 \ m$

$Slant \ height=l=2.8\ m$

$Curved \ Surface \ area \ of \ Cone=\pi rl$

$=\dfrac{22}{7} \times2\times2.8=17.6 \ m^2$

$Curved \ Surface \ area \ of \ Cylinder=2\pi rh$

$=2\times\dfrac{22}{7}\times2\times2.1=26.4 \ m^2$

$\therefore The \ Area \ of \ Canvas=17.6 + 26.4 = 44 m^2$

Now,

$Cost\ of\ the\ canvas\ of\ the\ tent\ for\ 1 \ m^2 = Rs \ 500$

$Cost\ of\ the\ canvas\ of\ the\ tent\ for\ 44 \ m^2 = 500\times44=Rs \ 22000$

$\therefore Total \ Cost=22000\ rupees$

1040928_1090907_ans_a57cdf10d58146f28acc1b9b66df06c1.png

A hemispherical depression is cut out from one face of a cuboidal block such that the diameter $l (\text{cm})$ of the hemisphere is equal to the edge of the cube. Find the surface area of the remaining solid.

Surface area of solid=area of cube + curved surface area of hemisphere -base area of hemisphere

$=6(side)^2+2 \pi r^2-\pi r^2$

$=6l^2+2 \pi\left ( \dfrac{l}{2} \right )^2- \pi \left ( \dfrac{l}{2} \right )^2$

$=6l^2+\pi\dfrac{l^2}{2}-\pi\dfrac{l^2}{4}$

$=6l^2+\pi\dfrac{l^2}{4}$

$=l^2\left ( 6+\dfrac{\pi}{4} \right )$

$=\dfrac{1}{4}l^2(\pi+24)$

1093674_1103189_ans_157d5624a5ee46429d763d26bdb0ee38.png

A toy is in the form of a cone mounted on a hemisphere of radius 3.5 cm . The total height of the toy is $15.5 cm$ . Find the total surface area and volume of the toy .

Given: toy in form of a cone mounted on a hemisphere

radius of hemisphere

$= 3.5$ cm

Total height of toy

$= 15.5$ cm

To find: Total surface area TSA = ? Volume, V = ?

Solution :

$h = 12$ cm

$r = 3.5$ cm

Assuming that the cone is completed mounted on hemisphere

TSA of toy = CSA of cone + CSA of hemisphere

$=\pi rl+2\pi r^{2}$ [

$r\rightarrow$ radius of both cone and hemisphere as it is here]

$= \pi r[l+2r]\\$

$=\dfrac{22}{7}\times \dfrac{35}{10}\left [ \sqrt{(12)^{2}+(3.5)^{2}}+2\times \dfrac{35}{10} \right ]\\$

$= 11\left [ \sqrt{144+12.25}+7 \right ]= 11\left [ \sqrt{156.25}+7 \right ]$

$=11(12.5+7)$

$= 11\times 19.5$

$= 214.5\, cm^{2}$

Volume of toy =

$\dfrac{1}{3}\pi r^{2}h+\dfrac{2}{3}\pi r^{3}$

$=\dfrac{1}{3}\pi r^{2}[h+2r]\\$

$= \dfrac{1}{3}\times \dfrac{22}{7}\times \dfrac{35}{10}\times \dfrac{25}{10}\left[12+2\times \dfrac{35}{10}\right]\\$

$=\dfrac{77}{6}[ 12+7]=\dfrac{77\times 19}{6}=\dfrac{1463}{6}$

$= 243. 83\, cm^{3}$

1205311_1115207_ans_afef723f59f94560ab993cb174a70983.png

Water flows at the rate of $10$ m per minute through a cylindrical pipe $5$ mm in diameter. How long would it take to fill a conical vessel whose diameter at the base is $40$ cm and depth $24$ cm?

R.E.F image

Volume of cone

$= \dfrac{1}{3}\times r^{2}h$

$= \dfrac{1}{3}\times 3.14\times 20\times 20\times 24$

$= 3.14\times 400\times 8$

$= 10048\, cm^{3}$

Volume flown in

$1$ min

$= \pi r^{2}h$

$= 3.14\times \dfrac{2.5}{10}\times \dfrac{2.5}{10}\times 1000\, cm^{3}$

$= 196.25\, cm^{3}$

$\therefore$ time taken

$= \dfrac{10048}{196.25} = 51.2$

$\therefore$ 51 minutes and 2 seconds

Two cylinders are 12 cm and 18 cm in height. Their diameters are 12 cm and 16 cm respectively. Both the cylinders are melted and the material is moulded into a solid right circular cone of height 33 cm. Find its diameter.

Diameter of cylinder

${d}_{1}=12$ cm

Height of cylinder

${h}_{1}=12$ cm

Volume of cylinder

${V}_{1}=\pi{r}_{1}^{2}{h}_{1}=\pi{\left(\dfrac{{d}_{1}}{2}\right)}^{2}{h}_{1}=\pi{\left(\dfrac{12}{2}\right)}^{2}\times 12=\dfrac{\pi{12}^{2}\times 12}{4}$ ...........

$\left(1\right)$ since radius

$=\dfrac{1}{2}\times$ diameter.

Diameter of cylinder

${d}_{2}=18$ cm

Height of cylinder

${h}_{2}=16$ cm

Volume of cylinder

${V}_{2}=\pi{r}_{2}^{2}{h}_{2}=\pi{\left(\dfrac{{d}_{2}}{2}\right)}^{2}{h}_{2}=\pi{\left(\dfrac{18}{2}\right)}^{2}\times 16=\dfrac{\pi{18}^{2}\times 16}{4}$ ...........

$\left(2\right)$ since radius

$=\dfrac{1}{2}\times$ diameter.

When both the cylinders are melted and moulded into a single cylinder, then the

Volume of single cylinder

$V=\pi{r}^{2}h=\pi{\left(\dfrac{d}{2}\right)}^{2}h=\dfrac{\pi{d}^{2}\times 33}{4}$ ...........

$\left(3\right)$ where

$h=33$ cm

Now,

$\dfrac{\pi{d}^{2}\times 33}{4}= \dfrac{\pi{12}^{2}\times 12}{4}+ \dfrac{\pi{18}^{2}\times 16}{4}$

$\Rightarrow 33{d}^{2}={12}^{3}+{18}^{2}\times 16$

$\Rightarrow {d}^{2}\left(3\times 11\right)=12\times 144+18\times 18\times 16$

$\Rightarrow 11{d}^{2}=576+1728=2304$ on simplification

$\Rightarrow {d}^{2}=\dfrac{2304}{11}=209.45$

$\therefore d=\sqrt{209.45}=14.47$ cm

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand .

Volume of cuboid

$=l\times b\times h$

$=15\times10\times 3.5=525cm^3$

Volume of cone

$=\dfrac{\pi r^2h}{3}$

$=\dfrac{3.14\times (0.5)^2\times 1.4}{3}=0.366cm^3$

Volume of Wood stand

$=$ Volume of Cuboid

$-$ 4

$\times$ Volume of cone

$=525-4(0.366)$

$=523.534cm^3$

$3$ cubes of edge $4 cm$ are joined end to end. Find surface area of resulting cuboid.

$3$ cubes of edge

$4$ cm are joined end to end, then length of cuboid

$=4+4+4=12$ cm

breadth and height

$=4$ cm

Surface area

$=2(lb+bh+lh)$

$=2(12\times 4+4\times 4+12\times 4)$

$=2(48+16+48)$

$=2(112)$

$=224cm^2$ .

Pragya wrapped a cord around a circular pipe of radius $4\ cm$ (adjoining figure) and cut off the length required of the cord. Then she wrapped it around a square box of side $4\ cm$ (also shown). Did she have any cord left? $(\pi=3.14)$

Now, the circumference of the circular pipe is

$2\pi r$

$=\left( 2\times 3.14\times 4 \right) cm=25.12cm$

Given, that there's a square box of side

$4cm$

So, perimeter of the square box

$=4\times$ side

$=4\times 4=16cm$

So, Pragya has chord left of length

$=(25.12-16)cm=9.12cm$

A metal cube of side $11\;cm$ is completely submerged in water contained in a cylindrical vessel with the diameter $28\;cm$ . Find the rise in the level of water.

Meta cube side

$=11\ cm$

Volume

$=11^3=1331\ cm^3$

Cylinder where it is submerged

Radius

$d/2=28/2=14\ cm$

Let level rise

$=h$

So,

$\pi r^2 h=1331$

$\implies h=\cfrac{1331}{22\times 2\times 14}=121\ cm$

How many shots each having diameter $4.2cm$ can be made from cuboidal lead solid of dimensions $66cm\times 42cm\times 21cm$ .

Volume of sphere

$=\dfrac{4}{3}\pi r^3$

Here

Diameter

$=4.2\ cm$

Radius

$=2.1\ cm$

Volume of sphere

$=\dfrac{4}{3}\times\dfrac{22}{7} \times(2.1)^3$

$=38.808\ cm^3$

Volume of cuboid

$=lbh$

Here

$l=66\ cm$

$b=42\ cm$

$h=21\ cm$

Volume of cuboid

$=66\times42\times21$

$=58212\ cm^3$

Hence

Number of shots

$=\dfrac{Volume\ of\ cuboid}{Volume\ of\ sphere}$

$=\dfrac{58212}{38.808}$

$=1500$

Hence 1500 shots can be made.

A tent is in the form of a cylinder whose height is $5\ m$ surmounted by a cone of equal base. The diameter of base is $105\ m$ . The slant height of cone is $50\ m$ . Find the cost of cloth at $Rs.\ 15$ per square meter to make the tent.

Tent cylinder height

$=5$ m

Diameter of base

$=105$ m

Slant height of cone

$=50m=l$

Cost of cloth let Rs.

$15$ per

$m^2$ .

Area of cylinder

$=2\pi rh$

$=2\times \dfrac{22}{7}\times \left(\dfrac{105}{2}\right)\times5$

$=1650m^2$

Area of cone

$=\pi r h$

$=\dfrac{22}{7}\times \left(\dfrac{105}{2}\right)\times 50$

$=8250m^2$

Total area

$=1650+8250=9900m^2$

Total cost

$=9900\times 15$

$=Rs\, 1,48,500$

1379761_1215463_ans_a5a692e08101423588880a8a9eb85044.png

A metal cylinder with height 10 cm and diameter of base 70 cm is melted to make a number of small discs with radius 1 cm and thickness 0.5 cm. How many such discs could be made from the cylinder?

Given

Radius of cylinder

$r=\dfrac{70}{2}=35\ cm$

Height of cylinder

$h=10\ cm$

Radius of disc

$r_1=1\ cm$

Height of disc

$h_1=0.5=\dfrac{1}{2}\ cm$

Volume of cylinder

$=\pi r^2h$

$=\dfrac{22}{7}\times35\times35\times10$

$=38500\ cm^3$

Volume of small disc

$=\pi r_1^2h_1$

$=\dfrac{22}{7}\times1\times1\times\dfrac{1}{2}$

$=\dfrac{11}{7}\ cm^3$

Number of discs

$=\dfrac{38500}{11/7}$

$=24500$

The diameter of base of a chimney of kiln is 80 cm and height is 12.5 m. What will be the cost of painting this chimney from outside at the rate of Rs. 140 per sq m?

Solution :-

Given: diameter of base of chimney

$= 80\,cm = 0.8\,m$

Height of chimney

$= 12.5\,m$

curved surface are

$= 2\pi rh$

$= 2\times \dfrac{22}{7}\times 0.4\times 12.5$

$= 31.42\, m^{2}$

cost of painting

$1$ sq m

$= Rs 140$

$\therefore$ Total cost

$= 31.42\times 140$

$Rs = 4398.8$

The decorative block shown in the figure is made of two solids, a cube, and a hemisphere. The base of the block is a cube with edge $5\;{ cm }$ and the hemisphere fixed on the top has a diameter of $4.2\;{ cm }$ . Find the total surface area of the block. (Take $\pi = \dfrac { 22 } { 7 } )$

The side of the Cube is

$5\; cm.$ So, the total surface area of the cube is,

$6a^2$

$=6\times 5\times 5$

$=150\; cm^2$

The radius of hemisphere is

$\dfrac {4.2}{2}=2.1\; cm.$

The base area of hemisphere is,

$\pi r^2 =\dfrac {22}{7}\times 2.1\times 2.1$

$=13.86\; cm^2$

The surface area of the cube considered is,

$150-13.86=136.14\; cm^2$

The curved surface area of hemisphere is,

$2\pi r^2=2\times \dfrac {22}{7}\times 2.1\times 2.1$

$=27.27\; cm^2$

The total surface area of solid is

$136.14+27.72=163.86\; cm^2$

Find the area of the field ABCDEF, as shown below:

Area of field ABCDEF

$=$ Area of rectangle ABDE

$+$ Area of

$\Delta$ BCD

$+$ Area of

$\Delta$ AEF

$=(120\times 150)cm^2+\left(\dfrac{1}{2}\times 90\times 120\right)cm^2+\dfrac{1}{2}\times40\times 150$

$=18,000+5400+3,000$

$=26,400cm^2$ .

1379285_1216433_ans_65502afc7a134de285bfbd93960e2452.JPG

A farmer connects a pipe of internal diameter $20\ cm$ form a canal into a cylindrical tank in her field, which is $10\ m$ in diameter and $2\ m$ deep. If water flows through the pipe at the rate of $3\ km\ h^{-1}$ , in how much time will the tank be filled ?

$R= 10 cm$

$R= 500 cm$

$h= 200 cm$

vol. of water flow in 1 hour =

$\pi \times 10\times 10\times 300000$

vol. of cylindrical

$tan x\Rightarrow \pi \times 500\times 500\times 200$

time taken to filled the tan x = vol. of cylindrical tank

$\div$ vol. of water flow in 1 hour

$\Rightarrow \dfrac{\pi \times 500\times 500\times 200}{\pi \times 10\times 10\times 300000}$

$\Rightarrow (5\div 3)\times 60$

$\Rightarrow 100$ min

The radii of ends of a frustum are $14\text{ cm}$ and $6\text{ cm}$ respectively and its height is $6\text{ cm}$ . Find its volume.

Here,

$R=14 \text{ cm}, r=6 \text{ cm}, h=6 \text{ cm}$

$\begin{aligned}{}V &= \frac{1}{3}\pi h\left( {{r^2} + Rr + {R^2}} \right)\\ &= \frac{1}{3} \times \frac{{22}}{7} \times 6\left[ {{6^2} + (14)(6) + {{(14)}^2}} \right]\\& = \frac{{22}}{7} \times 2\left[ {36 + 84 + 196} \right]\\& = \frac{{22}}{7} \times 632\\ &= 1986.2\text{ cm}^3\end{aligned}$

Shankar made the pictures shown above.
What shapes can they be broken into so that we can find their areas easily?

We can break down the given shapes as shown above.

That way, we can easily find out their areas.

1372449_1224985_ans_6b299c81a110425c9d2e7b258a0fafe2.png

The volume of a right circular cone is $4710\ cu.cm$ . if the radius and height of the cone are in the ratio $3:4$ , find its slant height, when $\pi=3 \cdot 14$ .

Given that

the volume of right circular

cone

$=4710 \ c.m^3$

and

radius:height

$=3:4$

let radius

$=r$

height

$=h$

let the ratio

$=x$

then

$h=4x$

$r=3x$

now,A.T.Q,

Volume

$=\dfrac{1}{3}\pi r^2h$

$4710=\dfrac{1}{3}\times 3.14\times (3x)^2\times 4x$

$4710\times 3=36x^3\times 3.14$

$x^3=125$

$x=5$

now,

radius

$=3x=3\times 5=15cm.$

height

$=4x=4\times 5=20cm.$

slant height(l)

$=\sqrt{h^2+r^2}$

$=\sqrt {20^2+15^2}$

$\sqrt {400+225}$

$=\sqrt 625$

$=25c.m.$

then,

slant height(l)

$=25 \ c.m.$

Hence, this is the answer.

A semicircular sheet of metal of diameter $28cm$ is bent into an open conical cup. Find the depth and capacity of cup. $(\sqrt{3}=1.73)$

Diameter of circular sheet is

$=28\ c.m.$

Radius

$=14\ cm.$

Let the radius

$=r=14\ c.m.$

Now, circumference of semicircular

Sheet

$=\dfrac{1}{2}\times 2 \pi r$

$= \pi r$

$=14 \pi\ c.m.$

Let the height of conical cup

$=l$

Radius of circular sheet

$=14\ c.m.$

$R=14\ c.m.$

According to the question

Circumference of the base of conical cup

$=$

Circumference of semicircular sheet

$2 \pi R=14 \pi$

$R=7\ c.m.$

Depth of conical cup

$=$ Height of cone

$=\sqrt{l^{2}-r^{2}}$

$\sqrt{14^{2}-7^{2}}$

$\sqrt{196-49}$

$\sqrt{147}=7\sqrt{3}\ c.m.$

Capacity of conical cup

$=\dfrac{1}{3} \pi R^{2} h$

$=\dfrac{1}{3} \times \dfrac{22}{7} \times 7 \times 7 \times 7 \sqrt{3}$

Volume

$+\dfrac{1078}{3} \sqrt{3}\ CM^{3}$

Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is $\dfrac{4r}{3}$ . Also show that the maximum volume of the cone is $\dfrac{8}{27}$ of the volume of the sphere.

Radius of sphere=r

Radius of cone=R

Height of cone=h

$BC=\sqrt { { r }^{ 2 }-{ R }^{ 2 } }$

$\therefore h=r+\sqrt { { r }^{ 2 }-{ R }^{ 2 } }$

$\therefore$ Volume of cone

$=\cfrac { 1 }{ 3 } \pi { R }^{ 2 }\left( r+\sqrt { { r }^{ 2 }-{ R }^{ 2 } } \right)$

$\cfrac { dV }{ dR } =\cfrac { 2 }{ 3 } \pi Rr+\cfrac { 2\pi }{ 3 } R\sqrt { { r }^{ 2 }-{ R }^{ 2 } } +\cfrac { \pi { R }^{ 2 } }{ 3 } .\cfrac { \left( -2R \right) }{ 2\sqrt { { r }^{ 2 }-{ R }^{ 2 } } }$

$=\cfrac { 2\pi Rr }{ 3 } +\cfrac { 2\pi R\left( { r }^{ 2 }-{ R }^{ 2 } \right) -\pi { R }^{ 3 } }{ 3\sqrt { { r }^{ 2 }-{ R }^{ 2 } } }$

$=\cfrac { 2\pi Rr }{ 3 } +\cfrac { 2\pi R{ r }^{ 2 }-3\pi { R }^{ 3 } }{ 3\sqrt { { r }^{ 2 }-{ R }^{ 2 } } }$

For max. volume

$\cfrac { dV }{ dR } =0$

$\Rightarrow \cfrac { 2\pi Rr }{ 3 } =\cfrac { 3\pi { R }^{ 2 }-2\pi { r }^{ 2 } }{ 3\sqrt { { r }^{ 2 }-{ R }^{ 2 } } }$

$\Rightarrow 2Rr=\cfrac { 3{ R }^{ 3 }-2{ r }^{ 2 }R }{ \sqrt { { r }^{ 2 }-{ R }^{ 2 } } } \Rightarrow 2r\sqrt { { r }^{ 2 }-{ R }^{ 2 } } =3{ R }^{ 2 }-2{ r }^{ 2 }$

$\Rightarrow 9{ R }^{ 4 }-8{ r }^{ 2 }{ R }^{ 2 }=0$

$\therefore { R }^{ 2 }=0$

$\cfrac { 8{ r }^{ 2 } }{ 9 }$

$\cfrac { { d }^{ 2 }V }{ d{ R }^{ 2 } } <0$

${ R }^{ 2 }=\cfrac { 8{ r }^{ 2 } }{ 9 }$

$\therefore$ For

${ V }_{ max },{ R }^{ 2 }=\cfrac { 8{ r }^{ 2 } }{ 9 }$

$\therefore h=r+\cfrac { r }{ 3 } =\cfrac { 4r }{ 3 }$

$V=\cfrac { 1 }{ 3 } \times \cfrac { 8{ r }^{ 2 } }{ 9 } \times \cfrac { 4r }{ 3 } =\cfrac { 8 }{ 27 } \cfrac { 4\pi { r }^{ 3 } }{ 3 }$

1189464_1221350_ans_1be260668e014dd1b4caf64e7e2ed352.JPG

Derive the formula for the volume of the frustum of a cone, given to you in Section $13.5$ using the symbols as explained.

let

$ABC$ be a cone. A frustum

$DECB$ is cut by a plane parallel to its base.

let

$r_1 \ and \ r_2$ be the radil of the ends of the frustum of the cone

and

$h$ be the height of the frustum of the cone

$\triangle ABG$ and

$\triangle ADR, DR \parallel BG$

$\therefore \triangle ABG\sim \triangle ADF$

$\dfrac{DF}{BG}=\dfrac{AF}{AG}=\dfrac{AD}{AB}$

$\dfrac{r_2}{r_1}=\dfrac{h_1-h}{h_1}=\dfrac{l_1-l}{l_1}$

$\dfrac{l-l}{l_1}=\dfrac{r_2}{r_1}$

$\dfrac{l}{l_1}=\dfrac{r_1-r_2}{r_1}$

$\dfrac{l_1}{l}=\dfrac{r_1}{r_1-r_2}$

$l_1=\dfrac{r_1 l}{r_1-r_2}$

$CSA$ of frustum

$DECB=CSA$ of cone

$ABC- CSA$ of cone

$ADE$

$=\pi r_1 l_1-\pi r_2(l_1-l)$

$=\pi r_1\left(\dfrac{lr_1}{r_1-r_2}\right)-\pi r_2\left[\dfrac{r_1l}{r_1-r_2-l}-l\right]$

$=\dfrac { { \pi { r_{ 1 } }^{ 2 }l } }{ { r_{ 1 }-r_{ 2 } } } -\pi r_{ 2 }\left[ \dfrac { r_{ 1 }l-r_{ 1 }l+r_{ 2 }l }{ r_{ 1 }-r_{ 2 } } \right]$

$\dfrac{\pi{r_1}^2l}{r_1-r_2}-\dfrac{\pi{r_2}^2l}{r_1-r_2}$

$=\dfrac{\pi r(({r_1})^2-{r_2}^2)}{r_1-r_2}$

CSA frustum

$=\pi l(r_1+r_2)$

Total surface area

$=CSA$ of frustum

$+$ Area of upper conedar end

$+$ Area of lower conedar end

$=\pi(r_1+r_2)l+\pi{r_2}^2+\pi {r_1}^2$

$=\pi\left((r_1+r_2)l+{r_2}^2+ {r_1}^2\right)$

1349813_1223386_ans_0a8dee8915f04610929f612e95541358.jpg

The given figure shows a solid formed of a solid cube of side
$40$ $\mathrm { cm }$ and a solid cylinder of radius 20 $\mathrm { cm }$ and height $50$ $\mathrm { cm }$
attached to the cube as shown.Find the volume and the total
surface area of the whole solid
ITake $\pi = 3 \cdot 14 ]$

1148364_1224783_ans_ec030c50841c44d19f37915f6d13dcdb.jpg

A field is $60\ m$ long and $20\ m$ wide. A tank $12\ m$ long, $10\ m$ broad and $3\ m$ deep is dug in the field. The earth taken out of it is spread evenly over field. How mach is the level of the field raised supporting the earth taken out increases by $\dfrac {1}{8}$ of the volume?

1208463_1248748_ans_25232a5384bd45c28b65eb0e9c48ecf3.jpg

What length of canvas $3\ m$ wide will be required to make a conical tent of height $8\ m$ and radius of base $6\ m$ . (use $\pi=3.14$ )

3 m wide canvas, height 8 m & radius 6 m

$\rightarrow$ curved surface Area of tent =

$\pi .rl.$

$r = 6 \,m, h = 8 \,m$

slant height

$l = \sqrt{h^{2}+r^{2}}= \sqrt{8^{2}+6^{2}}=\sqrt{64+36}= \sqrt{100}= 10 m$

curved surface Area

$= \pi .r l = 3.14\times 6\times 10 = 188.4 \,m^{2}$

Now,

Area of canvas = Area of tend

$L\times breadth = 188.4\,m^{2}$

$L\times 3\,m = 188.4\,m^{2}$

$L = \frac{188.4}{3}$

$L = 62.8 \,meter$

1166413_1247997_ans_d78a05e8d1324acfa5c6580d413b86a9.jpg

A toy is in the form of a cone of radius $3.5\text{ cm}$ mounted on a hemisphere of the same radius. The total height of the toy is $15.5\text{ cm}$ . Find the total surface area of the toy.

Radius of cone

$= 3.5\text{ cm},$ height of cone

$= 15.5 – 3.5 = 12\text{ cm}.$

Slant height of cone can be calculated as follows:

$l=\sqrt{{h}^{2}+{r}^{2}}$

$=\sqrt{{12}^{2}+{3.5}^{2}}$

$=\sqrt{144+12.25}$

$=\sqrt{156.25}$

$=12.5\text{ cm}$

Curved surface area of cone

$=\pi rl$

$=\dfrac{22}{7}\times3.5\times 12.5$

$=137.5\text{ cm}^{2}$

Curved surface area of hemispherical portion

$=2\pi{r}^{2}$

$=2\times\dfrac{22}{7}\times3.5\times 3.5$

$=77\text{ cm}^{2}$

Hence, total surface area of the toy is equal to

$137.5+77=214.5\text{ cm}^{2}$

1269661_1246345_ans_794f4f0d81774a60ad75a1562bdfdd8d.png

A cylinder, a cone and a hemisphere are of same base and of same height. Find the ratio of their volumes ?

We have,

A cylinder

$= A$ cone

$= A$ hemisphere

$=$ same height

then, we know that

Cylinder volume =

$\pi r^{2} h=V_{1}$

Cone volume =

$\dfrac {1}{3}\pi r^{2} h=V_{2}$

Hemisphere volume =

$\dfrac {3}{2}\pi r^{2} h=V_{3}$

when

$h=$ height

$r=$ radius of Base

According to given question,

$V_{1}: V_{2}: V_{3}=\pi r^{2} h: \dfrac {1}{3}\pi r^{2} h: \dfrac {2}{3}\pi r^{2}$

$V_{1}: V_{2}: V_{3}=3:1:2$

Hence, this is the answer.

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold Pens. The dimensions of the cuboid are 15 $\mathrm { cm }$ by 10 $\mathrm { cm }$ by 3.5 $\mathrm { cm }$ . The radius of each of the depression is 0.5 $\mathrm { cm }$ and the depth is 1 $\cdot 4 \mathrm { cm }$ .Find the volume of the wood in the entire stand,correct to 2 decimal places.

Dimensions of the pan stand are

$l=15$ cm

$b=10$ cm

$h=3.5$ cm

$\Rightarrow$ Volume

$=l\times b\times h$

$=15\times 10\times 3.5$

$=525cm^3$

$4$ conical depressions are taken out to make pen holders.

$\Rightarrow$ Volume taken out

$=4$ (volume of one cone)

$=4\left(\dfrac{1}{3}\pi r^2h\right)$

Radius

$(r)=0.5$ cm

Height

$(h)=1.4$ cm

$\Rightarrow$ Volume taken out

$=\dfrac{4}{3}\times \dfrac{22}{7}\times 0.5\times 0.5\times 1.4$

$=1.466cm^3$

$\Rightarrow$ Volume of wood in stand

$=525-1.466$

$=523.534cm^3$

$=523.53cm^3$ .

Figure 7.13 shows a toy. Its lower part is a hemisphere and the upper part is a cone. Find the volume and the surface area the toy from the measures shows in the figure. $(\pi =3.14)$

REF.image

Volume of the toy

= Volume of cone + Volume of hemisphere

$= \frac{1}{3} \pi\, r^{2} h + \frac{2}{3} \pi \, r^{3}$

$= \frac{\pi r^{2}}{3}\left [ h + 2r \right ] = \frac{3.14 \times3^{2}}{3} \left [ 4 + 2 \times 3 \right]$

$= 9.42 \times 10 = 94.2 cm^{3}$

Total surface area of the toy

= CSA of cone + CSA of hemisphere

$= \pi r l + 2 \pi r^{2}$

$= \pi r (l+2r)$

$= 3.14 \times 3 \left [ 5 + 2 \times 3 \right ]$

$= 103.62 \,cm^{2}$

1134027_1246732_ans_71d9c5ee3304469ea86d66ce9f53ae43.png

A cylinder and a cone have the same height and the same radius. The volume of the cylinder is $24\ \text{cm}^{3}$ . What will be the volume of the cone.

1198415_1243551_ans_a407865b39e7412ba999d34e5821560b.jpg

An ice-cream cone consisting of the cone is surmounted by a hemisphere.The common radius of a hemisphere & cone is $3.5\ cm$ & the total height of ice cream is $12.5\ cm$ . Calculate the volume of ice-cream in the solid shape .

Radius of cone

$= r$ of

$h.s$

$r = 3.5$

Total height

$= 12.5$

Total

$h$ of cone

$- h.s.$

$12.5 - 3.5$

$h$ of conical

$= 9\ cm$

$V$ . of ice cream in the cone

$=$ volume of conical part

$+ V.$ of

$h.s.$

$= \dfrac {1}{3} \pi r^{2} h + \dfrac {2}{3} \pi r^{3}$

$= \dfrac {1}{3} \pi r^{2} (h + 2r)$

$= \dfrac {1}{3} \times \dfrac {22}{7}\times (3.5)^{2} (9 + 2\times 3.5)$

$= \dfrac {1}{3} \times \dfrac {22}{7} \times (3.5)^{2} (9 + 7)$

$= \dfrac {1}{2} \times \dfrac {22}{7} \times (3.5)^{2} \times 16$

$V$ . of ice cream in the cone

$= 205.33\ cm^{2}$ .

An open metal bucket is in the shape of frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The diameter of the two circular end of buckets are $45$ cm and $25$ cm, the total vertical height of the bucket is $40$ cm and that of cylindrical base is $6$ cm. Find the area of the metalic sheet used to make the bucket, where we do not take into account the handle of the bucket. Also find the volume of water the bucket can hold. [Take $\pi=\dfrac{22}{7}$ ]

Curved bucket surface area

$BB'C'C$

$=$ curved surface area of cone

$ABC$

$-$ curved surface area of

$AB'C'$

slant height of cone

$ABC$

$=$

$\sqrt { { 90 }^{ 2 }+{ 22.5 }^{ 2 } }$

$=$

$92.77cm$

slant height of cone

$AB'C'=$

$\sqrt { { 50 }^{ 2 }+{ 12.5 }^{ 2 } }$

$=$

$50.12$

$cm$

so curved surface area of bucket

$BB'C'C=$

$\pi \times 22.5\times 92.77-\pi \times 12.5\times 50.12$

$=4587{ cm }^{ 2 }$

surface area of cylindrical base

$=2\times 3.14\times { 12.5 }^{ 2 }+2\times 3.14\times 12.5\times 6\\ =1452.25{ cm }^{ 2 }$

total surface area

$=1452.25+4587=6039.25{ cm }^{ 2 }$

volume of water

$=\frac { \pi { \times 22.5 }^{ 2 }\times 90 }{ 3 } -\frac { \pi \times { 12.5 }^{ 2 }\times 50 }{ 3 } \\ =39511.67{ cm }^{ 3 }\\ =\frac { 39511.67 }{ 1000 } ℓ\\ =39.5ℓ$

1200450_1249918_ans_1047f455e0784211bde9b3decd040d92.jpg

The radii of the ends of frustum of a cone 45 cm high are 28 cm and 7 cm. Find its volume, the curved surface area and the total surface area

Let '

$\ell$ ' & R be the radii of the bottom and top circular ends of a frustum of a cone, respectively and 'h' be height of cone.

Volume of the frustum of the cone

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 45 (28^2 + 7^2 + 28 \times 7) cm^3$

$= \dfrac{22}{7} \times 15 (784 + 49 + 196) cm^3$

$= \dfrac{22}{7} \times 15 \times 1029 cm^3 = 22 \times 15 \times 147 cm^3$

$= 330 \times 147 cm^3$

Now,

$l = \sqrt{h^2 + (R - r^2)} = \sqrt{(45)^2 + (28 - 7)^2} = \sqrt{2025 + 441} = \sqrt{2466}$

$= \sqrt{2466} = 49.65 cm$

curved surface area of the frustum of the cone

$= \pi (R + r) \ell = \left[ \dfrac{22}{7} * (28 + 7) * 49.65 \right] cm^2$

$= [22 \times 5 \times 49.65] cm^2$

$5461.5 cm^2$

total surface area of frustum of the cone

$\pi (R + r) \ell + \pi R^2 + \pi r^2$

$= [5461.5 + \dfrac{22}{7} \times (28)^2 + \dfrac{22}{7} \times (7)^2 ] cm^2$

$= [5261.5 + \dfrac{22}{7} \times (28)^2 + (7)^2] cm^2$

$= [5461.5 + 22 \times 1 + 9 ] cm^2$

$= 80 79.5 cm^2$

A cistern, internally measuring $150$ cm $\times 120$ cm $\times 110$ cm, has $129600cm^3$ of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being $22.5$ cm $\times 7.5$ cm $\times 6.5$ cm?

Volume of cistern = 150 x 120 x 110
= 1980000 cm

$^3$
Volume of water in cistern
= 1980000 - 129600
= 1850400 cm

$^3$
volume of one brick = 22.5 x 7.5 x 6.5
= 1096.875 cm

$^3$
Let 'N' no of bricks can be placed into the cistern
Hence volume of 'N' no of bricks
= N x 1096.875 cm

$^3$
One bricks absorb water

$\frac{1}{17}$ of his volume hence water absorb by 1 brick

$= \frac{1096.875}{17} cm^3$
Hence total N no. of brick absorb water

$= \frac{1096.875}{17} \times N cm^3$
Volume to be filled in tank + water absorbs by N no. of bricks = volume of N No of bricks.

$1850400 + \frac{1096.875 N}{17} = N \times 1097.875$

$1850400 = N \times 1096.875 - \frac{1096.875 N}{17}$

$= N \times 1096.875 [1 - \frac{1}{17}]$

$1850400 = \frac{N \times 1096.875 \times 16}{17}$

$N = \frac{1850400 \times 17}{16 \times 1096.875}$

$= \frac{115650 \times 17}{10968.875}$

$= 1792.4$

$\therefore$ No. of bricks can be put in cistern without over flowing water.

Find the surface area of biggest sphere that can fit inside a cube of side $4a\ cm$ .

The diameter of biggest sphere will be equal to side of sphere.

$\implies$ radius

$=\dfrac{4 a}{2}=2{a}$

Surface area of sphere is

$4\pi r^2=4\pi (2{a})^2=16{\pi}{a^2}$

A container shaped like a right circular cylinder having diameter $12cm$ and height $15cm$ is full of ice cream. The ice cream is to be filled into cones of height $12cm$ and diameter $6cm$ , having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

Diameter of right circular cylinder is

$12 cm$

$\rightarrow$

$R = 6 cm$ .
Height,

$H = 15 cm$
Diameter of Cone,

$6 cm$ .

$\rightarrow$

$r = 3 cm$
Height,

$h = 12 cm$
(i) Volume of each ice cream=

$= Volume \ of \ cone+ Volume\ of\ hemisphere.$

$= \dfrac{1}{3} \pi r^2 h + \dfrac{2}{3} \pi r^3$

$= \dfrac{1}{3} \times \pi \times (3)^2 \times 12 + \dfrac{2}{3} \times \pi \times (3)^3$

$=( 36 \pi + 18 \pi )cm^3$

$= 54 \pi \ cm^3$
(ii) Volume of cylindrical vessel, V

$= \pi R^2 H$

$= \pi \times (6)^2 \times 15\ cm^3$

$= \pi \times 36 \times 15\ cm^3$
Let number of cones be 'n'

$n \times 54 \pi = \pi \times 36 \times 15$

$\therefore 54n = 36 \times 15$

$\therefore n = \dfrac {36 \times 15}{54} = \dfrac{540}{54}$

$\therefore n = 10$
Ice cream can be filled in

$10$ cones.

A toy is in the shape of a right circular cylinder with a hemisphere on one end and a come on the other. The radius and height of the cylindrical pan are $5\ cm$ and $13\ cm$ respectively. The radii of the hemispherical and the conical parts are the same as that of the cylindrical part. Find the surface area of the toy, if the total height of the toy is $30\ cm$ .

Perimeter is the sum of the length of boundary of any shape

$H = 30\ cm \,\, h = 13\ cm$

$r = 5\ cm$

$h_c = H - h - r$

$= 30 - 13 - 5 = 12\ cm$

SA of toy

$= CSA$ of cone

$+ CSA$ of cylinder

$+ CSA$ of hemisphere

$= \pi rl + 2 \pi r h + 2 \pi r^2$

$= \pi r \sqrt{r^2 + h^2} + 2 \pi r (h + r)$

$= \pi [5(13) + 2 (5) (18)]$

$= 245 \pi$

$= 769.69\ cm^2$

1339107_1268372_ans_c3d23ab57d7c4bcdbb196df143437dda.png

Monica has a piece of canvas whose area is $551$ $m^2$ . She uses it to have a conical tent made, with a base radius of $7$ m. Assuming that all the stitching margins and the wastage incurred while cutting, amounts to approximately $1 m^2$ , find the volume of the tent that can be made with it.

Given :

area : 551 m, r = 7 m

wastage :

$1m^{2}$

as area of cone

$- \frac{1}{2}\times base\times height$

$551 = \frac{1}{2}\times 7\times height$

$= 157.4m$ = height

Volume of cone =

$\frac{1}{3}\pi r^{2} = h$

$= \frac{1}{3}\times \frac{22}{7}\times \frac{7}{7}\times 157.4\times 1$

$= 164.89$

1119774_1250428_ans_42f7e861b15c447387b0881551f015b9.jpeg

A toy is in the form of a cone mounted on a hemisphere with same radius. The diameter of the conical base is 7 cm and the height of the toy is 14.5 cm. Find the volume of the toy

Diameter of hemisphere

$=7$ cm

Radius of hemisphere

$r=d/2$

$=\dfrac{7}{2}=3.5$ cm

Height of hemisphere

$=$ Radius of hemisphere

$=3.5$ cm

Height of Toy

$=14.5$ cm

Height of cone

$=14.5-3.5=11$ cm

Radius of cone

$=3.5$ cm

$\therefore$ Volume of cone

$=$ Volume of cone

$+$ Volume of hemisphere

$=\dfrac{1}{3}\cdot \pi r^2h+\dfrac{4}{3}\cdot \pi r^3$

$=\dfrac{1}{3}\cdot \dfrac{22}{7}\cdot (3.5)^2\cdot 11+\dfrac{4}{3}\cdot \dfrac{22}{7}(3.5)^3$

Volume of toy

$=320.83cm^3$ .

1191582_1265227_ans_c165ed9bafd84e6abddecab1bf85b394.jpg

A field is 700 m long and 40 m broad, A trench which is 10 m long, 30 m broad and 14 m deep is dug in the field .The earth taken out of it is evenly spread over the remaining part of the filed. Find the rise in the level of the field.

1203578_1249043_ans_e80ca03538894c948a7c80248f2c3201.jpg

If the triangle ABC is revolved about the side $5$ cm, then find the volume of the solid so obtained.

Cone is formed when

$\Delta ABC$ is revolved about

$5 cm$

$r =12 cm$

$h = 5 cm$

Volume of cone is given by,

$V = \frac {1}{3} \pi r^2 h$

$\therefore V = \frac {1}{3} \times \frac {22}{7} \times (12)^2 \times 5$

$\therefore V=\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 144\times 5$

$\therefore V=\frac { 22\times 144\times 5 }{ 21 }$

$\therefore V=754.285\ cm^{ 3 }$

1953746_1323964_ans_718ef8b093254f599024d52c76dd54fd.png

Find the total volume of these three identical toy blocks.

Volume

$=3\times$ Volume of

$1$ Block

$=3\times [3\times 10\times 3]$ ........ [Dimension of

$1$ block

$=3, 10, 3$ ]

$=3\times 3\times 10\times 3$

$=270\ cm^3$ .

An umbrella has $8$ ribs which are equally spaced. Assuming umbrella to be flat circle of radius $45\ cm$ , find the area between the two consecutive ribs of the umbrella.

${\textbf{Step 1: }}{\mathbf{ Find}}\;{\mathbf{angle}}\;{\mathbf{protruded}}\;{\mathbf{by}}\;{\mathbf{each}}\;{\mathbf{rib}}$

${\text{No}}.{\text{of ribs}} =8.$

${\text{The center of an umbrella protrudes an angle of }}360^\circ$

${\text{Hence}},{\text{one rib protrudes an angle of}} = {\text{}}\dfrac{{{\text{}}360}}{8}$

${\text{}} = {\text{}}45^\circ$

${\textbf{Step 2: }}{\mathbf {Find}}\;{\mathbf{the}}\;{\mathbf{area}}\;{\mathbf{consecutive}}\;{\mathbf{between}}\;{\mathbf{two}}\;{\mathbf{ribs}}$

${\text{As we know that area of the sector = } \dfrac{\theta}{360}\times\pi \times{{\text{r}}^2}}$

$\therefore {\text{Area between two consecutive ribs}}$

$= \dfrac{45}{360}\times\pi \times{{\text{r}}^2}$

$= \dfrac{45}{360}\times \pi \times{{\text{45}}^2}$

$= \dfrac{{22275}}{{28}}$

${\textbf{Hence,}}{\textbf{ the area}}\;{\mathbf{between}}\;{\mathbf{two}}{\textbf{ consecutive}}\;{\textbf{ribs is}} \; \mathbf{\dfrac{{22275}}{{28}}}\;{\mathbf{c}}{{\mathbf{m}}^2}$

If a sphere is inscribed in a cube, then find the ratio of the volume of sphere to the volume of the cube.

$\textbf{Step -1: Finding the required volume of the cube and the volume of the sphere.}$

$\text{Let the side of the cube be }a \text{ units.}$

$\because \mathbf{\text{Diameter of the sphere inscribed in the cube }=\text{ Side of the cube.}}$

$\therefore \text{Radius of the sphere inscribed in the cube of side }a\text{ would be }\dfrac{a}{2}\text{ units.}$

$\text{As, }\mathbf{\text{Volume of cube}=\text{(Side)}^3}$

$\text{Here, Volume of the required cube}=a^3\text { cubic units.}$

$\text{and }\mathbf{\text{Volume of sphere}=\dfrac{4}{3}\pi\text{(Radius)}^3}$

$\therefore \text{The volume of the required sphere}=\dfrac{4}{3}\pi \left(\dfrac{a}{2}\right)^3=\dfrac{1}{6}\pi a^3\text{ cubic units.}$

$\textbf{Step -2: Finding the required ratio.}$

$\dfrac{\text{Volume of sphere}}{\text{Volume of cube}}=\dfrac{\dfrac{1}{6}\pi a^3}{a^3}=\dfrac{\pi}{6}$

$\textbf{Hence, the ratio of the volumes of the Sphere to the Cube is }\mathbf{\dfrac{\pi}{6}}$

Find the volume of largest sphere covered out of a cube of side $7\ cm$ .

$\begin{array}{l} Diameter\, \, of\, \, sphere\, \, =7\, \, cm \\ \Rightarrow volume\, \, of\, \, sphere\, \, =\dfrac { 4 }{ 3 } \pi { r^{ 3 } } \\ \Rightarrow Volume=\dfrac { 4 }{ 3 } \times \dfrac { { 22 } }{ 7 } \times \dfrac { 7 }{ 2 } \times \dfrac { 7 }{ 2 } \times \dfrac { 7 }{ 2 } =\dfrac { { 11\times 49 } }{ 3 } =179.66\, \, c{ m^{ 3 } } \\ \Rightarrow 179.66\, \, c{ m^{ 3 } } \end{array}$

A wooden pole is $7\ m$ high and $20\ cm$ in diameter. Find its weight if the wood weight $225\ kg$ per $m^3$ .

Radius

$= 10$ cm

$= 0.1$ m
height

$= 7$ m
volume $$= $\pi \times {r}^{2} \times h$ $$

$=\dfrac{22}{ 7} × 0.1 × 0.1 × 7$

$=0.22$ meter cube

Density

$=225$ kg/meter cube

$225 =\dfrac{ mass }{ 0.22}$
mass

$= 225 × 0.22 = 49.5$ kg.

A cylindrical tube of radius $12\ cm$ contains water to a depth of $20\ cm$ . A spherical form ball is dropped into the tube and thus the level of water is raised by $6.75\ cm$ . What is the radius of the ball?

Volume of spherical ball= volume of water with h=

$6.75\, \, cm$

$\begin{array}{l} \dfrac { 4 }{ 3 } \pi { r^{ 3 } }=\pi { r^{ 2 } }\times 6.75 \\ =\dfrac { 4 }{ 3 } \pi { r^{ 3 } }=\pi { \left( { 12 } \right) ^{ 2 } }\times 6.75 \\ ={ r^{ 3 } }=\dfrac { { { { \left( { 12 } \right) }^{ 2 } }\times 6.75\times 3 } }{ 4 } \\ ={ r^{ 3 } }=729 \\ =r=9\, \, cm\, \, \left( { radius\, \, of\, \, ball } \right) \end{array}$

A solid consisting of a right circular cone of height $120$ cm and radius $60$ cm standing on a hemisphere of radius $60$ cm is placed upright in a right circular cylinder full of water such that it touches the bottom. How many litres of water is left in the cylinder, If the radius of the cylinder is $60$ cm and its height is $180$ cm.

Volume of cylinder

$=\pi r^2 h$

$=\dfrac{22}{7}\times (60)^2\times 180$

$=\dfrac{22\times 3600\times 180}{7}=\dfrac{14256000}{7}\text{ cm}^3$

The volume of solid

$=$ Volume of cone

$+$ volume of the hemisphere

$=\dfrac{1}{3}\pi r^2h+\dfrac{2}{3}\pi r^3$

$=\dfrac{1}{3}\times \dfrac{22}{7}\times 60^2\times 120 +\dfrac{2}{3}\times \dfrac{22}{7}\times 60^3$

$=\dfrac{3168000}{7}+\dfrac{3168000}{7}=\dfrac{6336000}{7}\text{ cm}^3$

Volume of water left in the cylinder

$=$ Volume of the cylinder

$-$ Volume of the solid

$=\dfrac{14256000}{7}-\dfrac{6336000}{7}=\dfrac{7920000}{7}$

$=1131428.57142\text{ cm}^3$

$=1.131\text{ m}^3$

$(\because 1 \text{ m}^3=1000000\text{ cm}^3)$

1213361_1305895_ans_be37dadba4e9420cba3a725e9516680c.png

If a sphere in inscribed a cube, then find the ratio of the volume of the cube to the volume of the sphere.

$\textbf{Step -1: Finding the required volume of the cube and the volume of the sphere.}$

$\text{Let the side of the cube be }a \text{ units.}$

$\because \mathbf{\text{Diameter of the sphere inscribed in the cube }=\text{ Side of the cube.}}$

$\therefore \text{Radius of the sphere inscribed in the cube of side }a\text{ would be }\dfrac{a}{2}\text{ units.}$

$\text{As, }\mathbf{\text{Volume of cube}=\text{(Side)}^3}$

$\text{Here, Volume of the required cube}=a^3\text { cubic units.}$

$\text{and }\mathbf{\text{Volume of sphere}=\dfrac{4}{3}\pi\text{(Radius)}^3}$

$\therefore \text{The volume of the required sphere}=\dfrac{4}{3}\pi \left(\dfrac{a}{2}\right)^3=\dfrac{1}{6}\pi a^3\text{ cubic units.}$

$\textbf{Step -2: Finding the required ratio.}$

$\dfrac{\text{Volume of cube}}{\text{Volume of sphere}}=\dfrac{a^3}{\dfrac{1}{6}\pi a^3}=\dfrac{6}{\pi}$

$\textbf{Hence, the ratio of the volumes of the cube to the sphere is }\mathbf{\dfrac{6}{\pi}}$

A solid sphere of radius $15$ cm is melted and recast into solid right circular cones of radius $2.5$ cm and height $8$ cm. Calculate the number of cones recast.

Radius of sphere

$R=15\ cm$

Dimensions of cone

$r= 2.5\ cm \ h=8\ cm$
When a solid is converted from one form to another, the amount of volume remains the same. Let

$x$ number of cones can be melted.

Number of cones,

$x=\frac{\text{Volume of sphere}}{\text{Volume of 1 cone}}$

$\displaystyle =\frac{\frac{4}{3}\pi (15)^{3}}{\frac{1}{3}\pi (2.5)^{2}(8)}=\frac{4(15)^{3}}{(2.5)^{2}8}= 270\ cones$

A hollow sphere of external and internal diameters of $8\ cm$ and $4\ cm$ respectively is melted and made into another solid in the shape of a right circular cone of base diameter of $8\ cm$ . Find the height of the cone.

volume of hollow sphere

$=$ volume of cone

$\dfrac{4}{3}\pi (R^{3}-r^{3})=$

$\dfrac{1}{3}\pi (r_c^{2})h_c$

$\Rightarrow 4(8^{3}-4^{3})=4^2 h_c$

$\Rightarrow 4(8^{3}-4^{3})=4\times 4 h_c$

$\Rightarrow h_c=\dfrac{8^{3}-4^{3}}{4}$

$\Rightarrow h_c=\dfrac{512-64}{4}$

$\Rightarrow h_c=\dfrac{448}{4}=112\ cm$

height of cone

$h_c= 112\ cm$ .

How many planks each measuring $5\ m$ by $24\ cm$ by $10\ cm$ can be stored in a plank $115\ m$ long. $4\ m$ wide and $60\ cm$ deep?

No. of planks

$=\dfrac{Total\quad volume}{volume\quad of\quad planks}$

$=\dfrac{115\times 100\times 4\times 100\times 60}{5\times 24\times 10}$

$= 2,30000$ planks.

1187862_1353861_ans_da049def07cd468ea11577a9fa97dbcb.jpg

The dimensions of a cuboid are $44\ cm, 21\ cm, 12\ cm$ . it is melted and a cone of height $24\ cm$ is made. Find the radius of its base.

Volume of cuboid $=l\times b\times h$

$=44\times 21\times 12\ cm^3$

Volume of cone $=\dfrac{1}{3}\pi {r^2}h = \dfrac{1}{3}\pi {r^2}\left( {24} \right)$

Volume of cone $=$ volume of cuboid

$\dfrac{1}{3}\pi {r^2}\left( {24} \right)=44\times 21\times 12\ cm^3$

$\dfrac{1}{3}\times \dfrac{22}{7}\times {r^2}\left( {24} \right)=44\times 21\times 12\ cm^3$

$r^2=\dfrac{44\times 21\times 12\times 7\times 3}{22\times 24}$

$r^2=21\times 21$

$r=\sqrt{21\times 21}$

$r=21$

Thus radius of the base is $r=21\ cm$

A hollow sphere of internal and external radii $6$ cm and $8$ cm respectively is melted and recast into small cones of base radius $2$ cm and height $8$ cm. Find the number of cones.

No. of cones =

$\dfrac{Volume\ of\ sphere}{volume\ of\ 1\ cone}$

$\dfrac{\frac{4}{3}\pi (R^{3}-r^{3})}{\dfrac{1}{3}\pi r_{c}^{2}h_{c}}=\dfrac{4(8^{3}-6^{3})}{2\times 2\times 8}=37$ cones

What is the area of the glass?

Area of glass

$=$ Area of cuboid

$=2(lb+bh+hl)$

$=2[(30)(25)+(25)(25)+(25)(30)]$

$= 4250\ cm^{2}$

A solid wooden toy is in the shape of right circular cone mounted on a hemisphere of radius $4.2\ cm$ . The total height of the toy is $10.2\ cm$ find the volume of the toys.

Radius of hemisphere

$= 4.2 \; cm$

Height of toy

$= 10.2 \; cm$

$\Rightarrow$ Height of the cone

$= 10.2 - 4.2 = 6 \; cm$

And the radius of the cone

$= 4.2 \; cm$

The volume of the toy

$=$ Volume of the hemisphere

$+$ Volume of the cone

Volume of the toy

$= \cfrac{2}{3} \pi {r}^{3} + \cfrac{1}{3} \pi {r}^{2} h$

$= \cfrac{1}{3} \pi {r}^{2} \left( 2r + h \right)$

$= \cfrac{1}{3} \times \cfrac{22}{7} \times {\left( 4.2 \right)}^{2} \times \left( 2 \times 4.2 + 6 \right)$

$= 266.112 \; {cm}^{3}$

1264416_1330511_ans_4b49cdc72e324ad68de7f5fb9b1480c4.png

Total volume of three identical cones is the same as that of a bigger cone whose height is $9$ cm and diameter $40$ cm. Find the radius of the base of each smaller cone, if height of each is $108$ cm.

Volume of a cone

$= \dfrac{1}{3}\pi (r^{2})(h)$

By given data,

Volume of 3 cones

$=$ volume of 1 big cone

$\Rightarrow [\dfrac{1}{3}\pi (r^{2})(108)]\times 3=\dfrac{1}{3}\pi (20)(20)(9)$

$\Rightarrow r^{2}=\dfrac{20\times 20\times 9}{3\times 108}= 11.111$

$\Rightarrow$

$r= 3.333 cm$

Find the rate of change of volume of a cone w.r.t. its radius, when the height is kept constant.

$(V)$ volume of cone

$=\cfrac{1}{3}\pi r^2h$

$\cfrac{dv}{dr}=\cfrac{1}{3}\pi\left[(2r)h\cfrac{dr}{dv}+r^2\cfrac{dh}{dr}\right]$

$\cfrac{dv}{dr}=\cfrac{1}{3}\pi (2rh)$

A godown measures $40$ m $\times 25$ m $\times 10$ m. Find the maximum number of wooden crates each measuring $1.5$ m $\times 1.25$ m $\times 0.5$ m that can be stored in the godown.

No of wooden crates

$=\dfrac{volume\ of\ godown }{volume\ of \ crates}$

$\dfrac{40\times 25\times 10}{1.5\times 1.25\times 0.5} =10666.66$

$=10666$ boxes.

Suhail wants to paint the flour walls of a room having dimensions $20\ m\times 6\ m$ . From each can of paint, $96$ sq. m of the area is painted. How many cans of paint will he need to paint the room?

Total Area

$= 4\times 20\times \times 6 =480m^{2}$

No. of rooms

$= \frac{480}{96} = 5$ rooms

1189183_1354020_ans_3f1aa20644154e9f83e2ecbb54235192.jpg

On a map, a length of $1 cm$ represents $25,000 cm$ on land. If a road measure $5.75 cm$ on a map, what is its actual length in metres?

1 cm = 25000 cm on land

5.75 cm

$= 25000\times 5.75$

= 143750 cm on land

1189204_1354125_ans_1b967f12bebd4d558ba4182430e68289.jpg

A soft drink is available in two packs (i) a tin can with a rectangular base 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

${\textbf{Step-1: Write given values and use volumes formula.}}$

${\text{Given, length(l)}}$

$=5cm,$

${\text{width(b)}}$

$=4cm,$

${\text{height(h)}}$

$=15cm,$

$\text{As we know the volume of a cuboid = } l \times b \times h$

$\Rightarrow V = l \times b \times h$

$= 5 \times 4 \times 15 = 300cm^3$

${\text{And diameter}}$

$=7cm,$

${\text{Height(h)}}$

$=10cm,$

${\text{Radius(r)}}$

$= \dfrac{\text{Diameter}}{2}$

$= \dfrac{7}{2}$

$\text{As we know the volume of a cylinder = } \pi r^2h$

$\Rightarrow V = \pi r^2h$

$= \dfrac{22}{7} \times (\dfrac{7}{2})^2 \times 10$

$= \dfrac{2695}{7}$

$= 385cm^3$

${\textbf{Step-2: Find the difference of the resulted volumes.}}$

${\text{Difference}}$

$= 385cm^3 - 300cm^3 = 85cm^3$

${\text{Hence, cylinder container has greater capacity by 85 cubic cm.}}$

${\textbf{Thus, option A is correct. }}$

Height of a cylindrical barrel is $50$ cm and radius of its base is $20$ cm. When it was empty, Anurag started to fill the barrel with water, by a cylindrical mug. The diameter and height of the mug was $10$ cm and $15$ cm respectively. How many minimum number of mugs will be required for the barrel to overflow?

Given,

Height of the cylindrical barrel

$=50\ cm$

Radius of its base

$=20\ cm$

Height of the cylindrical mug

$=15\ cm$

Diameter of the cylindrical mug

$=10\ cm$

$\therefore$ Radius of the cylindrical mug

$=\dfrac{10}{2}\ cm=5\ cm$

Volume of the cylindrical barrel

$=\pi r^{2}h$

$= \pi\times (20)^{2}\times50\ cm^3$

$=20000\pi\ cm^3$

Volume of the cylindrical mug

$=\pi r^{2}h$

$= \pi\times (5)^{2}\times15\ cm^3$

$=375\pi\ cm^3$

$\therefore$ minimum number of mugs required to fill the barrel

$=\dfrac{volume\, of\, barrel }{volume\, of \,mug}$

$=\dfrac{20000\pi}{375\pi}$

$=\dfrac{20000}{375}$

$= 53.33$

Hence, the minimum number of mugs will be required for the barrel to overflow is

$54$ .

A bucket, in the shape of a frustum of a cone, has a height of $6\ cm$ . Its top and its bottom have radii of $24\ cm$ and $16\ cm$ respectively. Find its curved surface area.

The bucket is in the shape of a frustum.

The slant height of frustum

$=\sqrt{((Top \ radius - Bottom \ radius )^2 +(Height)^2}$

Therefore, slant height of the bucket

$l=\sqrt{(24-16)^2+6^2}=10$

$m$

Therefore, its curved surface area =

$\pi l(R+r)=\pi(10)(24+16)=400\pi$

$m^2$

A laboratory is $9\ m$ long, $8\ m$ broad and $6\ m$ high. It has $2$ doors each of size $3 \ m\times 1.5\ m$ and four windows each of size $1.5\ m\times 1\ m$ . find the cost of whitewashing the walls of the laboratory at the rate of $Rs. 1.75$ per $sq. m$ .

Area of walls

$= 2h (l+b) = 2\times 6(17) = 204 m^{2}$

Area of doors

$= 2\times 3\times 1.5 = 9m^{2}$

Area of window

$= 4\times 1.5\times 1 = 6 m^{2}$

Area to be whitewashed

$= 204-9-6 =189 m^{2}$

Total cost

$= 1.75\times 189$

= 330.75 RS

1189179_1354010_ans_b61270d85d5e47c7b8329e0a16f4bb67.jpg

The radii of the circular ends of a solid frustum of a cone are $33\ cm$ and $27\ cm$ and its slant height is $10\ cm$ . Find its total surface area.(take $\pi =3.14$ )

Here,

$r_{1}=33\ cm$

$r_{2}=27\ cm$

$l=10\ cm$

T.S.A. =

$\pi[(r_{1}+r_{2})l +{r_{1}}^{2}+{r_{2}}^{2}]$

$3.14[(33+27)10+1089+729]$

$3.14[600+1089+729]$

$3.14 \times 2418 = 7592.52\ cm^{2}$

A wooden toy rocket is in the shape of a cone mounted on a cylinder, as shown in fig. The height of the entire rocket is $26\ cm,$ while the height of the conical part is $6\ cm.$ The base of the conical portion has a diameter of $5\ { cm } ,$ while the base diameter of the cylindrical portion is $3\ { cm } .$ If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these colours. (Take $\pi = 3.14$ )

The height of cone is

$6\ cm$

The radius of cone is

$2.5\ cm$

The slant height is given as

$=\sqrt{2.5^2+6^2}$

$=\sqrt{6.25+36}$

$=\sqrt {42.25}=6.5$

The Surface area of cone is

$\pi r l=\pi 5 \times 6.5=32.5\pi$

Area of base is

$=\pi R^2-\pi r^2$

$=\pi(2.5^2-1.5^2)$

$=\pi(6.25-2.25)=4\pi$

The area that is painted with orange is

$=32.5\pi+4\pi=36.5\pi cm^2$

The height of cylinder

$=20\ cm$

The radius of cylinder

$=1.5\ cm$

The curved surface area is

$=\pi r h$

$=\pi( 1.5\times 20)=30\pi$

Area of base is

$=\pi 1.5^2=2.25\pi$

The area that is painted with Yellow is

$=30\pi+2.25\pi=32.25\pi\ cm^2$

Kanika had a piece of cloth 1 m long and 60 cm broad. She cut it into square shaped handkerchiefs, each side measuring 20 cm, and put a lace on the boundary. How much lace is needed for the handkerchiefs?

Given that, the cloth is

$1\ m$ long and

$60\ cm$ broad.

Also, the length of the square-shaped handkerchief is

$20\ cm$

The total area of the cloth is the combined area of all the handkerchiefs.

$\therefore \ \$ Number of handkerchiefs

$=\dfrac{\text{Total area of cloth}}{\text{Area of 1 handkerchief}}$

Using area of rectangle

$=l\times b$ and area of square

$=a^2$ , we get:

Number of handkerchiefs

$=\dfrac{100\times 60}{20\times 20} =15$

$[1\ m=100\ cm]$

Length of lace required for

$1$ handkerchief

$=$ perimeter of

$1$ handkerchief

Hence, total lace required for

$15$ handkerchiefs

$=$ perimeter of

$15$ handkerchief

$\Rightarrow 15\times (4\times 20)$ [Perimeter of square

$=4a$ ]

$= 15\times 80$

$= 1200 \ cm$

Hence,

$1200\ cm$ lace will be needed for all handkerchiefs.

A cone, a hemisphere and a cylinder stand on the same base and have equal height. Find the ratio of their volumes.

Let, radius of bases be '

$r$ '

and their height be '

$h$ '

For hemisphere,

$radius=height\implies r=h......(1)$

Now,

volume of cone,

$v_1 = \dfrac{1}{3} \pi r^{2}h$

volume of hemisphere,

$v_2 = \dfrac{2}{3} \pi r^{3}$

volume of cylinder,

$v_3 = \pi r^{2} h$

The ratio,

$v_1 : v_2 : v_3 = \dfrac{1}{3} \pi r^{2}h : \dfrac{2}{3} \pi r^{3} : \pi r^{2} h$

$= \dfrac{h}{3} : \dfrac{2r}{3} : h$

$= \dfrac{h}{3} : \dfrac{2h}{3} : h$ [From

$(1)$ ]

$= \dfrac{1}{3} : \dfrac{2}{3} : 1$

$= 1 : 2 : 3$

Hence, the required ratio is

$1:2:3$

The shape of a garden is rectangular in the middle and semicircular at the ends.Find the area of the garden?

To find the area of the given figure, it may be considered as the sum of the area of a rectangle and

$2$ semicircles.

Length of rectangle

$=20-\left(3.5+3.5\right)=20-7=13$ m

Width of rectangle

$=7$ m

Area of rectangle

$=13\times 7=91$ sq.cm

Area of semicircle

$A=\dfrac{1}{2}\left(\dfrac{22}{7}\times 3.5\times 3.5\right)$

Area of semicircle

$B=\dfrac{1}{2}\left(\dfrac{22}{7}\times 3.5\times 3.5\right)$

Area of

$A$ and

$B=\dfrac{1}{2}\left(\dfrac{22}{7}\times 3.5\times 3.5\right)+\dfrac{1}{2}\left(\dfrac{22}{7}\times 3.5\times 3.5\right)$

$=2\times \dfrac{1}{2}\left(\dfrac{22}{7}\times 3.5\times 3.5\right)$

$=\dfrac{22}{7}\times 3.5\times 3.5=38.5$ sq.m

Total area

$=$ Area of

$A+B+C$

$=91+38.5$ sq.m

$=129.5$ sq.m

A bucket is in the form of a frustum of a cone and holds $28.490$ litres of water. The radii of the top and bottom are $28$ cm and $21$ cm respectively. Find the height of the bucket.

Let the height of the bucket be h cm.

$r_{1}=28 \ cm$

$r_{2}=21 \ cm$

$V =$ Volume of the bucket

$=28.490 \ litres = 28.490 \times 1000 \ cm^3 = 28490 \ cm^3\\$

$V= 28490 \ cm^3\\$

$\dfrac{1}{3} \pi h(r_{1}^2 + r_{1}r_{2} +r_{2}^2)=28490$

$\dfrac{1}{3} \times \dfrac{22}{7} \times h \times (28^2 + 28 \times 21 +21^2)=28490$

$\dfrac{22}{21} \times h \times 1813 = 28490$

$h=15 \ cm$

from the given information in the figure find slant height of cone.

Using Pythagoras theorem, we have

${\left(slant\,\, height\right)}^{2}={\left(height\right)}^{2}+{\left(base\right)}^{2}$

$\Rightarrow {\left(slant\,\, height\right)}^{2}={\left(24\right)}^{2}+{\left(7\right)}^{2}$

$\Rightarrow$ Slant height

$=\sqrt{576‬+49}=\sqrt{625‬}=25$ cm

$\therefore$ slant height of the cone is

$25$ cm

The height of a cone is $20\text{ cm}$ . A small cone is cut off from the top by a plane parallel to the base. If its volume is $\dfrac{1}{125}$ of the volume of the original cone, determine at what height above the base the section is made.

Volume of a cone of radius

$r$ and height

$h$ is

$=\cfrac { \pi { r }^{ 2 }h }{ 3 }$

$\Rightarrow \cfrac { 1 }{ 3 } \pi { r }^{ 2 }h=\cfrac { 1 }{ 125 } \cfrac { 1 }{ 3 } \pi { R }^{ 2 }(20)\\ \Rightarrow { r }^{ 2 }h={ R }^{ 2 }\cfrac { (20) }{ 125 } \quad \quad -(1)$

Also

$BE||CD$ (given)

$\triangle ABE$ &

$\triangle ACD$

$\angle BAE= \angle CAD$ (common )

$\angle ABE = \angle ACD$ (corresponding angles)

$\triangle ABE\sim \triangle ACD$ by

$AA$ criterion

thereby the sides will be in proportion

$\Rightarrow \cfrac { BE }{ CD } =\cfrac { AB }{ AC } \\ \Rightarrow \cfrac { r }{ R } =\cfrac { h }{ 20 } \quad \quad -(3)$

Substituting

$(2)$ in

$(1)$ we get

$(\cfrac { { r }^{ 2 } }{ { R }^{ 2 } } )h=20\\ \Rightarrow { (\cfrac { h }{ 20 } ) }^{ 2 }h=\cfrac { 20 }{ 125 } \\ \Rightarrow { h }^{ 3 }=\cfrac { 20\times 20\times 20 }{ 5\times 5\times 5 }$

$\Rightarrow h=\cfrac { 20 }{ 5 } =4\text{ cm}$

Therefore, the section is made

$(20-4)\text{ cm}=16\text{ cm}$ above the base of original cone.

1222828_1382282_ans_48404cd44ed44546b431a05433f62305.png

Curved Surface area of a cone is $308 cm^{2}$ and its slant height is $14 cm$ . Find
(i) radius of the base
(ii) Total surface area of the cone.

Curved surface area of a cone of radius

$'r'$ and height

$'h'$ and slant height

$'l'$ is given by

$\pi rl$ .

$i)$ Curved surface area

$=308{ cm }^{ 2 }$

$l=14$ cm

$\pi rl=308\\ \Rightarrow \pi \times 14\times r=308\\ \Rightarrow \cfrac { 22 }{ 7 } \times 14\times r=308\\ \Rightarrow 22\times 2\times r=308$

$\Rightarrow r=\cfrac { 308 }{ 44 } 7$ cm

$ii)$ Total surface area

$=$ curved surface area

$+$ area of the base

$=\pi rl+\pi { r }^{ 2 }\\ =\cfrac { 22 }{ 7 } \times 7(14+7)\\ =22\times 21\\ =462{ cm }^{ 2 }$

The radii of the bases of two right circular solid cones of same height are $r_{1}$ and $r_{2}$ respectively. The cones are melted and recast into a solid sphere of radius R. Show that the height of each cone is given by $h=\dfrac{4R^3}{r_{1}^2+r_{2}^2}$ .

Let

$h$ be the height of each cone and

$r_1$ and

$r_2$ be their radius. Then,

Sum of the volumes of two cones = Volume of the sphere

$\dfrac{1}{3}\pi r_{1}^2h+\dfrac{1}{3} \pi r_{2}^2h=\dfrac{4}{3}\pi R^3$

$(r_{1}^2+r_{2}^2)h=4R^3$

$h=\dfrac{4R^3}{r_{1}^2+r_{2}^2}$

The radius of a cone is $3$ cm and vertical height is $4$ cm. Find the area of the curved surface.

Given,

$r=3$

$cm$

$h=4$

$cm$

$l^{2}=r^{2}+h^{2}$

$l=\sqrt{25}=5$

$cm$

Area of the curved surface =

$\pi rl$

$=\dfrac{22}{7} \times 3 \times 5 = 47.14$

$cm^{2}$

A cube is made by arranging $64$ cubes having side of $1\ cm$ , find total surface area of cube so formed.

We know that volume of

$64$ small cubes

$=$ volume of the big cube

Let

$a$ be the side of the big cube

$\Rightarrow 64(1)^{3}=a^{3}$

$\Rightarrow a=4\ cm$

Surface area of big cub

$=6a^{2}$

$=6\times 16$

$=96\ cm^{2}$

Write the formula for volume of frustum of right circular cone .

Let R and r be the radii of base and top of the frustum of a cone. Let h be the height, then

Volume of the frustum of right circular cone =

$\dfrac{\pi h}{3}(R^{2} + r^{2} + Rr) cm^{3}$

Water in a canal, 30 dm wide and 12 dm deep, is flowing with a velocity of 20 km per hour. How much area will it irrigate in 30 min, if 9 cm of standing water is desired ?

Velocity of water=

$20Km/h$

Water in the canal forms a cuboid of breadth =

$12 dm = \dfrac{12}{10}m = 1.2 m$

Height =

$30dm=\dfrac{30}{10}m=3m$

Length = Distance covered by water in 30 minutes

= Velocity of water

$\times$ time

$=20000\times \dfrac{30}{60}=10000$

$m$

Therefore,

Volume of water flown in 30 min =

$lbh=(1.2 \times 3 \times 10000)=36000$

$m^{3}$

Suppose area irrigated be A m

$^{2}$ . Then,

$A\times \dfrac{9}{100}$ = Volume flown in 30 minutes

$A\times \dfrac{9}{100}=36000$

$A=400000$

$m^{2}$

Hence, area irrigated

$= 400000 \ m$

$^{2}$

If $r, h, c, v$ are radius curved surface area and volume of a cone respectively, then prove that $3\pi vh^{3}-c^{2}h^{2}+9v^{2}=0$

The curved surface area of a cone

=πrl

=πr√(h²+r²)

where h, l and r are the height, slant height and radius of the cone.

Volume of the cone

=πr²h/3

∴3πVh³-C²h²+9V²

=3π{(πr²h)/3}h³-{πr√(h²+r²)}²h²+9(πr²h/3)²

=π²r²h⁴-π²r²(h²+r²)h²+9π²r⁴h²/9

=π²r²h⁴-π²r²h⁴-π²r⁴h²+π²r⁴h²

=0 (Proved)

The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

Slant height of a i frustum of a cone, I = 4 cm
Circumference of its circular ends = 18 cm. and 6 cm

$2 \pi r = 18 cm$

$\therefore R = \frac{9}{\pi}$

$2 \pi r = 6$

$r = \frac{3}{\pi}$ cm

$\therefore$ Lateral surface area of a frustum of a cone

$= \pi (R + r)l$

$= \pi \left ( \frac{9}{\pi} + \frac{4}{\pi} \right ) \times 4$

$= \pi \left ( \frac{12}{\pi} \right ) \times 4$

$= 48 cm^3$
1839332_1449913_ans_0f75ac072c17441b8a26f2f7b7914b1a.png

A metal cuboid of dimension $22 \ \text{cm} \times 15\ \text{cm} \times 7.5 \ \text{cm}$ was melted and cast into a cylinder of height $14$ cm. What is its radius?

Dimensions of the metal cuboid are

Length

$=l=22$ cm

Breadth

$=b=15$ cm

Height

$=h=7.5$ cm

Height of the cylinder

$(H)=14$ cm

Let the radius of the cylinder be

$r$ cm

Volume of cuboid

$=$ Volume of cylinder

$l\times b\times h=\pi r^2h$

$\Rightarrow 22\times 15\times 7.5=\dfrac{22}{7}\cdot r^2\cdot 14$

$\Rightarrow 2475=22.2.r^2$

$\Rightarrow r^2=\dfrac{2475}{22.2}$

$\Rightarrow r^2=56.25$

$\Rightarrow r=\sqrt{56.25}$

$\Rightarrow r=7.5$

$\therefore$ Radius

$=r=7.5$ cm.

A cylindrical glass of radius 7 cm is filled with mango juice up to a height of 17 cm when carefully, observed, the bottom of glass is covered by a transparent cone of height 6 cm and radius 7 cm. What is the actual volume of juice ?

Volume

$=$ volume

$-$ volume of juice of cylinder of cone.

$=\pi (7)^2(17)-\dfrac{1}{3}\pi (7)^2(6)$

$=\pi (49)(17)-\pi (49)(2)$

$=\pi (49)(17-2)$

$=\pi (49)(15)=2309.07 cm^3$ .

1195926_1392176_ans_dde445c430974517b0ef31285f25e298.jpg

The base radius and height of a right circular cone are $2$ cm and $8$ cm respectively. It is melted to change into sphere of diameter $2$ cm each. Find the number of sphere formed.

1226848_1397943_ans_2a66507d19cc4ad09cbee912d2d6f6ea.jpg

A cube of metal of $6$ cm edge is melted and cast into a cuboid whose base is $9$ cm $\times$ $8$ cm. Find the height.

1196672_1397871_ans_1a0c75b1ece741da9962d6520334fe9c.jpg

A right circular cylinder having radius $6$ cm and height $15$ cm is full or ice-cream. the ice-cream. the ice-cream is to be filled in cones of height $12$ cm and radius $3$ cm having a hemispherical top.Find the number of such cones that are required to empty the cylinder.
(Take, $\pi = \frac{{22}}{7}$ )

1212892_1435930_ans_cca926d6e1174a78bcef6162d7079ae3.jpg

A fight circular cone is $3.6\ cm$ high and radius is $1.6\ cm$ it is melted and recast into a cylinder with radius $1.2\ cm$ . This is height.

Volume of cone

$=$ Volume of cylinder

$\Rightarrow \dfrac{1}{3}\pi R^2H= \pi r^2h$

$\Rightarrow \dfrac{1}{3}\times 1.6\times 1.6\times 3.6=1.2\times 1.2\times h$

$h=\dfrac{1.6\times 1.6\times 3.6}{3\times 1.2\times 1.2}$

$h=2.13$ cm.

1205654_1449777_ans_d5c10c64f8404497ac406bcd4341a2ed.jpg

A pen stand is made of wood in the shape of cuboid with three concial depressions to hold the pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depression is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.

Volume of cuboid

$=$ lbh

$=15\times 10\times 3.5=525cm^3$ .

Volume of

$3$ conical depressions

$=3\times \dfrac{1}{3}\pi r^2h$

$=\pi (0.5)^2(1.4)=1.1cm^3$

Volume of wood

$=525-1.1$

$=523.9cm^3$ .

A rocket is in the shape of a cone mounted on a right circular cylinder, their common base diameter is $8 \text{ cm}.$ The height of cylindrical and conical shapes are $6 \text{ cm}$ and $3 \text{ cm}$ respectively. Find the volume of the rocket.

For conical part,

Diameter,

$d=8 \text{ cm}$

Radus,

$r=4 \text{ cm}$

Height,

$h=3 \text{ cm}$

$\begin{aligned}{}\text{Volume of conical part} &= \frac{1}{3}\pi {r^2}h\\ &= \frac{1}{3}\pi {\left( 4 \right)^2}\left( 3 \right)\\ &= 16\pi \text{ cm}^3\end{aligned}$

For cylindrical part,

Diameter,

$d=8 \text{ cm}$

Radus,

$r=4 \text{ cm}$

Height,

$h=3 \text{ cm}$

$\begin{aligned}{}\text{Volume of cylindrical part} &= \pi {r^2}h\\ &= \pi {\left( 4 \right)^2}\left( 6 \right)\\ &= 96\pi \text{ cm}^3\end{aligned}$

Hence, the total volume of the rocket will be,

Volume of rocket

$=$ Volume of conical part

$+$ Volume of cylindrical part

$=16\pi+96\pi$

$=112 \text{ cm}^3$

A solid metallic spherical ball of radius 28 cm is melted down and recast into small cones. If the diameter of the base of the cone is 28 cm and the height is 4 cm, find the number of such cones can be made?

1240716_1507847_ans_d63be922f5024f75a4df13e44fa5d1eb.PNG

A sphere of diameter $12$ cm is dropped in a right circular cylinder vessel partly filled with water . If the sphere is completely submerged in water,the water level of cylindrical vessel rises by $3\dfrac{5}{9}$ cm . Find the diameter of the cylindrical vessel .

1237271_1503457_ans_74cb9a9e3d574b6a84830bd98b864043.PNG

A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. 0.5 cm are dropped into the vessel one- fourth a the water flows out. Find the number of lead shots dropped in the vessel

Let the number of lead shots dropped

$=n$
The total number of number of lead shots

$=1/4$ volume of the conical vessel
Lead shots
Radius,

$r=0.5\ cm$
Volume,

$V=\dfrac{4}{3}\ \pi r^3$

$\dfrac{4}{3}.\dfrac{22}{7}.0.5 \times 0.5 \times 0.5 =\dfrac{4}{3} \times \dfrac{22}{7} \times 0.125$
Cone
Radius,

$r=5\ cm$
Height,

$h=8\ cm$
Volume,

$V=\dfrac{1}{3} \pi r^2 h$

$=\dfrac{1}{3}.\dfrac{22}{7} \times 5 \times 5 \times 8=\dfrac{1}{3} \times \dfrac{22}{7} \times 200$

$\dfrac{1}{4}th \ volume=\dfrac{1}{4} \times \dfrac{1}{3} \times \dfrac{22}{7} \times 200$
Total volume of number of shots

$=n \times \dfrac{4}{3} \times \dfrac{22}{7} \times 0.125$

$n \times \dfrac{4}{3} \times \dfrac{22}{7} \times 0.125=\dfrac{1}{4} \times \dfrac{1}{3} \times \dfrac{22}{7} \times 200$

$n=\dfrac{1}{4} \times \dfrac{1}{3} \times \dfrac{22}{7} \times 200 \times \dfrac{3}{4} \times \dfrac{7}{22} \times \dfrac{1}{0.125}$

$=\dfrac{200}{4 \times 4 \times 0.125}=\dfrac{200}{2}=100$

$\therefore$ The number of lead shots

$=100$

1762248_1513157_ans_9ee3c2d3f4384adf9bc7bf72fbfcd33c.png

A vessel is in the form of a hemispherical bowl mounted by a hollow cylinder. The diameter of the hemispherical is 14 cm and the total height of the vessel is 13 cm. Find

Height of cylinder $=$ Height of bowl $-$ Radius of hemisphere $= 13 - 7 = 6$ cm

Now, capacity of bowl $=$ Volume of hemisphere $+$ volume of cylinder

$= \dfrac{2}{3}\pi{r}^{3}+\pi {r}^{2}h$

$=\dfrac{ 2}{3} \times \dfrac{22}{7}\times {7}^{3} +\dfrac{22}{7}\times {7}^{2}\times 6$

$= \dfrac{(2\times 22\times 49)}{3} + 22\times42$

$= \dfrac{2156}{3} + 924$

$= 718.67 + 924$

$= 1642.67 {cm}^{3}$

Water flows through a cylindrical pipe of internal diameter $7 cm$ at $5 m$ per sec. Calculate
(i) the volume in litres of water discharged by the pipe in one minute.
(ii) the time in minutes, the pipe would take to fill an empty rectangular tank of size $4m \times 3 m \times 2.31 m$ .

Given,

Speed of water flows through cylindrical pipe

$=5\ m/sec$

Internal diameter of the pipe

$=7cm$

So, radius

$(r)=7.2cm$

Now, length of water flow in

$1$ minutes

$(h)=5\times 60=300m$

Volume of water

$=\pi { r }^{ 2 }h=\dfrac{22}7\times \dfrac72\times \dfrac 72\times 300\times 100\ { cm }^{ 3 }=1155000\ { cm }^{ 3 }=1155\ litres$

(i) the volume of water

$=1155000{cm}^{3}$

Volume of rectangular tank of size

$=4m\times 3m\times 2.31m=27.72{ m }^{ 3 }$

Also, given speed of water

$4\ m/sec$

Radius of pipe

$=\dfrac 72\ cm$

Volume of water in

$1\quad sec=\dfrac{22}7\times \dfrac72\times \dfrac72\times 5\times 100\ { cm }^{ 3 }=19250{\ cm }^{ 3 }$

Hence

(ii) Time taken to empty the tank

$=\dfrac{27.72{ m }^{ 3 }}{19250}=\dfrac{(2772\times 100\times 100\times 100)}{(100\times 19250)}=1440\ sec=\dfrac{1440}{60}=24\ minutes$

Two cylindrical vessels are filled with milk. The radius of one vessel is $15\,cm$ and height is $40\,cm$ , and the radius of other vessel is $20\,cm$ and height is $45\,cm$ , Find the radius of another cylindrical vessel of height $30\,cm$ which may just contain the milk which is in the two given vessels.

Given,
Radius of one cylinder

$\left( { r }_{ 1 } \right) =15cm\quad$
and height

$\left( { h }_{ 1 } \right) =40cm$
Radius of second cylinder

$\left( { r }_{ 2 } \right) =20cm$
and height

$\left( { h }_{ 2 } \right) =45cm$

Now
Volume of first cylinder

$\pi { r }_{ 1 }^{ 2 }{ h }_{ 1 }=\dfrac{22}7\times 15\times 15\times 40{ cm }^{ 3 }=\dfrac{198000}7{\ cm }^{ 3 }$

Volume of second cylinder

$=\pi { r }_{ 2 }^{ 2 }{ h }_{ 2 }=\dfrac{22}7\times 20\times 20\times 45=\dfrac{39600}7{\ cm }^{ 3 }$

So, total volume

$=\left( \dfrac{198000}7+\dfrac{396000}7 \right) {\ cm }^{ 3 }=\dfrac{594000}7{\ cm }^{ 3 }$

Now, height

$=30cm$
Thus,
Radius

$=\sqrt [ 3 ]{ \cfrac { 594000 }{ 7 } \times \cfrac { 7 }{ 22 } \times \cfrac { 1 }{ 30 } } =\sqrt { 900 } =30cm$

Hence, Radius of the third cylinder

$=30cm$

A cylinder whose height is two-third of its diameter has the same volume as a sphere of radius $4$ cm. Calculate the radius of the base of the cylinder.

Height of cylinder =

$\dfrac{2}{3}\times$ Diameter

Diameter =

$2\times$ radius

$\therefore h=\dfrac{2}{3}\times 2r=\dfrac{4}{3}r$

The volume of cylinder= volume of the sphere.

$\pi r^{2}h=\dfrac{4}{3}\pi r^{3}$

$\pi r^{2}\times \dfrac{4}{3}r=\dfrac{4}{3} \pi (4)^{3}$

$r^{3}=4^{3}$

$r=4 \, cm$

$\therefore$ Radius of base of cylinder

$= 4 \, cm$

The diagram shows a section of a rocket firework. If this section can be completely filled with gunpowder what is the volume of gunpowder required?

We observe that

$\sin 60^{\circ}= \dfrac PH$

$\Rightarrow \dfrac {\sqrt{3}}{2}=\dfrac r6$

$\Rightarrow r=3\sqrt{3}\ cm$

In the cone,

$6^{2}=h^{2}+r^{2}$

$\Rightarrow 6^{2}=h^{2}+(3\sqrt{3})^{2}$

$=h^{2}+27$

$\Rightarrow h^{2}=36-27=9$

$\therefore h=3\ cm$

Volume of gunpower = volume of cone + volume of cylinder

$=\dfrac{1}{3}\pi r^{2}h+\pi r^{2}h$

$=\pi r^{2}\bigg(\dfrac{1}{3}h+h\bigg)$

$=\dfrac{22}{7}\times 27\times \dfrac{4h}{3}$

$=\dfrac{22}{7}\times 27\times \dfrac{4}{3}\times 3$

$= 1782\ cm^{3}$

A vessel is in the form of hemispherical bowl surmounted by a hollow cylinder of same diameter.The diameter of the hemispherical bowl is $14$ cm and the total height of the vessel is $13$ cm.Find the total surface area of the vessel. (Use $\pi=\dfrac{22}{7}$ )

REF.Image

Let the radius and height of cylinder be r cm and h cm respectively.

Diameter of the hemispherical bowl = 14 cm

$\therefore$ Radius of the hemispherical bowl =

Radius of cylinder =

$r=\frac{14}{2} cm = 7 cm$

Total height of the vessel = 13 cm.

$\therefore$ Height of the cylinder, h=13 cm - 7 cm = 6 cm

Total surface area of the vessel=

2(CSA of cylinder + CSA of hemisphere)

Since, the vessel is hollow

$=2(2\pi rh+2\pi r^{2})$

$=4\pi r (h+r)$

$=4\times \dfrac{22}{7}\times 7\times (6+7)cm^{2}=1144 cm^2$

1240701_1507510_ans_9101995edd064ec381692c944646931f.png

Assume that a drop of water is spherical and its diameter is one tenth of a cm. A conical glass has height equal to its diameter of the rim. If 2048000 drops of water fill the glass completely then find the height of the glass.

Given that,

A drop of water is spherical in shape and has a diameter equal to one-tenth of a cm.

A conical glass has its height equal to the diameter of its rim.

The glass can contain

$2048000$ drops of water.

To find out,

The height of the conical glass.

Let the height of the glass be

$h\ cm$ .

Since, the glass is completely filled by

$2048000$ drops of water,

Volume of glass

$=$ Volume of

$2048000$ drops of water.

A water drop is spherical in shape.

We know that, volume of a sphere

$=\dfrac43 \pi r^3$

The diameter of a drop is one-tenth of a cm.

$\therefore \ d=\dfrac{1}{10}cm$

Hence,

$r=\dfrac d2=\dfrac{1}{20}\ cm$

So, volume of one drop of water

$=\dfrac43 \pi \left(\dfrac{1}{20}\right)^3$

Hence, volume of

$2048000$ drops of water

$=2048000\times \dfrac43 \pi \left(\dfrac{1}{20}\right)^3$

$=\dfrac{2048000\times 4\pi}{3\times 20\times 20\times 20}$

$=\dfrac{1024\pi}{3}\quad \quad ...(1)$

We also know that, volume of a cone

$=\dfrac13 \pi r^2h$

Here,

$d=h$

Hence,

$r=\dfrac d2=\dfrac h2$

So, volume of the conical glass

$=\dfrac13 \pi \left(\dfrac h2\right)^2 h$

$=\dfrac{\pi h^3}{12}\quad \quad ...(2)$

From

$(1)$ and

$(2)$ , we get:

$\dfrac{1024\pi}{3}=\dfrac{\pi h^3}{12}$

$\Rightarrow h^3=\dfrac{12\times 1024}{3}$

$\Rightarrow h^3=4\times 1024$

$\Rightarrow h^3=2^2\times 2^{10}$

$\Rightarrow h^3=2^{12}$

$\Rightarrow h^3=\left({2^4}\right)^3$

$\therefore\ h=2^4$

$=16\ cm$

Hence, the height of the conical glass is

$16\ cm$ .

A swimming pool is 40 m long and 15 m wide. Its shallow and deep ends are 1.5 m and 3 m deep respectively. If the bottom of the pool slopes uniformly. find the amount of water in litres required to fill the pool.

Area of cross section of the solid

$=\dfrac{1}{2} (1.5+3) \times (40)cm^2$

$=\dfrac{1}{2} (4.5) \times (40)cm^2$

$=90\ cm^2$
Volume of solid

$=$ Area of cross section

$\times$ length

$=90 \times 15\ cm^3$

$=1350\ cm^3$
We know that,

$1cm^3=1000\ lt$
Then,

$1350cm^3=1350000\ litres$

A metal cuboid of dimension $22 \,cm \times 15 \,cm \times 7.5 \,cm$ was melted and casted into a cylinder of height $14\, cm$ . Find the radius of cylinder.

Given: volume of cuboid=volume of cylinder

Volume of cuboid

$=lbh=22\times 15\times 7.5$

Volume of cylinder

$=\pi { r }^{ 2 }h=\pi \times { r }^{ 2 }\times 14$

$22\times 15\times 7.5=\cfrac { 22 }{ 7 } \times { r }^{ 2 }\times 14$

$15\times 7.5=2\times { r }^{ 2 }$

${ r }^{ 2 }=56.25$

$r=7.5$ cm

A cuboid measuring $15cm \times 12cm \times 6cm$ is melted. How many new cubes of side $3cm$ can be made?

1869493_1528964_ans_aa2c8666e1104cc6acfde5fb467f8c41.jpg

The picture shows the shape of a boiler. The total height of the boiler is $12m$ and the diameter is $6 m$ , also the height of the cylindrical shape is $6 m$ .
a. What is the height of the cone?
b. How many litres can the boiler hold? $(1m^3 = 1000 litre)$

Given, total height

$=12m$
Radius

$= \dfrac{6}{2} = 3m$
Height of the cylinder part

$(h) = 6m$

Height of the hemispherical part

$=r = 3m$
Therefore, Height of the cone

$=12 - (6+3) = 3m$

The volume of the boiler

$=$ Volume of the cone + volume of the cylinder + volume of the hemisphere.

$= \dfrac{1}{3} \pi r^2h + \pi r^2 h + \dfrac{2}{3} \pi r^3$

$=\dfrac{1}{3} \times \pi \times 3\times 3 \times 3+ \pi \times 3\times 3\times 6 + \dfrac{2}{3}\times \pi \times 3\times 3\times 3$

$=9\pi + 54 \pi + 18 \pi = 81\pi m^3$

$=81\times 3.14 = 254.34m^3$

Hence, the volume in liters

$= 254.34 \times 1000 = 254340$ liters.

There are $25$ persons in a tent which is conical in shape. Every person needs an area of $4$ sq. m. of the ground inside the tent. If height of the tent is $18$ m, find the volume of the tent.

Let the radius of the base of the cone be

$r\ m$
Height of the conical tent,

$h=18\ m$
Area occupied by each person on the ground

$=4\ m^2$

$\therefore$ Area occupied by

$25$ persons on the ground

$=25 \times 4\ m^2=100.m^2$

$\Rightarrow$ Area of the base of the cone

$=\pi r^2=100\ m^2...(1)$

$\therefore$ Volume of the conical tent

$=\dfrac{1}{3} \pi r^2 h=\dfrac{1}{3} \times 100 \times 18=600\ m^3$
[Using

$(1)$ ]
Thus, the volume of the tent is

$600\ m^3$

The radii of the circular ends of a solid frustum of a cone are $33$ cm and $27$ cm and its slant height is $10$ cm. Find the volume. [Take $\pi =\dfrac{22}{7}$ ]

$l^2-(r_1-r_2)^2+h^2\Rightarrow h=\sqrt{100-36}=8cm$

Volume of

$=\cfrac{\pi}h{3}(r^2_1+r_2^2+r_1r_2)\\=\cfrac{176}{21}(100+891)=\cfrac{174416}{21}$

A bucket open at the top is in the form of a frustum of a cone with a capacity of $12308.8 \mathrm { cm } ^ { 3 }$ . The radii of the top and bottom circular ends are $20 \,cm$ and $12 \,cm$ respectively. Find the height of the bucket and the area of metal sheet used in making it. (Use $\pi = 3.14$ )

Given,

Volume of bucket

$(V) = 12408.8 cm^3$

$R = 20 cm$ (top radii)

$r = 12 cm$ (bottom radii)

Now, volume of frustum

$= \dfrac{\pi}{3} h (R^2 + r^2 + Rr)$

$\Rightarrow 12408.8 = \dfrac{\pi}{3} \times h (20^2 + 12^2 + 20 \times 12)$

$\Rightarrow h = \dfrac{12408.8 \times 3}{\pi \times 784} = \dfrac{37226.4}{2485.28} cm$

$\Rightarrow h = 14.98 cm \approx 15 cm$

$\therefore$ Area of metal sheet used

$= TSA$ of the frustum

$= CSA +$ area of bottom

$= \pi (R + r) \sqrt{(R - r)^2 + h^2} + \pi r^2$

$= \pi (20 + 12) \sqrt{(20 + 2)^2 + 15^2} + \pi \times 12^2$

$= \pi \times 32 \times \sqrt{289} + 144 \pi$

$= (544 + 144) \pi cm^2$

$= 688 \times 3.14 = 2160.32 cm^2$

So, Height of the bucket is

$15 cm$ and area of metal sheet used in making it is

$2160.32 cm^2$ .

1847664_1534372_ans_89be370ddb1f4186a51db98d29e8c790.png

A rectangle is $16\mathrm { m }$ by $9\mathrm { m } .$ Find the side of a square whose area equals the area of the rectangle. By how much does the perimeter of the rectangle exceed the perimeter of the square?

Area of rectangle

$=(16\times 9){ m }^{ 2 }=144{ m }^{ 2 }$

Given condition,

Area of square

$=$ Area of rectangle

$\Rightarrow (side)^2=144\\ \therefore side=\sqrt { 144 } =12m$

Now,

Perimeter of square

$=4\times side =4\times 12=48m$

Perimeter of rectangle

$=2(l+b)=2(16+9)=50m$

Hence, difference in their perimeters

$=50-48=2m$

A metalic containers in the shape of a frustum of a cone.if the diameter of the cicrular ends are 20cm and 16cm, respectively and the vertical height is 42cm,find the volume of the container.also find the cost milk it can hold,if the rate of milk is Rs.40 per litre.

1305426_1607964_ans_26144c398cd4469d90475c557a1d816a.jpg

Three cubes, whose edge are x cm, 8 cm and 10 cm respectively, are melted and recasted into a single cube of edge 12 cm. Find 'x',

Volume of a melted single cube

$=x^3+8^3+10^3\ cm^3$

$=x^3 +512+1000\ cm^3$

$=x^3+1512\ cm^3$
According to the question

$12\ cm$ is the edge of the single cube.

$12^3=x^3+1512\ cm^3$

$x^3=12^3-1512$

$x^3=1728-1512$

$x^3=216$

$x^3=6^3$

$x=6\ cm$

A rectangle field is $48\ m$ long and $20\ m$ wise. How many right triangular flower beds, whose sides containing the right angle measure $12\ m$ and $5\ m$ can be laid in this field?

Length of the rectangular field

$=48 \mathrm{m}$

Breadth of the rectangular field

$=20 \mathrm{m}$

Area of the rectangular field

$=$ Length

$\times$ Breadth

$=48 \space\mathrm{m} \times 20\space \mathrm{m}=960 \space\mathrm{m}^{2}$

Area of one right triangularf flower bed

$=\dfrac{1}{2}\times\text{Base}\times\text{Height}=\dfrac{1}{2} \times 12 \space\mathrm{m} \times 5\space \mathrm{m}=30 \space\mathrm{m}^{2}$

$\therefore$ Required number of right triangular flower beds

$=\dfrac{960 \mathrm{m}^{2}}{30 \mathrm{m}^{2}}=32$

A cylindrical tube of radius $12\ cm$ contains water to a depth of $20\ cm.$ A spherical ball is dropped into the tub and thus the level of water is raised by $6.75\ cm$ . What is the radius of the ball?

Let

$r\ cm$ be the radius of the ball.

Then, the volume of ball = volume of water raised

$\Rightarrow \dfrac{4}{3}\pi r^{3} = \pi r^2 h$

$\Rightarrow \dfrac{4}{3}\pi r^{3} = \pi \times (12)^{2}\times 6.75\Rightarrow r = 9\,cm$

If the volumes of two cones are in the ratio of 1:4 and their diameters are in the ratio of 4:5, them find the ratio of their heights.

Let

$V_1$ be the volume of first cone and v2 be the volume of second cone.
Then,

$V_1: V_2 = 1.:4$ (Given) ...(2)

Let

$h_1$ be the height of first cone and

$h_2$ be the height of second cone.

We know that, volume of cone

$= v = \dfrac1 3$

$\pi (\dfrac{d^2}4) h$

$\dfrac{V_1}{V_2} = \dfrac { \dfrac { 1 }{ 2 } \pi \dfrac { d^ 2_ 1 }{ 4 } h_ 1 }{ \dfrac { 1 }{ 3 } \pi \dfrac { d^ 2_ 2 }{ 4 } h_ 2 }$

$\dfrac{V_1}{V_2} = \dfrac { d^ 2_ 1h_ 1 }{ d^ 2_ 2h_ 2 }$

$\dfrac{1}{4} = \dfrac { d^ 2_ 2h_ 1 }{ d^ 2_ 2h_ 2 }$
(from (1) )

$\dfrac{1}{4} = \left( \dfrac { d_ 1 }{ d_ 2 } \right) ^ 2\times\dfrac { h_ 1 }{ h_ 2 }$

$\dfrac{1}{4} = \left( \dfrac { 4 }{ 5 } \right) ^ 2\times \dfrac { h_ 1 }{ h_ 2 }$

$\dfrac{h_1}{h_2} = \dfrac{25}{64}$

Using equation (2)
Therefore, ratio of height of two cones is

$25:64$ .

A bucket in the form of a frustum of a cone height $30\text{ cm}$ with radii of its lower and upper ends as $10\text{ cm}$ and $20\text{ cm}$ , respectively. Find the capacity of the bucket. Also, find the cost of milk which can completely fill the bucket at the rate of $\text{Rs. }40$ per litre. $\left(\text{Use} \,\,\pi=\dfrac{22}{7}\right)$

We know that volume/capacity of frustum of a cone is given by

$= \dfrac{1}{3} \pi h (R^2 + r^2 + R\times r)$ .

Here,

$h$ is the height of the frustum and

$R,r$ is the radius of both the ends.

By using above formula,

$\text{Capacity} = \dfrac{1}{3} \times \pi \times 30 (20^2 + 10^2 + 20 \times 10)$

$= 10 \times (400 + 100 + 200)$

$= 7000 \pi \text{ cm}^3$

$= 7000 \times \dfrac{22}{7}$

$= 22000\text{ cm}^3$

Cost of milk will be equal to cost of per litre milk times volume of the bucket in litres.

$\text{Cost}= 40 \times 22000 \times 10^{-3}$

$\{ \because 1\text{ cm}^3 = 10^{-3}\text{litre}\}$

$= \text{Rs. }880$

1501322_1700989_ans_1d6db68bd6c84c0897c513a62ce98ef3.png

A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 16 cm and 12 cm. Find the capacity of the glass.

1569036_1714255_ans_e5a44345a2e8467f8c000eae2889b5db.PNG

The radii of the circular ends of a solid frustum of a cone are 18 cm and 12 cm and its height is 8 cm. Find its total surface area.

Height of frustum

$= h = 8 cm$
Radius of lower circular end

$= r = 12 cm$
Radius of upper circular end

$= R = 18 cm$
Let

$l =$ slant height
Slant height

$= l^2 = (R-r)^2 + h^2$

$l^2= 36 + 64$

$l^2=100$

$l=10cm$
We know, Total surface of frustum

$= \pi r^2 + \pi R^2 + \pi (R+r) l cm^2$

$= \pi \times 12^2 + \pi \times 18^2 + \pi (18+12) \times 10 cm^2$

$= 3.14 (144 + 324 + 300)$

$=2411.52$

The total surface area of the frustum is

$2411.52 cm^2$

A metallic bucket,open at the top, of height 24 cm is in the from of the frustum of a cone, the radii of whose lower and upper circular ends are 7 cm and,respectively.Find (i) the volume of water which can completely fill the bucket;
(ii) the care of the metal sheet used to make the bucket.

Radius of lower circular end

$= r = 7 cm$
Radius of upper circular end

$= R = 14 cm$
height of bucket

$= h = 24 cm$

Slant height of frustum:

$l = \sqrt { \left( R-r \right) ^ 2+h^ 2 }$

$=\sqrt { \left( 14-7 \right) ^ 2+24^ 2 }$

$=\sqrt {7^2+24^2}$

$= \sqrt{625}$

$= 25 cm$

(i) Volume of frustum of cone

$= \dfrac{1}{3} \pi h (R^2 + r^2 + Rr)cm^3$

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 24 (14^2 + 7^2 + 14 \times 7)$

$=\dfrac{1}{3} \times \dfrac{22}{7} \times 24 \times (196 + 98+49)$

$=\dfrac{1}{3} \times \dfrac{22}{7} \times 24 \times (343)$

$= 8624 cm^3$

(ii) Curved surface area

$= x (R+r)$

$=22/7 \times 25 \times (14 + 7)$

$= 1650 cm^2$

Area of the base of bucket =

$\pi r^2$

(consider the lower base of a bucket)

$= 22/7 \times 7 \times 7$

$= 154 cm^2$

Area of metal sheet used to make the bucket = curved surface area + Area of the base

$= 1650 + 154$

$= 1804$
Area pf metal sheet used to make the bucket is

$1804 cm^2$

A container,open at the top,is in the form of a frustum of a cone of height 24 cm with radii of its lower and upper circular ends as 8 cm and 20 cm, respectively.Find the cost of milk which can completely fill the container at the rate of Rs.21 litre.

radius of lower circular end

$= r = 8 cm$
Radius of upper circular end

$= R = 20 cm$
cost of 1 litre milk

$= Rs. 21$
Height of frustum container

$= h = 24 cm$

Volume of milk completely fill the container = volume of frustum of cone

$V=\dfrac 13 \pi \times 24 (20^2 + 8^2 + 20\times8)$

$=15689.14 cm^3$

$= 15.68914 litres$

Since 1 litre is

$1000 cm^3$

cost of 1 litre milk

$= Rs. 21$

Cost of 15.68914 litre milk

$= Rs. 21 \times 15.68914 = Rs.329.47$

A cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the tub and thus level of water is raised by 9 cm. what is the radius of the ball (in cm) ?

Let r cm be the radius of iron ball. Then,

$\Rightarrow \dfrac{4}{3}\pi r^3 = \pi \times \left ( 16 \right )^2 \times 9$

$\Rightarrow r^{3} = 64 \times 9 \times 3$

$\Rightarrow r^{3} = 4^{3} \times 3^{3} \Rightarrow r = 4 \times 3 = 12 \,cm$

An open metal bucket is in the shape of a frustum of cone of height $21 cm$ with radii of its lower and upper ends are $10 cm$ and $20 cm$ respectively. Find the cost of milk which can completely fill the bucket at the rate of $Rs. 40$ per litre.

We know,

Volume of frustum

$= \dfrac{1}{3} \pi h (R^2 + r^2 + R \times r)$

Hence,

$R = 20, r = 10$

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 21 (400 + 100 + 20 \times 10)$

$\dfrac{1}{3} \times \dfrac{22}{7} \times 21 \times 700$

$= 22 \times 7 \times 100$

$= 15400 cm^3$

$= 15400 \times 10^{-3}ltr$

$= 15.4ltr$

$\therefore$ cost of milk

$= 15.4 \times 40$

$= 616 Rs.$

1501182_1700933_ans_c88149c4501a47f3a698590da33a20c0.png

A night camp was organized for class X students for two days and their accommodation was planned in tents. Each tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are $2.1\ \text{m}$ and $4\ \text{m}$ respectively and the slant height of the top is $2.8\ \text{m}$ , find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of $\text{Rs. } 500$ per $\text{m}^2$ . ( Note that the base of the tent will be covered with canvas.) Is camping helpful to students in their development? Justify your answer.

We have,

Radius of cylindrical base

$=\dfrac{4}{2}=2\ \text{m}$

Height of cylindrical portion

$=2.1\ \text{m}$

$\therefore$ Curved surface area of cylindrical portion

$=2\pi rh$

$=2\times\dfrac{22}{7}\times 2\times 2.1$

$=26.4\ \text{m}^2$

Radius of conical base

$(r)=2\ \text{m}$

Slant height of conical portion

$(l)=2.8\ \text{m}$

$\therefore$ Curved surface area of conical top

$=\pi rl$

$=\dfrac{22}{7}\times 2\times 2.8$

$=17.6\ \text{m}^2$

So, total area of the canvas

$=(26.4+17.6)\ \text{m}^2=44\ \text{m}^2$

$\therefore$ Total cost of the canvas used

$=\text{Rs. } 500\times 44=\text{Rs. } 22,000$

Yes, it provides the feeling of self-confidence, sharing, caring and other social values.

Due to a heavy flood in a state, thousands were rendered homeless. 50 schools collectively offered to the state to provide place and the canvas for 1500 tents to be fixed by the government and decided to share the whole expenditure equally. The lower part of each tent is cylindrical of base radius 2.8 m and height 3.5 m, with conical upper part of the same base radius but of height 2.1 m. If the canvas used to make the tents costs Rs. 120 per square metre, find the amount shared by each school to set up the tents.
What value is generated by the above problem $\left( \text{use}\ \pi=\dfrac{22}{7}\right)$

Given,

Height of conical part

$=2.1\ m$

Radius of conical part

$=2.8\ m$

Radius of cylindrical part

$=2.8\ m$

Height of cylindrical part

$=3.5\ m$

Thus,

Slant height of conical part

$=\sqrt{(2.8)^2+(2.1)^2}$

$=3.5\ m$

We know that, curved surface area of a cylinder

$=2\pi rh$

And curved surface area of a cone

$=\pi rl$

$\therefore$ Area of canvas/tent

$=2\pi rh+\pi rl$

$=2\times \dfrac{22}{7}\times 2.8\times 3.5+\dfrac{22}{7}\times 2.8\times 3.5\ m^2\quad \quad \left[\because \ \pi=\dfrac{22}{7}\right]$

$=\dfrac{22}{7}\times 2.8\times 3.5(2+1)$

$=3\times \dfrac{22}{7}\times 2.8\times 3.5$

$=92.4\ m^2$

The area of one tent

$=92.4\ m^2$

Also, the cost of

$1\ m^2$ of tent

$=Rs.\ 120$

Hence, the cost of

$1500$ tents

$=1500\times 92.4\times 120=Rs. 1,66,32,000$

$50$ schools contribute this amount.

So, the share of each school

$=\dfrac{1}{50}\times 16632000=Rs. 3,32,640$

Hence, the share of each school is

$Rs.\ 3,32,640$ .

And the values generated from the given problem is, 'Helping the needy people'.

A manufacturer involved ten children in colouring playing top (lattu) which is shaped like a cone surmounted by a hemisphere. The entire top is $5\ cm$ in height and the diameter of the top is $3.5\ cm$ . Find the area they had to paint if $50$ playing tops were given to them. $\left( \text{Take}\ \pi =\dfrac{22}{7}\right)$
How is child labour an abuse for society?

Given: The top is shaped like a cone surmounted by a hemisphere.

$\therefore$ Total surface area of the top

$=$ Curved surface area of hemisphere

$+$ Curved surface area of the cone

Now, the curved surface area of hemisphere

$=2\pi r^2$

$=\left( 2\times \dfrac{22}{7}\times \dfrac{3.5}{2}\times \dfrac{3.5}{2}\right)\ cm^2$

Also, the height of the cone,

$h =$ Height of the top

$-$ Height ( radius ) of the hemisphere

$=\left( 5-\dfrac{3.5}{2}\right) cm-3.25\ cm$

So, the slant height of the cone,

$l\ =\sqrt{r^2+h^2}$

$=\sqrt{\left( \dfrac{3.5}{2}\right)^2+(3.25)^2\ cm}-3.7\ cm$ ( approx)

Therefore, curved surface area of cone

$=\pi r l$

$=\left( \dfrac{22}{7}\times \dfrac{3.5}{2}\times 3.7\right) \ cm^2$

Thus, the surface area of the top

$=\left( 2\times \dfrac{22}{7}\times \dfrac{3.5}{2}\times \dfrac{3.5}{2}\right) \ cm^2 +\left( \dfrac{22}{7}\times \dfrac{3.5}{2}\times 3.7\right)\ cm^2$

$=\dfrac{22}{7}\times \dfrac{3.5}{2}(3.5+3.7)\ cm^2$

$=\dfrac{22}{7}\times \dfrac{3.5}{2}\times 7.2\ cm^2$

$=39.6\ cm^2$ ( approx).

Hence, Surface area to be painted

$=50\times 39.6\ cm^2=1980\ cm^2$

A circus tent is cylindrical to a height of $3$ metres and conical above it. If its diameter is $15\ m$ and the slant height of the conical portion is $53\ m$ , calculate the length of the canvas $5\ m$ wide to make the required tent.

It is given that
Diameter of the cylinder

$=105\ m$
Radius of the cylinder

$=105/2=52.5\ m$
Height of the cylinder

$=3\ m$
Slant height of the cylinder

$=53\ m$

We know that

Area of canvas

$=2\pi RH+\pi R1$

By substituting the values

Area of canvas

$=(2\times (22/7)\times 52.5\times 3)+((22/7)\times 52.5\times 53)$

On further calculation

Area of canvas

$=990+8745=9735\ m^{2}$

We know that

Length of canvas =area/width

$=9735/5=1947\ m$

Therefore, the length of canvas required to make the tent is

$1974\ m$

How many bricks each of size $25\ cm\times 13.5\ cm\times 6\ cm$ , will be required to build a wall $8\ m$ long, $5.4\ m$ high and $33\ cm$ thick?

We know that volume of cuboid = length

$\times$ breadth

$\times$ height

So,

Volume of the brick

$=25\times 13.5\times 6=2025\ cm^{3}$

Volume of the wall

$=800\times 540\times 33=14256000\ cm^{3}$

Therefore, total number of bricks required

$= \dfrac{Volume\ of\ the\ wall}{ Volume\ of\ each\ brick}$

$=1425000/2025$

$=7040$ bricks

A solid cone of radius r and height h is placed over a solid cylinder having same base radius and height as that of a cone. What is the total surface area of the combined solid?

According to the question,

The total surface area of the combined solid

$=$ curved surface area of cone

$+$ curved surface area of cylinder

$+$ Area of the base.

$= \pi r l + 2\pi r h + \pi { r }^{ 2 }+ \pi r(l+2h+r)\\=\pi r (\sqrt{r^2+h^2} + 2h + r )$

A manufacturer involved ten children in colouring playing top ( lattu) which is shaped like a cone surmounted by a hemisphere. The entire top is $5\ cm$ in height and the diameter of the top is $3.5\ cm$ .
What steps can be taken to abolish child labour?

Following steps can be taken to abolish child labour:

(i) Spreading awareness against child labour in the society.

(ii) Abolishing the use of products involving child labour.

(iii) Providing free education at elementary level to poor children.

(iv) Enforcing the law to abolish child labour.

The radii of two circles are $8\ cm$ and $6\ cm$ respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.

Let

$r$ be the radius of required circle. Then, we have

$\pi r^{2}=\pi (8)^{2}+\pi (6)^{2}$

$\Rightarrow \pi r^{2}=64\pi +36\pi \Rightarrow \pi r^{2}=100\pi$

$\therefore r^{2}=\dfrac{100\pi}{\pi}=100\Rightarrow r=10\ cm$

Hence, radius of required circle is

$10\ cm$ .

The size of matchbox is $4\ cm\times 2.5\ cm\times 1.5\ cm$ . What is the volume of a packet containing $144$ matchboxes? How many such packets can be placed in a carton of size $1.5\ m\times 84\ cm\times 60\ cm$ ?

We know that volume of cuboid = length

$\times$ breadth

$\times$ height
Volume occupied by single matchbox

$=4\times 2.5\times 1.5 \ cm^{3}=15\ cm^{3}$
Volume of a packet containing

$144$ matchbox

$=15\times 144\ \ cm^{3}=2160\ cm^{3}$
Volume of carton is

$150\ cm\times 84\ cm\times 60\ cm=756000\ cm^{3}$

Therefore total number of packets that can be accommodated in a carton is

$=\dfrac{Volume\ of\ the\ carton} {Volume\ of\ the\ box}$

$=75600/2160$ packets

$=350$ packets

Two identical solid hemispheres of equal base radius $r$ cm are struck together along their bases. What will be the total surface area of the combination?

The resultant solid will be a sphere of radius

$r.$ Hence, its total surface area will be

$4\pi r^2.$

The radii of the ends of a frustum of a cone $40$ cm high are $20$ cm and $11$ cm. Find its slant height.

Given,

Heightof frustum,

$h=40$ cm

Radius of larger circular end,

$r_1=20$ cm

Radius of smaller circular end,

$r_2=11$ cm

$\therefore$ Slant height,

$l=\sqrt{h^2+(r_1-r_2)^2}$

$=\sqrt{40^2+(20-11)^2}$

$=\sqrt{1681}=41\,cm$

Hence, the slant height

$=41$ cm

A rectangular water tank of base $11 m \times 6 m$ contains water upto a height of $5 m$ . If the water in tank is transferred to a cylindrical tank of radius $3.5cm$ , find the height of the water level in the tank.

Rectangular water tank, below figure 1

$l = 11 m$ ,

$b = 6 m$ ,

$H = 5 m$

Cylindrical tank, below figure 2

$r = 3.5 m$

$h$ = ?

Water is transferred from cuboid to cylinder, so, the volume of water in both vessels will be same.

$\therefore \pi r^{2}h=1\times b\times H$

$\Rightarrow \dfrac{22}{7}\times3.5\times3.5\times h=11\times6\times5$

$\Rightarrow h=\dfrac{11\times6\times5\times7\times100}{22\times35\times35}=\dfrac{60}{7}$

$\Rightarrow h=8.6m$ (approx.)

Hence, the height of water level in cylindrical tank is $8.6 m.$

1763534_1796422_ans_946b9293c93547ce927e1a83d5399855.png

Priyanka took a wire and bent it to form a circle of radius $14 \,cm$ . Then she bent it into a rectangle with one side $24\, cm$ long. What is the length of the wire? Which shape encloses more area, the circle or the rectangle?

Given, Radius of circle

$= 14\,cm$ and Length of rectangle

$l= 24\,cm$

Hence,

$\text{ length of the wire= circumference of circle}$

$=2\pi r=2 \times \cfrac {22}{7}\times 14 = 88\,cm$

So, length of wire is

$88\ cm$

Let

$b$ the width of rectangle.

Since, the wire is bent again in the form of rectangle

Hence,

$\text{perimeter of rectangle = circumference of circle=length of wire}$

$\Rightarrow 2(l+b)=88$

$\Rightarrow 2(24+b)=88$

$\Rightarrow 24+ b =44$

$\Rightarrow b = 44-24$

$\Rightarrow b = 20\, cm$

Now, Area of circle

$\pi r^2 =\dfrac {22}{7}\times 14 \times 14 = 616\, cm^2$

And, area of rectangle

$lb=24 \times 20= 480\, cm^2$

Therefore, the circle encloses more area than rectangle.

How many cubic centimetres of iron is required to construct an open box whose external dimensions are $36 cm, 25 cm$ and $16.5 cm$ provided the thickness of the iron is $1.5 cm$ . If one cubic centimetre of iron weighs $7.5 g$ , find the weight of the box.

External diamensions	Internal diamensions
$l_{2}=36cm$	$l_{1}=36-1.5-1.5=36-3=33cm$
$b_{2}=25cm$	$b_{1}=25-1.5-1.5=25-3=22cm$
$h_{2}=16.5cm$	$h_{1}=16.5-1.5=15cm$

Volume of iron in the open box

$= l_{2}b_{2}h_{2}– l_{1}b_{1}h_{1}$

$= (36\times25\times16.5) – (33\times22\times15)$

$= 14850 – 10890$

$= 3960 cm^{3}$

Volume of iron is $3960 cm^{3}$ .

$1 cm^{3}$ of iron weighs = $7.5 gm$

So, $3960 cm^{3}$ of iron will weight = $\dfrac{3960\times75}{10}=396\times75gm$

$=\dfrac{396\times75}{1000}kg=\dfrac{297}{10}kg$

Hence, the weight of the box = $29.7 kg$ .

The barrel of a fountain pen, cylindrical in shape, is $7cm$ long and $5mm$ in diameter. A full barrel of ink in the pen is used up on writing $3300$ words on an average. How many words can be written in a bottle of ink containing onefifth of a litre?

Let $n$ times the barrel of pen is filled.

Radius of barrel, $r=\dfrac{5mm}{2}=\dfrac{5}{20}cm=\dfrac{1}{4}cm$

Height of barrel, $h = 7 cm$

$\therefore$ $n \times$ Volume of barrel = Volume of Ink

$\Rightarrow n\times\pi r^{2}h=\dfrac{1}{5}\times$ one liter

$\Rightarrow n\pi r^{2}h=\dfrac{1}{5}\times1000cm^{3}$

$\Rightarrow n\times\dfrac{22}{7}\times\dfrac{1}{4}\times\dfrac{1}{4}\times7=200cm^{3}$

$\Rightarrow n=\dfrac{200\times7\times4\times4}{22\times7}=\dfrac{1600}{11}$

Ink in one full barrel can write words = $3300$

So, $n$ barrels can write words = $3300n$

$=3300\times\dfrac{1600}{11}=4,80,000$

Hence, the required number of words = $4,80,000$ words.

A square sheet of paper is converted into a cylinder by rolling it along its side. What is the ratio of the base radius to the side of the square?

${\textbf{Step-1: Find ratio.}}$

${\text{Cylinder is made by rolling a square sheet of paper along its side.}}$

${\text{Base radius of cylinder =r.}}$

${\text{Side of squares =a.}}$

${\text{Perimeter of base circle= side of square.}}$

$\Rightarrow 2 \pi r=a$

$\Rightarrow \dfrac{r}{a} = 1:2 \pi$

${\textbf{Hence the ratio of the base radius to the side of the square}}$

$\mathbf{1:2 \pi.}$

A cube of side 5 cm is cut into as many 1 cm cubes as possible. what is the ration of the surface area of the original cube to that of the sum of the surface areas of the Smaller cubes?

Surface area of the cube is given by,

$A=6{ a }^{ 2 }$

1) Surface area of the original cube (

$a=5$ ) is given by,

${ A }_{ 1 }=6\times { \left( 5 \right) }^{ 2 }$

$\therefore { A }_{ 1 }=6\times { 25 }$

$\therefore { A }_{ 1 }=150$ sq.cm.

2) Let this cube is divided into n number of smaller cubes of side 1 cm.

Let

${ V }_{ 1 }$ = volume of the original cube

${ V }_{ 2 }$ = volume of the smaller cube

Volume of original cube = total volume of all the smaller cubes

$\therefore { V }_{ 1 }=n{ V }_{ 2 }$

$\therefore { 5 }^{ 3 }=n\left( { 1 }^{ 3 } \right)$

$\therefore 125=n\left( 1 \right)$

$\therefore n=125$

Thus, total surface area of the smaller cubes is,

${ A }_{ 2 }=125\times 6{ \left( 1 \right) }^{ 2 }$

$\therefore { A }_{ 2 }=125\times 6$

$\therefore { A }_{ 2 }=750$ sq.cm.

Thus,

$\dfrac { { A }_{ 1 } }{ { A }_{ 2 } } =\dfrac { 150 }{ 750 }$

$\therefore \dfrac { { A }_{ 1 } }{ { A }_{ 2 } } =\dfrac { 1 }{ 5 }$

If the slant height of the frustum of a cone is $10$ cm and the perimeters of its circular base are $18$ cm and $28$ cm respectively. What is the curved surface area of the frustum?

Let

$r$ and

$R$ be the radii of the two circular ends of the frustum of the cone

Then.

$2\pi r= 18$ and

$2\pi R = 28$ (Given)

$\Rightarrow r=\dfrac{18}{2\pi}$ and

$R=\dfrac{28}{2\pi}$

$\Rightarrow r=\dfrac{9}{\pi}$ and

$R=\dfrac{14}{\pi}$

$\therefore$ The curved surface area of the frustum

$=\pi(r + R)l$

$=\pi\Bigg(\dfrac{9}{\pi}+\dfrac{14}{\pi}\Bigg)\times 10\\=23\times10=230\,cm^2$

Hence, the curved surface area of the frustum is

$230\ cm^2$

Use the Fig showing the layout of a farmhouse:
(a) What is the area of land used to grow hay?
(b) It costs $Rs. 91/m^2$ to fertilize the vegetable garden. What is the total cost ?
(c) A fence is to be enclosed around the house. The dimensions of the house are $18.7\, m \times 12.6\, m$ . At least how many meters of fencing are needed ?
(d) Each banana tree required $1.25\, m^2$ of ground space. How many banana trees can there be in the orchard ?

$(a)$ For hay,

length of land =

$17.8\ m$ $

Breadth of land

$=10.6\ m$

So, Area of land used to grow hay

$= 17.8\times 10.6 =188.68\, m^2$

$(b)$ For, vegetable garden,

Length

$=49\ m$

Breadth

$=15.2\ m$

So, Area of vegetable garden

$=49\times 15.2= 744.80\, m^2$

Since, cost to fertilize

$1\, m^2$ vegetable garden

$= Rs.\,91$

Hence, cost to fertilize

$744.8\ m^2$ vegetable garden

$=91 \times 744.80= Rs. \,67776.8$

$(c)$

House is of dimension

$(18.7\ m\times 12.6\ m)$ .

fence is to be enclosed around the house.

Perimeter of the house

$=2(l+b)$

Hence, total lengths of the fence

$2(18.7+12.6)= 2\times 31.3 = 62.6\,m$

$(d)$

Area covered by banana orchard

$= (20 \times 15.7)\,m =314 \,m^2$

Since,

$1.25$ sq. area required by

$1$ banana tree.

Hence, number of banana trees can be fitted in all area of

$314m^2$

$=\cfrac{314}{1.25}= 251.2 \approx 251 \,trees$

Find the area of flower bed (with semi-circular ends) shown in figure.

The figure has two semi-circles and one rectangle.
the radius of the semi-circles

$= \dfrac{10}{2} \,cm$

$\Rightarrow r = 5 \,cm$
The length of the rectangle

$= l = 38 \,cm$
Breadth

$= b = 10 \,cm$
Area of the flower bed = Area of rectangle + Area of two semi-circles

$= l \times b + 2 \times \dfrac{\pi r^2}{2} = l \times b + \pi r^2$

$= 38 \times 10 + \dfrac{22}{7} \times 5 \times 5 = (380 + 25 \pi)cm^2$
Hence, the area of flower bed is

$(380 + 25 \pi) \,cm^2.$

1764163_1796490_ans_83cbc055a1334321899dd45649043d63.png

A solid right circular cone of height $120cm$ and radius $60cm$ is placed in a right circular cylinder full of water of height $180cm$ such that it touches the bottom. Find the volume of water left in cylinder, if the radius of the cylinder is equal to the radius to the cone.

Cone

$r=60cm$

$h=120cm$

Cylinder

$R=r=60cm$

$H=180cm$

Cone is placed inside the cylindrical vessel full of water. So, the volume of water from cylinder will over flow equal to the volume of cone.

Hence, the water left in cylinder = Vol. of cylinder – Vol. of cone (i)

Volume of water left after immersing the cone into cylinder full of water

= Volume of cylinder – Volume of cone,

$=\pi R^{2}H-\dfrac{1}{3}\pi r^{2}h$

$\therefore$ Required volume of water in cylinder,

$=\pi r^{2}H-\dfrac{1}{3}\pi r^{2}h$ [ $\because R=r$ ]

$=\pi r^{2}\left [ H-\dfrac{1}{3}h \right ]=\dfrac{22}{7}\times60\times60\left [ 180-\dfrac{120}{3} \right ]$

$=\dfrac{22}{7}\times60\times60\times140cm^{3}$

$=\dfrac{22\times60\times60\times140}{7\times100\times100\times100}=\dfrac{22\times72}{1000}=\dfrac{1584}{1000}m^{3}$

$\therefore$ Volume of water in cylinder = $1.584 m^{3}$ .

Hence, required volume of water left = $1.584 m^{3}$ .

1764187_1796491_ans_fe68cfd212794dbe887a291eaa550423.png

Find the area of the shaded field shown in the given figure.

Redraw the figure and divide it into well known shapes. It is clear from the figure that there is one semicircle and one rectangle.
Rectangle Circle

$l = 8 \,m$

$r = 6 - 4 = 2$

$b = 4 \,m$
Area of shaded region = Area of rectangle + Area of semicircle

$= l \times b + \dfrac{\pi r^2}{2} = 8 \times 4 + \pi \times \dfrac{2 \times 2}{2} = (32 + 2 \pi) \,m^2$
Hence, the required area of shaded region

$= 38.28 \,m^2.$
1764219_1796495_ans_8eae9e48f3444da3a124fad4ffe68bf6.png

A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contain $41\frac{19}{21}m^{3}$ of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building.

Cylinder

Radius = $r$
Dome (hemisphere)

Radius = $r$

$H = 2r$ [given]

$\Rightarrow h + r = 2r$

$\Rightarrow h = 2r – r = r$

Volume of building = $41\dfrac{19}{21}=\dfrac{800}{21}m^{3}$

$\Rightarrow$ Volume of cylinder + Volume of hemisphere = $\dfrac{880}{21}m^{3}$

$\Rightarrow \pi r^{2}h+\dfrac{2}{3}\pi r^{3}=\dfrac{880}{21}$

$\Rightarrow \pi r^{2}r+\dfrac{2}{3}\pi r^{3}=\dfrac{880}{21}$ [ $\because h=r$ ]

$\Rightarrow \dfrac{5}{3}\pi r^{3}=\dfrac{880}{21}$

$\Rightarrow \dfrac{5}{3}\times\dfrac{22}{7}\times r^{3}=\dfrac{880}{21}$

$\Rightarrow r^{3}=\dfrac{880\times3\times7}{21\times5\times22}$

$\Rightarrow r^{3}=8$

$\Rightarrow r=2m$

Hence, height of the building is $2\times2 = 4 m$ . [ $\because H = 2r$ (Given)]

Water flows at the rate of $10m$ per minute through a cylindrical pipe $5mm$ in diameter. How long would it take to fill a conical vessel whose diameter at the base is $40cm$ and depth $24cm$ ?

When a fluid (water) flows through a pipe of a area of cross–section $A$ with velocity $v$ , then volume of water coming from pipe in time $t$

= Area of cross–section \times Length

= $A\times v.t$ [ $\because$ $V$ = Area of base \times Height]

Cylinder

Cone

$A=\pi r^{2}$

$R=\dfrac{40}{2}cm=0.2m$

$r=\dfrac{5mm}{2}=\dfrac{5}{2000}m$

$H=24cm=0.24m$

$r=\dfrac{1}{400}m$

$v=10m/min$

$\Rightarrow v=\dfrac{10}{60}m/s=\dfrac{1}{6}m/s$

$\therefore$ Volume of lowing water = Volume of cone

$\Rightarrow$ Area of base \times height (distance)= $\dfrac{1}{3}\pi R^{2}H$

$\Rightarrow A\times v.t=\dfrac{1}{3}\pi R^{2}H$

$\Rightarrow \pi r^{2}.v.t=\dfrac{1}{3}\pi R^{2}H$

$\Rightarrow r^{2}.v.t=\dfrac{1}{3}R^{2}H$

$\Rightarrow \dfrac{1}{400}\times\dfrac{1}{400}\times\dfrac{1}{6}t=\dfrac{1}{3}\times0.2\times0.2\times0.24$

$\Rightarrow t=\dfrac{2\times2\times24\times400\times400\times6}{3\times10000}$

$\Rightarrow t=4\times24\times4\times4\times2sec$

$=\dfrac{4\times24\times4\times4\times2}{60}min=\dfrac{512}{10}=51.2min$ .

$= 51 min + 0.2 min = 51 min + 0.2\times60sec$

$\Rightarrow t = 51 min$ and $12 sec$ .

Hence, conical tank will fill in $51 min$ and $12 sec$ .

$16$ glass sphere each of radius $2cm$ are packed into cuboidal box of internal dimensions $16cm8cm8cm$ and then the box is filled with water. Find the volume of water. filled in the box.

Cuboidal box $16$ spheres,

$l = 16 cm$ $r = 2cm$

$b = 8 cm$

$h = 8 cm$

Volume of water filled in the box = Volume of box – Volume of $16$ glass spheres,

$\therefore$ Volume of water filled in the box = $lbh-16.\dfrac{4}{3}\pi r^{3}$

$=16\times8\times8-16\times\dfrac{4}{3}\times\dfrac{22}{7}\times2\times2\times2$

$=16\times8\times8-16\times8\times8\times\dfrac{1}{3}\times\dfrac{11}{7}$

$=16\times8\times8[1-\dfrac{11}{21}]=16\times8\times8\times\dfrac{10}{21}$

$=\dfrac{16\times8\times80}{21}=\dfrac{10240}{21}=487.6cm^{3}$

A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of cylinder. The diameter and height of cylinder are $6cm$ and $12cm$ , respectively. If the slant height of the conical portion is $5cm$ , then find the total surface area and volume of rocket. (Use $\pi = 3.14$ )

Cylinder

$r=\dfrac{6}{2}=3cm$

$H=12cm$

Cone
$l=5cm$

$r=3cm$

$\therefore l^{2}=r^{2}+h^{2}$ or $h^{2}=l^{2}-r^{2}$

$=5^{2}-3^{2}=25-9=16$

$\Rightarrow h=\sqrt{16}=4cm$

Now, volume of rocket

= Volume of cylinder + Volume of cone

$=\pi r^{2}H+\dfrac{1}{3}\pi r^{2}h=\pi r^{2}\left [ H+\dfrac{1}{3}h \right ]$

$=3.15\times3\times3\left [ 12+\dfrac{1}{3}\times4 \right ]$

$=3.14\times9\left [ \dfrac{40}{3} \right ]=3.14\times3\times40=376.8cm^{3}$ .

$\therefore$ Volume of Rocket = $376.8cm^{3}$

Total surface area of rocket = Curved surface area of cylinder + Curved surface area of cone + Area of base of cylinder [As it is closed (Given)]

$=2\pi rH+\pi rl+\pi r^{2}=\pi r[2H+l+r]$

$=3.14\times3[2\times12+5+3]$

$=3.14\times3\times32$

$301.44cm^{2}$

Hence, the surface area of the rocket is $301.44cm^{2}$ .

1764003_1796479_ans_b8e55318b2374f8f96303857702ad12a.png

A milk container of height $16cm$ is made of metal sheet in the form of a frustrum of a cone with radii of its lower and upper ends as $8cm$ and $20cm$ respectively. Find the cost of milk at the rate of $22$ per L, which the container can hold.

Here, $r_{1} = 8cm$

$r_{2} = 20cm$

$h = 16cm$

$\therefore$ Volume of milk = Volume of frustrum as it is filled completely

$= \dfrac{1}{3}\pi h(r_{1}^{2}+r_{2}^{2}+r_{1}r_{2})$

$=\dfrac{1}{3}\times\dfrac{22}{7}\times16[8^{2}+20^{2}+8\times20]$

$=\dfrac{22\times16}{21}[64+400+160]$

$=\dfrac{352}{21}\times624=\dfrac{352\times208}{7}=\dfrac{73216}{7}$

$=10459.428cm^{3}=10.459$ liter

Volume of milk = $10.459$ litre

$\therefore$ Cost of milk = ₹ $22\times10.459$ = ₹ $230.098$

Hence, the cost of milk = ₹ $230.098$

1763849_1796472_ans_f91aacc47be748cdbf29dbfe0e951990.png

A hemispherical bowl of internal radius $9cm$ is full of liquid. The liquid is to be filled into cylindrical shaped bottles each of radius $1.5cm$ and height $4cm$ . How many bottles are needed to empty the bowl?

$n$ cylinder bottle

$r=1.5cm$

$h=4cm$

Hemisphere

$R=9cm$

As the volume of liquid does not change

So, Volume of n bottles = Volume of hemisphere

$\Rightarrow n\pi r^{2}h=\dfrac{2}{3}\pi R^{3}$

$\Rightarrow nr^{2}h=\dfrac{2}{3}R^{3}$

$\Rightarrow n\times1.5\times1.5\times4=\dfrac{2}{3}\times9\times9\times9$

$\Rightarrow n=\dfrac{2\times9\times9\times9\times100}{3\times15\times15\times4}=54$

Hence, $54$ bottles are needed.

1764153_1796488_ans_cba6665e83fb4373941e14aeb1724279.png

How many spherical lead shots each of diameter $4.2cm$ can be obtained from a solid rectangular lead piece with dimensions $66 cm, 42 cm$ and $21 cm$ ?

Let $n$ spherical shots (as shown in below figure 1) can be obtained,

$r=\dfrac{4.2}{2}=2.1cm$

$\Rightarrow r=2.1cm$

Cuboid (as shown in below figure 2)

$l = 66 cm$

$b = 42 cm$

$h = 21 cm$

Spherical lead shots are recasted from cuboid of lead. So, volume of $n$ spherical lead shots is equal to the volume of cuboid.

$\therefore$ Volume of $n$ spherical lead shots = Volume of lead cuboid

$\Rightarrow n\times\dfrac{4}{3}\pi r^{3}=l\times b\times h$

$\Rightarrow n\times\dfrac{4}{3}\times\dfrac{22}{7}\times2.1\times2.1\times2.1=66\times42\times21$

$\Rightarrow n=\dfrac{66\times42\times21\times3\times7\times1000}{4\times22\times21\times21\times21}$

$\Rightarrow n=3\times500=1500$

Hence, the number of lead shots are $= 1500$ .

1764442_1796825_ans_df963bf6f037451282649e1c4eee5b03.png

Two cones with same base radius $8cm$ and height $15cm$ are joined together along their bases. Find the surface area of the shape so formed.

When two identical cones are joined base to base, the total surface area of new solid becomes equal to the sum of curved surface areas of both the cones.

So, total surface area of solid = $\pi rl + \pi rl = 2\pi rl$

In two cones, $r = 8 cm, h = 15 cm$

Now, $l^{2} = r^{2} + h^{2} = 8^{2} + 15^{2} = 64 + 225 = 289$

$\Rightarrow l^{2} = (17)^{2}$

$\Rightarrow l = 17cm$

$\therefore$ Total surface area of solid = $2\pi rl$

$= 2 \times \pi \times 8 \times 17$

$= 272\pi cm^{2}$

$= 854.857 cm^{2}$

Hence, the surface area of new solid $= 854.857 cm^{2}$ .

How many spherical lead shots of diameter $4 cm$ can be made out of a solid cube of lead whose edge measures $44cm$ ?

Let $n$ spherical shots can be made.
$Cube$
$a=44cm$
Spherical lead shots
$r=\dfrac{4}{2}=2cm$

$\because$ Solid cube is recasted into $n$ spherical lead shots.

$\therefore$ Volume of $n$ spherical lead shots = Volume of cube

$\Rightarrow n\times\dfrac{4}{3}\pi r^{3}=a^{3}$

$\Rightarrow n\times\dfrac{4}{3}\times\dfrac{22}{7}\times2\times2\times2=44\times44\times44$

$\Rightarrow n=\dfrac{44\times44\times44\times3\times7}{4\times22\times2\times2\times2}=121\times21$

$\Rightarrow n=2541$

Hence, the number of lead shots are $2541$ .

A wall $24m$ long, $0.4m$ thick and $6m$ high is constructed with the bricks each of dimensions $25 cm * 16 cm * 10 cm$ . If the mortar occupies $\frac{1}{10}$ th of the volume of the wall, then find the number of bricks used in constructing the wall.

Wall is $24m$ long, $0.4m$ thick and $6m$ high.

So, volume of wall = $24 m \times 0.4 m \times 6 m = 57.6 m^{3}$ .

Since $\dfrac{1}{10}$ th of the volume of the wall is occupied by mortar, so the volume of bricks in the wall,

$=\left ( 1-\dfrac{1}{10} \right )$ part of the wall,

$=\dfrac{9}{10}$ th part of the wall

$=\dfrac{9}{10}\times57.6m^{3}=51.84m^{3}$

Volume of one brick = $25 cm \times 16 cm \times 10 cm$

$\dfrac{25}{100}\times\dfrac{16}{100}\times\dfrac{10}{100}m^{3}=0.004m^{3}$

$\therefore$ Required number of bricks = (Volume of bricks in the wall)/(Volume of one brick)

$=\dfrac{51.84}{0.004}=12960$

So, $12960$ bricks are used in constructing the wall.

A cone of radius $8cm$ and height $12cm$ is divided into two parts by a plane through the midpoint of its axis parallel to its base. Find the ratio of the volumes of two parts.

$Cone$ $I$	$Cone$ $II$	$Frustrum$
$(ABO)$	$(ODE)$	$(DEBA)$
$r_{2}=8cm$	$r_{1}=\dfrac{r_2}{2}=4cm$	$r_{1}=4cm$
$h=12cm$	$h_{1}=\dfrac{h}{2}=\dfrac{12}{2}\ cm$	$r_{2}=8cm$
	$\Rightarrow h_{1}=6cm$	$h_{2}=6cm$

Let, $OF=h_1$ and $FC=h_2$

$\Rightarrow h_{1}=$ $h_{2}=\dfrac{12}{2}cm =6cm$

$\Delta OEF\sim \Delta OBC$ [By AA criterion of similarity]

$\therefore \dfrac{r_{1}}{r_{2}}=\dfrac{h_{1}}{h}\Rightarrow \dfrac{r_{1}}{8}=\dfrac{h/2}{h}=\dfrac{1}{2}$

$\Rightarrow r_{1}=4cm$

Required ratio $=\dfrac {Volume\ of frustrum (DEBA)}{Volume \ of \ cone (ODE)}$

$=\dfrac{\dfrac{1}{3}\pi h_{2}[r_{1}^{2}+r_{2}^{2}+r_{1}r_{2}]}{\dfrac{1}{3}\pi r_{1}^{2}h_{1}}$

$=\dfrac{6[4^{2}+8^{2}+4\times8]}{4\times4\times6}=\dfrac{[16+64+32]}{4\times4}=\dfrac{112}{16}=7$

$\therefore$ Volume of frustrum : Volume of smaller cone = $7 : 1$ .

1764714_1796880_ans_882595cc63b44999ac80176eab2754b1.png

The slant height of the frustum of a cone is 4 cm, and perpendicular of its circular bases are 18 cm and 6 cm respectively. Find the curved surface area of the frustum.

We know that,

Curved surface area of frustum =

$\pi(r_1+r_2)l$

Here,

$l=4cm$

We need to find

$r_1$ and

$r_2$

For

$r_1$

Circumference of

$1^{st}$ end

$=18$ cm

$2 \pi r_1=18$

$r_1=\dfrac{18}{2}\times \dfrac{1}{\pi}\\$

$r_1=\dfrac{9}{\pi}$

For

$r_2$

Circumference of

$1^{st}$ end

$=6$ cm

$2\pi r_2=6$

$r_2=\dfrac{6}{2}\times \dfrac{1}{\pi}\\$

$r_2=\dfrac{3}{\pi}$

Curved surface area of frustum =

$\pi(r_1+r_2)l$

$=\pi(r_1+r_2)l$

$=\pi \left(\dfrac 9{\pi}+\dfrac 3{\pi}\right)4$

$=48$ Sq. m

From a solid cube of side $7cm$ , a conical cavity of height $7cm$ and radius $3cm$ is hollowed out. Find the volume of remaining solid.

$Cone$ $(cavity)$

$h=7cm$

$r=3cm$

$Cube$

$Side (a) = 7cm$

Volume of remaining solid = Volume of cube – Volume of cone

$=a^{3}-\dfrac{1}{3}\pi r^{2}h$

$=7\times7\times7-\dfrac{1}{3}\times\dfrac{22}{7}\times3\times3\times7$

$=343-66=277cm^{3}$

Hence, the volume of remaining solid $= 277 cm^{3}$ .

In the given figure, find the area of the path (shown shaded) which is 2 m wide all around.

Length of field

$= 100 m$
Breadth of field

$= 60 m$
Area of the field

$= L$

$\times$

$B = 100$

$\times$

$60 = 6000$

$m^2$
Length of field exclude path

$= 100 - (2 + 2) = 96 m$
Breadth of field exclude path

$= 60 - (2 + 2) = 56$
Area of field exclude path

$= L$

$\times$

$B = 96 m$

$\times$

$56 m = 5376$

$m^2$
Area of path = Area of field - Area of field exclude path

$= 6000 - 5376 = 624 sq. m$

The perimeters of the ends of the frustum of a cone are $96\ cm$ and $68\ cm$ . If the height of the frustum be $20 cm$ , find itd radii, slant height , volume and total surface. [Take $\pi=\dfrac{22}{7}$ ]

Let radii of the circular ends of the frustum are

$R$ and

$r$ respectively.
Given that perimeter of one end

$=96\ cm$

$\Rightarrow 2\pi R=96$

$\Rightarrow R=\dfrac{96}{2\pi}$

$\Rightarrow R=\dfrac{96\times 7}{2\times 22}$

$\Rightarrow R=15.27\ cm$
Perimeter of other end

$=68\ cm$

$\Rightarrow 2\pi r=68$

$\Rightarrow r=\dfrac{68}{2\pi}$

$\Rightarrow r=\dfrac{68\times 7}{2\times 22}$

$r=10.82\ cm$
It is given that height of the frustum

$=20\ cm$
So, Slant height,

$l=\surd {h^{2}+(R-r)^{2}}$

$=\surd {(20)^{2}+(15.27-10.82)^{2}}$

$=\surd 400+(4.45)^{2}$

$=\surd 400+19.80$

$=\surd 419.80$

$=20.49\ cm$
Now,
Volume of the frustum

$=\dfrac{1}{3}\pi h(R^{2}+r^{2}+Rr)$

$=\dfrac{1}{3}\times \dfrac{22}{7}\times 20[(15.27)^{2}+(10.82)^{2}+15.27\times 10.82]$

$\dfrac{22\times 20}{21}[233.1729+117.0724+165.2214]$

$\dfrac{22\times 20\times 515.4667}{21}$

$10800.25\ cm^{3}$
Total Surface area of the frustum

$\pi R^{2}+\pi r^{2}+\pi l(R+r)$

$=\pi[(15.27)^{2}+(10.82)^{2}+20.49\times (15.27+10.82)]$

$=\pi[233.1729+117.0724+534.5841]$

$=\dfrac{22}{7}\times 884.83$

$=2780.89\ cm^{2}$
1795642_1815069_ans_af7555f8b7774638b42d22b1d26e6186.png

What length of solid cylinder $2$ cm in diameter must be taken to recast into a hollow cylinder of external diameter $20$ cm, $0.25$ cm thick and $15$ cm long?

Given,

External diameter of hollow cylinder

$= 20\, cm$

So, it’s radius

$= 10\,cm = R$

Thickness

$= 0.25\,cm$

Hence, the internal radius

$= (10 – 0.25) = 9.75\,cm = r$

Length of cylinder (h)

$= 15\,cm$

Now,

Volume

$= \pi h (R^2 – r^2) = \pi \times 15(102 – 9.752) = 15 \pi (100 – 95.0625)$

$= 15 \pi \times 4.9375\,cm^3$

Now,

Diameter of the solid cylinder

$= 2\,cm$

so, radius (r)

$= 1\,cm$

Let h be the length of the solid cylinder,

$Volume = \pi r^2h = \pi 1^2h = \pi h\,cm^3$

Then, according to given condition in the question

$\pi h = 15 \pi \times 4.9375$

$h = 15 \times 4.9375$

$h = 74.0625$

Thus, the length of solid cylinder

$= 74.0625\,cm$

A tent is made in the form of a conic frustum, surmounted by a cone. The diameters of the base and top of the frustum are $14\$ and $7\ m$ and its height is $8\ m$ . The height of the tent is $12\ m$ . Find the quantily of canvas required. [Take $\pi=\dfrac{22}{7}$ ]

For the lower portion of the tent:
Diameter of the base

$=14\ m$
Radius,

$R$ , of the base

$=7\ m$
Diameter of the top end of the frustum

$=7\ m$
Radius of the top end of the frustum

$=r=\dfrac{7}{2}m =3.5\ m$
height of the frustum

$=h=8\ m$
slant height,

$l=\surd{h^{2}-(R-r)^{2}}$

$=\surd {(8)^{2}+(7-3.5)^{2}}$

$=\surd 64+(3.5)^{2}$

$=\surd 64+12.25$

$=\surd 76.25\ m$

$=8.73\ m$
For the conical part
Radius of the cone base

$=r=3.5\ m$
Height of the cone

$=$ height of the tent - height of frustum

$12-8=4\ m$
Slant height of cone

$=L=\surd {(4)^{2}+(3.5)^{2}}$

$=\surd 16+12.25$

$\surd 28.25=5.3\ m$
Total quantity of canvas

$=CSA$ of frustum

$+CSA$ of conical top

$=\pi l(R+r)+\pi rL$

$=\pi[8.73(7+3.5)+3.5\times 5.3]$

$=\dfrac{22}{7}[91.66+18.6]$

$=\dfrac{22}{7}[110.26]$

$346.5\ m^{2}$
1795575_1815127_ans_f134ec5ec14841c3ab8f44a93a27a903.png

A godown building is in the form as shown in the adjoining figure. The vertical cross-section parallel to the width side of the building is a rectangle of size $7\ m\times 3\ m$ mounted by a semicircle of radius $3.5\ m$ . The inner measurement of the cuboidal portion are $10\ m \times 7\ m\times 3\ m$ . The inner measurement of the cuboidal portion are $10\ m\times 7\ m\times 3\ m$ Find the
volume of the godown.

Volume of godown = Volume of cuboid

$+\dfrac 12$ volume of cylinder
Volume of cuboid

$=l \times b\times h=10\times 7\times 3=210\ m^3$
Volume of cylinder

$=\pi r^2h=\dfrac{22}{7}\times (3.5)^2 \times 10=385\ m^3$

$\therefore$ Volume of godown

$=210+\dfrac 12 \times 384.65$

$=210+192.5$

$=402.5\ cm^3$

From a rectangular solid of metal $42$ cm by $30$ cm by $20$ cm, a conical cavity of diameter $14$ cm and depth $24$ cm is drilled out. Find the surface area of the remaining solid.

Given,

Dimensions of rectangular solid $l = 42$ cm, $b = 30$ cm and $h = 20$ cm

Conical cavity’s diameter $= 14$ cm

So, its radius $= 7$ cm

Depth (height) $= 24$ cm

Total surface area of cuboid $= 2 (lb + bh + lh)$

$= 2 (42 \times 30 + 30 \times 20 + 20 \times 42)$

$= 2 (1260 + 600 + 840)$

$= 2 (2700)$

$= 5400\,cm^2$

Diameter of the cone $= 14\,cm$

So, its radius $= 14/2 = 7\,cm$

Area of circular base $= \pi r^2 = \dfrac{22}7 \times 7 \times 7 = 154\,cm^2$

We know that,

$l = (7^2 + {24}^2)^{1/2} = (49 + 576)^{1/2} = (625)^{1/2} = 25$

Area of curved surface area of cone $= \pi rl = \dfrac{22}7 \times 7 \times 25 = 22 \times 25 = 550\,cm^2$

Surface area of remaining part $= 5400 + 550 – 154 = 5796\, cm^2$

1798290_1833573_ans_e0cbc14c1d6a4780b603f6e0e1e2eddc.png

A student was asked to make a model in his workshop, which shaped like a cylinder with two cones attached at its two ends, using a thin aluminium sheet. The diameter of the model is $3\ cm$ and its length is $10\ cm$ . If each cone has a height of $2\ cm$ find the volume of air contained in the model. ( Consider the outer and inner dimensions of the model to be nearly the same).

Given that diameter of cylinder

$=3\ cm$
So,
Radius

$=\dfrac 32=1.5\ cm$
Given: Height of cone

$=2\ cm$ and
Height of the cylinder +height of cone +height of cone =

$10\ cm$
Height of the cylinder

$+2+2=10$
Height of the cylinder

$=6\ cm$
So, Volume of the cylinder

$=\pi r^2h$

$=\dfrac{22}{7}\times \dfrac{3}{2}\times \dfrac{3}{2}\times 6$

$=42.428\ cm^3$

$=42.43\ cm^3$ ( approx).
Volume of

$1st$ cone

$=\dfrac 13 \pi r^2h$

$=\dfrac 13 \times \dfrac{22}{7}\times \dfrac{3}{2}\times \dfrac{3}{2}\times 2$

$=4.71\ cm^3$
Volume of

$1^{st}$ cone = Volume of

$2^{nd}$ cone =

$4.71\ cm^3$
So, Total volume of the model
= Volume of cylinder + Volume of

$1^{st}$ cone +volume of

$2^{nd}$ cone

$=42.43+4.71+4.71$

$=51.85\ cm^3$

1795636_1815081_ans_d304bfb3c58e405d85a004a61713e5d4.png

An oil funnel of tin sheet consists of a cylindrical portion $8\ cm$ along attached to a frustum of a cone. It the total height be $15\ cm$ , the diameter of the cylindrical portion $1\ cm$ and diameter of the top of the funnel $10\ cm$ , find the area of the tin required. [Take $\pi=22/7$ ]

Diameter of top of funnel

$=10\ cm$
So, Radius

$(R)=5\ cm$
Diameter of cylindrical portion

$1\ cm$
So,
radius

$(r)=\dfrac{1}{2}=0.5\ cm$
Height of frustum = Total height - Height of cylindrical part

$=15-8=7\ cm$
and Slant height,

$l=\surd{h^{2}-(R-r)^{2}}$

$=\surd {(7)^{2}+(5-0.5)^{2}}$

$=\surd 49+(4.5)^{2}$

$=\surd 49+20.25$

$=\surd 69.25$

$=8.32\ cm$
Now,

$CSA$ of frustum

$=\pi l(R+r)$

$=\dfrac{22}{7}\times 8.32\times (5+0.5)$

$=\dfrac{22}{7}\times 8.32\times (5.5)$

$=143.81\ cm^{2}$
Height of cylinder,

$H=8\ cm$
Now,

$CSA$ of cylinder

$2\pi r H$

$=2\times \dfrac{22}{7}\times 0.5\times 8$

$=25.14\ cm^{2}$
Area of tin required

$=CSA$ of Frustum

$+CSA$ of cylinder

$=143.81+25.14$

$=168.95\ cm^{2}$
1795567_1815163_ans_263c6914927b42ffbdd57ba10bbc77d9.png

A hollow sphere of internal and external diameters $4$ cm and $8$ cm respectively is melted into a cone of base diameter $8$ cm. Find the height of the cone.

Internal diameter of the hollow sphere is

$4$ cm
Internal radius of the hollow sphere r,

$\dfrac{4}{2}=2\,cm$
External diameter of the hollow sphere is

$8$ cm
External radius of the hollow sphere r,

$\dfrac{8}{2}=4\,cm$
Volume

$= \dfrac{4}{3}\pi (4^3-2^3)$

$= \dfrac{4}{3}\pi (64-8)$

$= \dfrac{4}{3}\pi (56)cm^2$
Diameter of the cone

$= 8\,cm$
Radius of the cone r_c

$=\dfrac{8}{2}=4\,cm$
Let the height of the cone be h cm.
Volume of the cone = Volume of the metal

$\Rightarrow \dfrac{1}{3}\pi r_c^{2}h=\dfrac{4}{3}\pi (56)$

$\Rightarrow r_c^{2}h=4\times 56$

$\Rightarrow 4^2\times h=4\times 56$

$\Rightarrow h=\dfrac{56}{4}$

$\Rightarrow h=14\,cm$
Hence, the height of the cone is

$14$ cm.

A godown building is in the form as shown in the adjoining figure. The vertical cross-section parallel to the width side of the building is a rectangle of size $7\ m\times 3\ m$ mounted by a semicircle of radius $3.5\ m$ . The inner measurement of the cuboidal portion are $10\ m \times 7\ m\times 3\ m$ . The inner measurement of the cuboidal portion are $10\ m\times 7\ m\times 3\ m$ Find the
total internal surface area excluding the floor.

Internal surface area of a godown excluding the flooe
= CSA of cuboid

$+\dfrac 12$ the total surface area of the cylinder

$=2h(l+b)+\dfrac 12 \times 2\pi r(r+h)$

$=2\times 3(10+7)+\dfrac{22}{7}\times 3.5(3.5+10)$

$=6\times 17+22\times 0.5 \times 13.5$

$=102+148.5$

$=250.5\ cm^2$
Hence, internal surface area of godown excluding floor is

$250.5\ cm^2$

A cone of height 15 cm and diameter 7 cm is mounted on a hemisphere of same diameter. Determine the volume of the solid thus formed.

Given,

Height of the cone $= 15\,cm$

Diameter of the cone $= 7\,cm$

So, its radius $= 3.5\,cm$

Radius of the hemisphere $= 3.5\,cm$

Now,

Volume of the solid $=$ Volume of the cone $+$ Volume of the hemisphere

We know that, volume of a cone $=\dfrac13 \pi r^2h$

And, volume of a hemisphere $=\dfrac23 \pi r^3$

$\therefore \$ Volume of the solid $= \dfrac13 \times \dfrac{22}7 \times (3.5)^2 \times 15 + \dfrac23 \times \dfrac{22}7 \times (3.5)^3\quad \quad \left[\because \ \pi=\dfrac{22}{7}\right]$

$= \dfrac13 \times\dfrac {22}7 \times (3.5)^2 (15 + 2\times3.5)$

$= \dfrac{847}3$

$= 282.33\,cm^3$

Hence, the volume of the solid is $282.33\ cm^3$

1798288_1833552_ans_1ca05aba56fe406cb5f1011382b531a7.png

A sphere, a cylinder and a cone have the same radius and same height. Find the ratio of their volumes.
[Diameter of the sphere is equal to the height of the cylinder and the cone]

Volume of the sphere

$=V_{1}=\dfrac{4}{3}\pi r^{3}$
Volume of the cylinder

$=V_{2}=\pi r^{2}h$
Volume of the cone

$=V_{3}=\dfrac{1}{3}\pi r^{2}h$
Ratio of their volumes

$=V_{1} : V_{2} : V_{3}$

$\Rightarrow \dfrac{4}{3}\pi r^{3} : \pi r^{2} h:\dfrac{1}{3}\pi r^{2}h$

$(\because h=2r)$

$\Rightarrow \dfrac{4}{3}\pi r^{3} : \pi r^{2}2r: \dfrac{1}{3}\pi r^{2}2r$

$\Rightarrow 4\pi r^{3} : 3\times 2\pi r^{3} : 2\pi r^{3}$ (Multiply by

$3$ )

$\Rightarrow 4r^{3} : 6r^{3} : 2r^{3}$

$\Rightarrow 4 : 6 : 2$

$\Rightarrow 2 : 3 : 1$

1762453_1836529_ans_7be7b172f332436581048ccf46372f01.png

From a rectangular solid of metal $42$ cm by $30$ cm by $20$ cm, a conical cavity of diameter $14$ cm and depth $24$ cm is drilled out. Find the volume of remaining solid.

Volume of the rectangular solid $= (42 \times 30 \times 20)\,cm^3 = 25200\, cm^3$

Radius of conical cavity (r) $= 7\,cm$

Height (h) $= 24\,cm$

Volume of cone $= \dfrac13 \pi r^2 h$

$= \dfrac13 \times \dfrac{22}7 \times 7 \times 7 \times 24$

$= 1232\,cm^3$

1798293_1833574_ans_1e23ee977eb74eb3b369cf2394c650f4.png

A circus tent is cylindrical to a height of $8$ m surmounted by a conical part. If total height of the tent is $13$ m and the diameter of its base is $24$ m; calculate area of canvas, required to make this tent allowing $10\%$ of the canvas used for folds and stitching.

Area of canvas used in stitching = total area of canvas

Total area of canvas $= 7656/7$ + total area of canvas/10

Total area of canvas – Total area of canvas/10 $= 7656/10$

Total area of canvas $(1 – \dfrac1{10}) = \dfrac{7656}7$

Total area of canvas $\times \dfrac9{10} = \dfrac{7656}7$

Total area of canvas $= \dfrac{7656}7 \times \dfrac{10}9 = \dfrac{76560}{63}$

Therefore, the total area of canvas $= 1215.23\,m^2$

1798305_1833613_ans_7f5ed64add3546dabcc1d6a12511c2c8.png

From a rectangular solid of metal $42$ cm by $30$ cm by $20$ cm, a conical cavity of diameter $14$ cm and depth $24$ cm is drilled out. Find the weight of the material drilled out if it weighs $7$ gm per $cm^3$ .

Volume of the rectangular solid $= (42 \times 30 \times 20)\,cm^3 = 25200\, cm^3$

Radius of conical cavity (r) $= 7\,cm$

Height (h) $= 24\,cm$

Volume of cone $= \dfrac13 \pi r^2 h$

$= \dfrac13 \times \dfrac{22}7 \times 7 \times 7 \times 24$

$= 1232\,cm^3$

Weight of material drilled out

$= 1232 \times 7g = 8624g = 8.624\,kg$
1798294_1833575_ans_766afb860ae549659023b300c9cc152a.png

A cylindrical boiler, $2\text{ m}$ high, has $3.5\text{ m}$ diameter. It has a hemispherical lid. Find the volume of its interior, including the part covered by the lid.

Given,

Diameter of cylindrical boiler $= 3.5\text{ m}$

So, the radius (r) $= \dfrac{3.5}{2} = \dfrac{35}{20} = \dfrac{7}{4}\text{ m}$

Height (h) $= 2\text{ m}$

Radius of hemisphere (R) $= \dfrac{7}{4}\,\text{m}$

Total volume of the boiler $=$ Volume of cylinder $+$ volume of hemisphere $= \pi r^2h + \dfrac23 \pi r^3$

$= \pi r^2(h + \dfrac23r)$

$= \dfrac{22}7 \times \dfrac74 \times \dfrac74 \left(2 + \dfrac23 \times \dfrac74\right)$

$= \dfrac{77}{8} \left(2 + \dfrac76\right)$

$= \dfrac{77}8 \times \dfrac{19}6 = \dfrac{1463}{48}$

$= 30.48\,\text{m}^3$

1798311_1833633_ans_4abd507249fa4a5a87e1b64b3f525545.png

The lateral surface area of the cylinder is equal to the curved surface area of a cone. If their bases are the same, find the ratio of the height of the cylinder to the slant height of the cone.

A cylinder and cone have bases of equal radii and are of equal heights. Show that their volumes are in the ratio $3:1$ .

Radius of the cone =r

Height of the cone =h

Volume of teh cone =h

Radius of the cylinder =r

Height of the cylinder =h

Volume of the cylinder =πr2h

Ratio of volume of cylinder and cone

πr2h:13πr2h⇒1:13⇒3:1

∴ Their volumes are in the ratio 3:1

A toy made in the form of hemisphere surmounting by a right cone whose circular base is joined with the plane surface of the hemisphere. The radius of the base of the cone is $6cm$ and its volume is $\dfrac{3}{2}$ of the hemisphere.
Calculate the height of the cone and the surface area of the toy correct to $2$ places of decimal. $\left(\pi=3\dfrac{1}{7}\right)$

Radius of cone,

$r=7cm$

Volume of cone

$=\dfrac{3}{2}$ volume of the hemisphere

$\dfrac{1}{3}\pi r^2h=\dfrac{3}{2}\times \dfrac{2}{3}\pi r^3$

$\Rightarrow h=3r=3\times 7=21cm$

Surface area of the toy

$=$ curved

Surface area of the cone

$+$ curved

Surface area of the hemisphere.

Radius of cone,

$r=7cm$

Height of cone,

$h=21cm$

Slant height

$=l=\sqrt{r^2+h^2}$

$=\sqrt{7^2+21^2}=\sqrt{49+441}=\sqrt{490}=22.135cm$

Curved surface area

$=\pi rl$

$=\dfrac{22}{7}\times 7\times 22.135=486.99cm^2$

Radius of hemisphere

$=7cm$

Curved surface area

$=2\pi r^2$

$=2\times \dfrac{22}{7}\times 7\times 7=308cm^2$

Curved surface area of the toy

$=486.99+308=794.99cm^2$

1762487_1836521_ans_6f1789c25407472bbec5a877946fc4d5.png

What is the least number of solid metallic spheres, each of $6$ cm diameter, that should be melted and recast to form a solid metal cone whose height is $45$ cm and diameter is $12$ cm?

Given,

Diameter of solid metallic sphere

$= 6$ cm

So, its radius

$= 3$ cm

Height of solid metal cone

$= 45$ cm

Diameter of metal cone

$= 12$ cm

So, its radius

$= 6$ cm

Now,

Let the number of solid metallic spheres be

$n$

Volume of

$1$ sphere

$= \dfrac43 \pi (3)^3$

Volume of metallic cone

$= \dfrac13 \pi \times 6^2 \times 45$

Hence,

$n = \dfrac{\text{Volume of metal cone}}{ \text{Volume of 1 sphere}}$

$n = \dfrac{\dfrac13 \pi\times 6^2 \times 45}{ \dfrac43 \pi (3)^3}$

$= \dfrac{(6 \times 6 \times 45)}{ (4 \times 3 \times 3 \times 3)}$

$\implies n = 15$

$\therefore$ 15 is the least number of solid metallic spheres, each of

66 cm diameter, that should be melted and recast to form a solid metal cone whose height is

4545 cm and diameter is

1212 cm

In the following diagram a rectangular platform with a semicircular end on one side is $22\ m$ long from one end to the other end. If the length of the half circumference is $11\ m$ , find the cost of constructing the $1.5\ m$ high platform at the rate of $Rs.4$ per cubic meters.

Given,

Length of the platform

$= 22\ m$

Half circumference of the semi-circle

$= 11\ m$

Let, the radius of the semicircular part be

$r$

now, half circumference

$=11\ m$

$\Rightarrow \pi r=11$

$\Rightarrow \dfrac{22}{7}\times r=11$

$\Rightarrow r=\dfrac{11\times 7}{22}$

$\Rightarrow r=3.5$

Now, diameter of the semicircular part

$=$ breadth of the rectangular part

$= 2 \times 3.5\ m = 7\,m$

And, length of rectangular part

$= (22 – 3.5)\ m =18.5\,m$

Now, area of base platform

$=area\ of\ rectangular\ base +area\ of\ semicircular\ base$

$= l \times b + \dfrac{1}{2} \pi r^2$

$= (18.5 \times 7 + \dfrac{1}{2} \times \dfrac{22}{7} \times 3.5 \times 3.5) \ m^2$

$= (129.5 + 19.25)\,\ m^2$

$= 148.75\,\ m^2$

Also given, height of the platform

$= 1.5\,m$

So, the volume of the platform

$= (148.75 \times 1.5)\ m^3 = 223.125\,m^3$

Rate of construction

$= Rs.\ 4$ per cubic meter

So, total cost

$= Rs.(4 \times 223.125) = Rs.\ 892.50$

Hence, the total cost is

$Rs.\ 892.50$ .

A solid cube of edge 14 cm is melted down and recasted into smaller and equal cubes each of edge 2 cm; find the number of smaller cubes obtained

We know that
Edge of the big solid cube

$= 14 cm$
Volume of the big solid cube

$= 14 \times 14 \times 14 cm = 2744cm ^{3}$
Similarly
Edge of the small solid cube

$= 2 cm$
Volume of the small cube

$= 2 \times 2 \times 2cm=8cm^3$
So we get
Number of smaller cubes obtained

$= \dfrac{volume \,\,of\,\, big\,\, cube}{Volume \,\,of \,\,one \,\,small\,\, cube} = \dfrac{2774}{8} = 343$

Four cubes , each of edge 9 cm , are joined as shown below.
Write the dimension of the resulting cuboid obtained . Also fine the total surface area and the volume of the resulting cuboid.

Edge of each cube

$= 9 cm$
We know that
Length of the cuboid formed by 4 cubes

$(1) = 9 \times 4 = 36 cm$
Breadth

$(b) = 9 cm$
and height

$(h) = 9 cm$
Total surface area of the uboid

$= 2 (lb + bh + hl)$
Substituting the values

$= 2 \left ( 36 \times 9 + 9 \times 9 + 9 \times 36 \right )cm^{2}$
By further calculation the values

$= 2\left (324 + 81 + 324 \right )cm^{2}$
So we get

$= 2 \times 729 cm ^{2}= 1458 cm^{2}$
Volume

$= l \times b \times h = 36 \times 9 \times 9 cm^{2} = 2916 cm ^{3}$

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure If the height of the cylinder is $10\ cm$ . and its radius of the base us $3.5\ cm$ , find the total surface area of the article.

The external diameter of a closed wooden box are $27\ cm, 19\ cm$ and $11\ cm$ . If the thickness of the wood in the box is $1.5\ cm$ , find:
Volume of the wood in the box;

External volume of the box

$=27\times 19\times 11\ cm^2=5643\ cm^3$
Since, external dimensions are

$27\ cm. 19\ cm, 11\ cm$
Thickness of the wood is

$=1.5\ cm$
Internal simensions

$=(27-2\times 1.5)cm, (19-2\times 1.5)cm, (11-2\times 1.5)cm$

$=24cm, 16cm, 8cm$
Hence internal volume of box

$=(24\times 16\times 8)\ cm^3=3072\ cm^3$
Volume of wood in the box

$=5643\ cm^3 -3072\ cm^3 =2571\ cm^3$

The dimensions of a classroom are: length = 15 m, breadth = 12 m and height = 7.5 m. Find, how many children can be accommodated in this classroom, assuming 3.6 m $^{3}$ of air is needed for each child.

It is given that,
Length of the room

$=15\ m$
Breadth of the room

$=12\ m$
Height of the room

$=7.5 \ m$
We know that
Volume of the room

$= L \times B \times H$

$= 15 \times 12 \times 7.5\ m^{3}$

$= 1350\ m ^{3}$

Volume of air required for each child

$= 3.6 \ m ^{3}$
So we get,
Number of children who can be accommodated in the classroom

$= \dfrac{Volume \,\,of\,\, class\,\, room}{Volume\,\, of\,\, air\,\, needed \,\,for\,\, each\,\, child}$

$= \dfrac{1350\ m^3}{3.6 \ m^3}$

$= 375$

Hence, the classroom can accommodate

$375$ students.

Spherical marbles of diameter $1.4cm$ are dropped into a cylindrical beaker of diameter $7cm$ , which contains some water. Find the number of marbles that should be dropped into the beaker, so that water levek rises by $5.6cm$ .

Radius of beaker

$=\dfrac {d}{2}=\dfrac {7}{2}=3.5\ cm$
Height of the rise,

$h=5.6\ cm$
Volume of teh water rise

$=V=\pi r^2 h$

$=\dfrac {22}{7}\times 3.5\times 3.5\times 5.6=215.6\ cm^3$
Volume of each spherical marble dropped

$=\dfrac 43 \pi r^3$
Radius of marble

$=r=\dfrac {d}{2}=\dfrac {1.4}{2}=0.7\ cm$
Volume

$=\dfrac {4}{3}\times \dfrac {22}{7}\times 0.7\times 0.7\times 0.7=1.4373\ cm^3$
Volume of the water level rise = Total volume of the marbles
If the number of marble

$'n'$

$n\times$ volume of each marble = volume of the water level rise

$n\times 1.4373=215.6$

$\Rightarrow n=\dfrac {215.6}{1.4373}=150$

$\therefore$ Number of marbles

$=150$

1762445_1836535_ans_4475ce501e104d78b8c2a53e6f8641eb.png

The external diameter of a closed wooden box are $27\ cm, 19\ cm$ and $11\ cm$ . If the thickness of the wood in the box is $1.5\ cm$ , find:
The cost of the box, if wood costs $Rs. 1.20\ cm^3$ .

External volume of the box

$=27\times 19\times 11\ cm^2=5643\ cm^3$
Since, external dimensions are

$27\ cm. 19\ cm, 11\ cm$
Thickness of the wood is

$=1.5\ cm$
Internal simensions

$=(27-2\times 1.5)cm, (19-2\times 1.5)cm, (11-2\times 1.5)cm$

$=24cm, 16cm, 8cm$
Hence internal volume of box

$=(24\times 16\times 8)\ cm^3=3072\ cm^3$
Cost of wood

$=1.20\times 2571=Rs. 3085.2$

In the figure, the height of a solid cylinder is 10 cm and diameter is 7 cm. Two equal conical holes of radius 3 cm and height 4 cm are cut off as shown in the figure. Find the volume of the remaining solid.

Given that, the height of a solid cylinder is

$10\ cm$ and diameter is

$7 \ cm$ . Two equal conical holes of radius

$3 \ cm$ and height

$4 \ cm$ are cut off from the cylinder.

To find out: The volume of the remaining solid.

Volume of the remaining solid

$=$ Volume of the given solid

$-$ Total volume of the two conical holes

Radius of the cylinder,

$=r=\dfrac {d}{2}$

$=\dfrac {7}{2}=3.5\ cm$

Height of the cylinder,

$h=10\ cm$

We know that, volume of a cylinder,

$V=\pi r^2 h$

$\therefore \$ Volume of the given solid

$=\dfrac {22}{7}\times 3.5\times 3.5\times 10\quad \quad \left[\because \ \pi=\dfrac{22}{7}\right]\\$

$\Rightarrow \dfrac {2695}{7}=385\ cm^3\\$

Radius of each conical hole,

$r=3\ cm$

Height of each conical hole,

$h=4\ cm$

We know that, volume of a cone,

$V=\dfrac{1}{3}\pi r^2 h$

$\therefore \$ Volume of each conical hole

$V=\dfrac 13 \times \dfrac {22}{7}\times 3\times 3\times 4\quad \quad \left[\because \ \pi=\dfrac{22}{7}\right]$

$\Rightarrow \dfrac {792}{21}=\dfrac {264}{7}\ cm^3$

$\therefore \$ The total volume of two holes

$=2\times \dfrac {264}{7}=\dfrac {528}{7}\ cm^3$

$\therefore \$ The remaining volume of the solid

$=385-\dfrac {528}{7}=\dfrac {385\times 7-528}{7}$

$\Rightarrow \dfrac {2695-528}{7}=\dfrac {2167}{7}\\$

$\Rightarrow 309.571\ cm^3$

Hence, the volume of the remaining solid is

$309.571\ cm^3$ .

A cylindrical tube of radius $5\text{ cm}$ and height $9.8\text{ cm}$ is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed into the mug. The radius of the hemisphere is $3.5\text{ cm}$ and the height of the conical part $5\text{ cm}$ . Find the volume of water left in the cylindrical tube.

The tube is in cylindrical shape

Radius of cylinder

$=r=5\ cm$

Height of cylinder

$=h=9.8\ cm$

Volume of cylinder

$=V=\pi r^2 h$

$=\dfrac {22}{7}\times 5\times 5\times 9.8=\dfrac {5390}{7}=770\ cm^3$

$\therefore$ Volume of the tube

$=770\ cm^3$

Radius of hemisphere

$=r=3.5\ cm$

Volume of hemisphere

$=r=\dfrac 23 \pi r^3$

$=\dfrac {2}{3}\times \dfrac {22}{7}\times 3.5\times 3.5\times 3.5=\dfrac {269.5}{3} cm^3$

Radius of the cone

$=r=3.5\ cm$

Height of the cone

$=h=5\ cm$

Volume of the cone

$=V=\dfrac 13 \pi r^2 h$

$=\dfrac 13 \times \dfrac {22}{7} \times 3.5\times 3.5\times 5=\dfrac {192.5}{3}cm^3$

Volume of the solid

$=$ Volume of the hemisphere

$+$ Volume of the cone

$=\dfrac {269.5}{3}+\dfrac {192.5}{3}\\=\dfrac {462}{3}\\=154\ cm^3$

The volume of the water replaced

$=154\ cm^3$

$\therefore$ The volume of the water left in the tube $$=

$volume of tube$ -$$ volume of solid

$=770-154=616\ cm^3$

A pen stand is made of wood in the shape of cuboid with three conical depressions to hold the pens. The dimensions of the cuboid are $15cm$ by $3.5cm$ . The radius of each of the depression is $0.5cm$ and the depth is $1.4cm$ . Find the volume of the wood in the entire stand.

The following figure shows a square cardboard ABCD of side 28 cm. Four identical circles of the largest possible sizes are cut from this card as shown below. Find the area of the remaining card-board.

Area of the square is given by

$\begin{aligned}A &=28^{2} \\&=784 \mathrm{cm}^{2}\end{aligned}$

Since there are four identical circles inside the square.

Hence radius of each circle is one fourth of the side of the square.

$\begin{aligned}\text { Area of one circle } &=\pi \times 7^{2} \\&=154 \mathrm{cm}^{2}\end{aligned}$

Area of four circle

$=4 \times 154 \mathrm{cm}^{2}$

$\begin{aligned}&=616 \mathrm{cm}^{2} \\\text { Area remaning } &=784-616 \\&=168 \mathrm{cm}^{2}\end{aligned}$

Area remaining in the cardboard is

$=168 \mathrm{cm}^{2}$

The radii of ends of a frustum are $14$ cm and $6$ cm respectively and its height is $6$ cm. Find its total surface area.

Here, $r_1 = 14$ cm, $r_2 = 6$ cm and $h = 6$ cm.
Slant height of the frustum, $l = [\sqrt{h^2 + \left( r_2 - r_1 \right)^2} = \sqrt{6^2 + \left( 14 - 6 \right)^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100}]= 10\,cm$

Total surface area of frustrum

$[= \pi\left( r_1 + r_2 \right)l + \pi r_1^2 + \pi r_2^2 ]$
$[ = 3 . 14 \times \left( 14 + 6 \right) \times 10 + 3 . 14 \times {14}^2 + 3 . 14 \times 6^2 ]$
$[ = 3 . 14 \times 20 \times 10 + 3 . 14 \times 196 + 3 . 14 \times 36]$
$[ = 628 + 615 . 44 + 113 . 04]$
$[ = 1356 . 48 { cm}^2]$

A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter I of the hemisphere is equal to the edge of the cube. Determine the surface area of the cube. Determine the surface area of the remaining solid.

Let, Length of each edge of cubical block = Diameter of hemisphere = I

$\therefore$ Radius of hemisphere,

$r = \frac{1}{2}$ unit.
Total surface area of newly formed cube = Curved surface area of cube - upper part of cube + area of hemisphere which is depressed

$= 6l^2 - \pi \left ( \frac{1}{2} \right )^2 + 2 \pi \left ( \frac{1}{2} \right )^2$

$= 6l^2 + \pi \left ( \frac{1}{2} \right )^2$

$= \frac{1}{4} (24l^2 + \pi l^2)$

$= \frac{1}{4} l^2 (24 + \pi)$
1839343_1868621_ans_3b5ef39093854da387edbf234401f077.png

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

(i) Surface area of cylindrical capsule,

$= 2 \pi rh$

$= 2 \pi \times (\frac{5}{2}) \times 9$

$= 45 \pi mm^2$
(ii) Surface area of both hemispheres,

$= 2 \{ 2 \pi \times (\frac{5}{2})^2 \} mm^2$

$= 25 \pi mm^2$

$\therefore$ Total surface area of capsule

$= 45 \pi + 25 \pi mm^2$

$= 70 \pi mm^2$

$= 70 \times \frac{22}{7}$

$= 220 mm^2$
1839346_1868629_ans_ac155991b1c545c5bc5d189f62fcbd6c.png

The circumferences of circular faces of a frustum are $132$ cm and $88$ cm and its height is $24$ cm. To find the curved surface area of the frustum complete the following activity. $([\pi = \frac{22}{7}])$

$Circumference_1 = 2[\pi] r_1 = 132$

$r_1 = [\dfrac{132}{2\pi} = \dfrac{132 \times 7}{2 \times 22} = 21 cm]$

$Circumference_2 = 2 [\pi] r_2 = 88$

$r_2 = [\dfrac{88}{2\pi} = \dfrac{88 \times 7}{2 \times 22} = 14 cm]$

Slant height of frustum,

$[l = \sqrt{h^2 + \left( r_1 - r_2 \right)^2}]$

$[ = \sqrt{{24}^2 + \left( 21 - 14 \right)^2}]$

$[ = \sqrt{{24}^2 + 7 {}^2}\text{ textcolor} {\text{ white} }{}]$

$[ = \sqrt{576 + 49}]$

$[ = \sqrt{625}]$

$[ = 25\,cm]$
Curved surface area of the frustum

$= [\pi] (r_1 + r_2 ) l$

$[= \pi \times \left( 21 + 14 \right) \times 25]$

$[ = \pi \times 35 \times 25]$

$[ = \dfrac{22}{7} \times 35 \times 25]$

$[ = 2750 sq . cm .]$

The radii of ends of a frustum are $14$ cm and $6$ cm respectively and its height is $6$ cm. Find its curved surface area.

Curved surface area of frustrum

$[= \pi\left( r_1 + r_2 \right)l]$
$[ = 3 . 14 \times \left( 14 + 6 \right) \times 10]$
$[ = 3 . 14 \times 20 \times 10]$
$[ = 628 {cm}^2]$

∴ the curved surface area of frustum is $628$ sq. cm

A circle of largest area is cut from a rectangular piece of cardboard with dimensions $55 \mathrm{cm}$ and 42 cm. Find the ratio between the area of the circle cut and the area of the remaining card-board.

${\textbf{Step -1: Find the area of circle}}{\textbf{.}}$

$\text{A circle of the largest area is cut from a rectangular piece of cardboard }$

$\text{with dimensions 55cm and 42 cm.}$

${\text{The largest area of circle is possible when diameter is 42cm}}{\text{.}}$

${\text{For the largest circle, the radius of the circle will be half of the sorter side of rectangle}}{\text{.}}$

${\text{Radius = }}\dfrac{{42}}{2} = 21{\text{ cm}}$

${\text{Area of circle = }}\pi {\left( {21} \right)^2} = 1386{\text{ c}}{{\text{m}}^2}$

${\textbf{Step -2: Find the area of rectangle and then subtract from area of circle to get area of remaining}}$

${\textbf{cardboard and then the ratio.}}$

${\text{Area of rectangle = }}\left( {42 \times 55} \right){\text{cm = 2310 c}}{{\text{m}}^2}.$

${\text{Area of remaining cardboard = }}\left( {42 \times 55} \right) - \pi {\left( {21} \right)^2}$

$= 2310 - 1386{\text{ c}}{{\text{m}}^2}$

$= 924{\text{ c}}{{\text{m}}^2}$

${\text{Ratio of area of circle and area of remaining cardboard = }}\dfrac{{1386}}{{924}}{\text{ c}}{{\text{m}}^2}.$

$= 3:2$

${\textbf{Hence, the ratio of area of circle and area of remaining cardboard is 3:2}}{\textbf{.}}$

The given figures shows a rectangle ABCD inscribed in a circle as shown alongside. If $A B=28 \mathrm{cm}$ and $\mathrm{BC}=21 \mathrm{cm},$ find the area of the shaded portion of the given figure.

Given,

$\begin{aligned} A B &=28 \mathrm{cm} \\ B C &=21 \mathrm{cm} \\ A C &=\sqrt{A B^{2}+B C^{2}} \\ &=\sqrt{28^{2}+21^{2}} \end{aligned}$

$=35 \mathrm{cm}$

$\begin{aligned}\text { Area } &=\pi \times\left(\dfrac{35}{2}\right)^{2} \\&=962.5 \mathrm{cm}^{2}\end{aligned}$

Area of the rectangle

$=28 \times 21$

$=588 \mathrm{cm}^{2}$

Therefore, the area of the shaded portion,

$A=962-588=374.5 \mathrm{cm}^{2}$

In the given figure, a cylindrical wrapper of flat tablets is shown. The radius of a tablet is $7$ mm and its thickness is $5$ mm. How many such tablets are wrapped in the wrapper?

Radius of a tablet, $r = 7\,mm$
Thickness of a tablet, $h = 5\,mm$
Radius of cylindrical wrapper, $R = [\dfrac{14}{2}] = 7\,mm$

Height of wrapper (H)

$= 10\,cm$

$= 10 \times 10\,mm$

$= 100\,mm$
Let n be the number of tablets wrapped in the wrapper.

$[n = \dfrac{\text{ Volume of the cylindrical wrapper} }{\text{ Volume of each tablet } }]$

$[ \Rightarrow n = \dfrac{\pi R^2 H}{\pi r^2 h}]$

$[ \Rightarrow n = \dfrac{\left( 7 \right)^2 \times 100}{\left( 7 \right)^2 \times 5} = 20]$

Thus, $20$ tablets are wrapped in the wrapper.

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in figure. If the height of the cylinder is 10 cm. and its base is of radius 3.5 cm. find the total surface area of the article.

Radius of base of cylinder, r = 3.5 cm
Height, h = 10 cm.
Total area of article = Curved surface of cylinder + 2 x Area of Hemisphere

$= 2 \pi rh + 2 \times 2 \pi r^2$

$= 2 \times \frac{22}{7} \times 3.5 \times 10 + 2 \times 2 \times \frac{22}{7} \times (3.5)^2$

$= 3.5 \times 34 \times \frac{22}{7}$

$= 374$ sq.cm.
1840905_1868662_ans_0506469a456c4fd49c231e1387cf58eb.png

A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

(i) Radius of cylinder GHEF, r = 60 cm.
Height, h = 120 + 60 = 180 cm
Volume of cylinderical vessel

$= \pi r^2 h = \pi \times (60)^2 \times 180$
(ii) Volume of hemisphere + volume of cone

$= \frac{2}{3} \pi r^3 + \frac{1}{3} \pi r^2 h$

$= \frac{2}{3} \pi \times (60)^3 + \frac{1}{3} \times \pi \times (60)^2 \times 120$

$= \frac{4}{3} \pi \times (60)^3 cm^3$
(iii) Quality of water remained in cylinder.

$= \pi \times (60)^2 \times 180 - \frac{4}{3} \pi \times (60)^3 cm^3$

$= (60)^3 \times \{ \frac{3}{4}\pi - \frac{4}{3} \pi \}$

$= 216000 \times \frac{5}{3} \pi$

$= 360000 \pi cm^3$

$= 360000 \times \frac{22}{7} cm^3$

$= \frac{360000 \times 22}{(100)^3 \times 7} m^3$

$= 1.131 m^3$
1841108_1869033_ans_39f035fbcb0e4426bd898e9cbe1377f2.png

From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollow out. Find the total surface area of the remaining solid to the nearest cm $^2$ ?.

Height of the cylinder, h = 2.4 m.
Diameter of the cylinder, d = 1.4 m.

$\therefore$ Radius, r = 0.7 m.
(i) Outer Curved surface area of cylinder

$= 2 \pi rh$

$= 2 \times \frac{22}{7} \times (0.7) \times 2.4$

$= 44 \times 0.24$

$= 10.56 cm^2$
(ii) Area of base of

$= \pi r^2$

$= \frac{22}{7} \times (0.7)^2$

$= \frac{22}{7} \times 0.49$

$= 1.54 cm^2$
(iii) Inner surface area of cylinder,

$= \pi rl$

$= \frac{22}{7} \times 0.7 \times \sqrt{(0.7)^2 + (2.4)^2}$

$= \frac{22}{7} \times 0.7 \times 2.5$

$= 5.5 cm^2$

$\therefore$ Total surface area of newly formed cube = 10.56 + 1.54 + 5.5 = 17.6 cm

$^2$

$\therefore$ Nearest value is 18 sq.cm
1840900_1868644_ans_d9a36c058c37418896e9ccc8649b9f9d.png

Write the formula to find slant height of an frustum cone.

$l = \sqrt{h^2 + (r_1 - r_2)^2}$

Write the formula to find the lateral surface area of frustum of cone.

lateral surface area of frustum of cone

$= \pi l (R + r)$

A solid hemisphere of wax of radius 12 cm is melted and made into a cone of base radius 6 cm. Calculate the height of the cone.

Radius of the hemisphere,

$r_1 = 12$ cm
Radius of the base of the cone,

$r_2 = 6$ cm
Height of the cone = h = ?
volume of wax cone = volume of wax hemisphere.

1840957_1868834_ans_7d89d275a26f4da0a2d03268ef3f79d4.PNG

21 spheres of equal radii are melted to form a cylinder of radius 14 cm and height 49 cm. Find the radius of sphere.

radius of cylinder

$= r_1 = 14 cm$
height h = 49 cm and radius of sphere

$= r_2$
Total volume of 21 spheres = volume of cylinder.
1840966_1868836_ans_74e55ef94d40435392de561289e041e3.PNG

A Gulab Jamun contains sugar syrup upto about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.

${\textbf{Step 1: Volume of gulab jamun = volume of cylinder + volume of 2 hemispheres}}$

${\text{Volume = 3}}\pi {{\text{r}}^2}h + \dfrac{4}{3}\pi {r^3}$

${\text{total h = 5}}$

${\text{length of the cylinder h = 5 - 2}}{\text{.8 = 2}}{\text{.2cm}}$

${\text{volume = }}\dfrac{{22}}{7} \times 1.4 \times 1.4\dfrac{{\left( {3 \times 2.2 + 4 \times 1.4} \right)}}{3}$

$= \dfrac{{6.16(12.2)}}{3}$

$= \dfrac{{75.152}}{3} = 25.05c{m^3}$

${\text{Total number of gulab jamun = 45}}$

${\text{total volume of 45 gulab jamuns = 45}} \times 25.05$

$= 1127.2830c{m^3}$

${\text{Quantity of syrup = 30% of volume of gulab jamun}}$

$\dfrac{{30}}{{100}} \times 1127.28 = 338.184c{m^3}$

$\boldsymbol{{\textbf{Thus, the quantity of syrup is 338}}{\textbf{.184c}}{{\textbf{m}}^3}}$

The cores of three cubes are $8 \mathrm{cm}, 6 \mathrm{cm}$ and $1 \mathrm{cm}$ respectively. Having melted these cubes a new cube is recastesd. Find the total surface area of the new cube recanted.

Volume of the cube with core of

$8 \mathrm{cm}=(\mathrm{core})^{3}$

$=8^{3}=512 \mathrm{cm}^{3}$
Volume of the cube with core

$6 \mathrm{cm}=(\mathrm{core})^{3}$

$=(6)^{3}=216 \mathrm{cm}^{3}$
Volume of the cube with core

$1 \mathrm{cm}=(\text { core })^{3}$

$=(1)^{3}=1 \mathrm{cm}^{3}$
The total volume of three cubes

$512+216+1=729 \mathrm{cm}^{3}$
Having melted these cubes, a new cube is recasted

$\therefore$ The volume of the cube recasted

$=729 \mathrm{cm}^{3}$

$\Rightarrow(\text { core })^{3}=729$

$\Rightarrow$ core

$=\sqrt[3]{729}$

$=\left(9^{3}\right)^{1 / 3}=9 \mathrm{cm}$
Total surface area of cuboid recasted

$=6(\text { core })^{2}$

$6 \times 9 \times 9=486 \mathrm{cm}^{2}$
Hence the surface area of the new cube

$=486 \mathrm{cm}^{2}$

A conical vessel with base radius $10 \mathrm{cm}$ . contains some water in it. The water level in the vessel is $18 \mathrm{cm}$ high. It is pored into another cylindrical vessel with a radius $5 \mathrm{cm}$ . Find the water level in the cylindrical vessel.

Base radius of conical vessel

$(\mathrm{R})=10 \mathrm{cm}$
And the height

$(\mathrm{H})=18 \mathrm{cm}$
Volume of conical vessel

$=\dfrac{1}{3} \pi R^{2} H$

$=\dfrac{1}{3} \times \pi \times(10)^{2} \times 18$

$=\pi \times 100 \times 6$

$=600 \pi \mathrm{cm}^{3}$
Let the water lavel in the cylindrical vessel be h.
Radius

$(\mathrm{r})=5 \mathrm{cm}$
Now according to the problem Volume of water in the cylindrical vessel = Volume of water in conical vessel

$\pi r^{2} h=600 \pi$
or

$r^{2} h=600$

$\Rightarrow(5)^{2} h=600$

$\Rightarrow 25 \mathrm{h}=600$

$h=\dfrac{600}{25}$

$=24 \mathrm{cm}$
Hence the water level in the cylindrical vessel

$=24 \mathrm{cm}$ .

How many silver coins, 1.75 cm. in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm x 10 cm x 3.5 cm?

Diameter of silver coins is 1.75 cm
Thickness = 2 mm
Measurement of cuboid: 5.5 x 10 x 3.5 cm

$\therefore$ Let the required silver coin be 'n'.
Volume of 'n' coins = volume of cuboid

$\pi r^2 h$ = I x b x h

$n \times \{ \pi \times (\frac{1.75}{2})^2 \times 0.2 \} = 5.5 \times 10 \times 3.5$

$n \times \frac{22}{7} \times \frac{175}{200} \times \frac{175}{200} \times \frac{2}{10} = 55 \times \frac{35}{10}$

$n \times \frac{22}{7} \times \frac{7}{8} \times \frac{7}{8} \times 2 = 55 \times 35$

$n = \frac{55 \times 35 \times 16}{7 \times 11}$

$n = 5 \times 5 \times 16$

$\therefore n = 400$

$\therefore 400$ Silver coins are required.

A cylinder is made of glass, whose radius and height are $4 \mathrm{cm}$ and $10 \mathrm{cm}$ respectively. By melting it, how many sphere with radii $2 \mathrm{cm}$ each can be recasted?

Given, for a cylinder Radius

$=4 \mathrm{cm}$
And height

$=10 \mathrm{cm}$
Volume

$=\pi r^{2} h$

$=\pi \times 4 \times 4 \times 10 \mathrm{cm}^{3}$
Again the radius of a sphere recasted (

$r$ )

$=2 \mathrm{cm}$ .
Volume of sphere

$=\dfrac{4}{3} \pi r^{2}$

$=\dfrac{4}{3} \times \pi \times(2)^{3}$
Let the number of sphere recasted be

$n$ , then Volume of the cylinder

$=n \times$
Volume of sphere

$n=\dfrac{\text { Volume of cylinder }}{\text { Volume of a sphere }}$

$=\dfrac{\pi \times 4 \times 4 \times 10}{\dfrac{4}{3} \times \pi \times 2 \times 2 \times 2}$

$\begin{array}{l}=\dfrac{4 \times 4 \times 10 \times 3}{4 \times 4 \times 2} \\=15\end{array}$
Hence, the number of sphere recasted = 15

How many cones with radius $3 \mathrm{cm}$ and height $6 \mathrm{cm}$ can be formed by melting a metallic sphere with radius $9 \mathrm{cm}$ .

Given,
Radius of sphere

$(\mathrm{r})=9 \mathrm{cm}$
Volume of the sphere

$=\dfrac{4}{3} \pi r^{3}$

$=\dfrac{4}{3} \times \dfrac{22}{7} \times 9 \times 9 \times 9 \mathrm{cm}^{3}$
For a cone,
Radius

$(\mathrm{R})=3 \mathrm{cm}$
Height

$(\mathrm{h})=6 \mathrm{cm}$
Volume of the cone

$=\dfrac{1}{3} \pi r^{2} h$

$=\dfrac{1}{3} \times \dfrac{22}{7} \times 3 \times 3 \times 6 \mathrm{cm}^{3}$
Let the number of cones recasted be n.

$\therefore$ Volume of sphere

$=n \times$
volume of a cones

$\Rightarrow \dfrac{4}{3} \times \dfrac{22}{7} \times 9 \times 9 \times 9=n \times \dfrac{1}{3} \times \dfrac{22}{7} \times 3 \times 3 \times 6$

$\begin{aligned}n &=\dfrac{\dfrac{4}{3} \times \dfrac{22}{7} \times 9 \times 9 \times 9}{\dfrac{1}{3} \times \dfrac{22}{7} \times 3 \times 3 \times 6} \\&=54\end{aligned}$
Hence, the number of cones recasted = 54

Twenty seven solid iron spheres, each of radius $r$ and surface area $S$ are melted to from a sphere with surface area $S_1$ . Find the
(i) radius $r$ of the new sphere,
(ii) ratio of $S$ and $S_1$

(i) Volume of $1$ sphere $V = \dfrac {4}{2} \pi r^3$

Volume of $27$ solid sphere

$= 27 \times \dfrac {4}{3} \pi r^3$

Let $r_1$ is the radius of the new sphere

Volume of new sphere = volume of $27$ solid sphere

$\dfrac {4}{3} \pi r^3_1 = 27 \times \dfrac {4}{3} \pi r^3$

$\dfrac { r^3_1}{r^3} = 27$

$\dfrac {r_1}{r} = \sqrt {[27]}$

$\dfrac {r_1}{r} = \dfrac {3}{1}$

$\therefore r_1 = 3r$

(ii) $\dfrac { Surface\quad area\quad of\quad new\quad sphere\quad S_1 }{ Surface\quad area\quad of\quad old\quad sphere\quad S }$

$\dfrac {S_1}{S} = \dfrac { 4 \pi r^2_1}{4 \pi r^2 }$

$\dfrac {S_1}{S} = \dfrac { (3r)^2}{r^2}$

$\dfrac {S_1}{S} = \dfrac {9r^2}{r^2}$

$\dfrac {S_1}{S} = \dfrac {9}{1}$

$\therefore S_1 : S = 9 : 1$

$\therefore S : S_1 = 1 : 9$

Three metal cubes whose edges measure $3\ cm, 4\ cm$ and $5\ cm$ respectively are melted to form a single cube find side-length

Volume of a cube

$=l^3$
Volume of a cube

$1=3^3=27 cm^3$
Volume of a cube

$2=4^3=64 cm^3$
Volume of a cube

$3=5^3=125 cm^3$
Volume of a cube

$3=216 cm^3$
Volume of a New cube formed

$=216 cm^3$

$l^3=216$

$l=3 \sqrt {216}=6 cm$

$\therefore$ side -length

$=6 cm$

An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top pf the funnel is 18 cm, find the area of the tin sheet required to make the funnel.

$r_1 = \frac{18}{2} = 9$ cm,

$r_2 = \frac{8}{2} = 4$ cm
Height h of frustum = 22 - 10 = 12 cm
Slant height of the frustum cone =

$l = \sqrt{h^2 + (r_1 - r_2)^2}$

$l = \sqrt{(12)^2 + (9 - 4)^2}$

$l = \sqrt{144 + 25} = \sqrt{169}$

$l = 13$ cm
radius of cylindrical portion

$= \frac{8}{2}$
= 4 cm
Area of tin sheet required = CSA of cylinder + CSA frustum portion

$= 2 \pi r h + \pi (r_1 + r_2) l$

$= 2 \pi (4)(10) + \pi (9 + 4) 13$

$= 80 \pi + \pi \times 13 \times 13 = 80 \pi + 169 \pi$

$= 249 \pi$

$= 249 \times \frac{22}{7}$

$= \frac{5478}{7}$

$= 782 \frac{4}{7} cm^2$

A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of glass.

Height of drinking glass, h = 14 cm
Diameter of circular ends
D = 4 cm

$\therefore$ r = 2 cm
d = 2 cm

$\therefore$ r = 1 cm
Capacity of glass = ?
Volume of glass = Radius with 2 cm + Volume of frustum with radius 1 cm.

$= \frac{1}{3} \pi h (R^2 + r^2 + Rr)$

$= \frac{1}{3} \times \frac{22}{7} \times 14 (2^2 + 1^2 + 2 \times 1)$

$= \frac{1}{3} \times \frac{22}{7} \times 14 (4 + 1 + 2)$

$= \frac{1}{3} \times \frac{22}{7} \times 14 \times 7$

$= \frac{308}{3}$

$= 102 \frac{2}{3} cm^3$
1841112_1869116_ans_7a691c5c713648c8a060219f1b2ec539.png

Three metal cubes whose edges measure $3\ cm, 4\ cm$ and $5\ cm$ respectively are melted to form a single cube find total surface are of a new cube. What is the difference between total surface area of the new cube and the sum of total surface areas of the original cubes?

T.S.A of new cube

$=6 l^2$

$=6 \times 6^2=6 \times 36=216 cm^2$
T.S.A of cube

$1=6 \times l^2=6 \times 3^2=6 \times 9=54 cm^2$
T.S.A of cube

$2=6 \times l^2=6 \times 42=6 \times 16=96 cm^2$
T.S.A of cube

$3=6 \times l^2=6 \times 5^2=6 \times 25=150 cm^2$
Total T.S.A of

$3$ cubes

$=300 m^2$
Difference between the surface areas

$=300-216=84 cm^2$

The radius of a solid metallic sphere is $r$ . A solid metallic cone of height $h$ has base radius $r$ . The two are melted together and recast into a solid right circular cone with base radius $r$ . Prove that the height of the resulting cone is $4 r + h.$

Volume of resulting cone

$=$ Volume of sphere

$+$ Volume of cone which are melted.
Volume of a sphere of radius

$'r' = \cfrac { 4 }{ 3 } \pi { r }^{ 3 }$
Volume of a cone

$= \cfrac { 1 }{ 3 } \pi { r }^{ 2 }h$

where,

$r$ is the radius of the base of the cone and

$h$ is the height.
Let the height of the resulting cone be

$H$ .
Hence,

$\cfrac { 1 }{ 3 } \pi { r }^{ 2 }H=\cfrac { 4 }{ 3 } \pi { r }^{ 3 }+\cfrac { 1 }{ 3 } \pi { r }^{ 2 }h$

$=> H = 4r + h$
Hence, height of the resulting cone is

$4r + h$

A solid iron pole consists of a cylinder of height $220$ cm and base diameter $24$ cm, which is surmounted by another cylinder of height $60$ cm and radius $8$ cm. Find the mass of the pole, given that 1 $cm^3$ of iron has approximately $8$ g mass. (Use $\pi$ = 3.14)

Height (

$h_1$ ) of larger cylinder

$= 220$ cm
Radius (

$r_1$ ) of larger cylinder =

$\dfrac{24}{2} = 12$ cm
Height (

$h_2$ ) of smaller cylinder =

$60$ cm
Radius (

$r_2$ ) of smaller cylinder =

$8$ cm
Total volume of pole = Volume of larger cylinder + Volume of smaller cylinder

$=\pi r_{1}^2 h_1 +\pi r_{2}^2 h_2$

$=(\pi (12)^2\times 220) + (\pi (8)^2 \times 60)$

$=\pi[144 \times 220 + 64 \times 60]$

$=3.14[31,680 + 3,840]$

$=3.14 \times 35520 = 111,532.8$ cm

$^3$

Mass of 1

$cm^3$ iron =

$8$ g

Mass of

$111532.8$ cm

$^3$ iron =

$11532.8 \times 8 = 892262.4$ g

A farmer connects a pipe of internal diameter 20 cm form a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 kilometer per hour, in how much time will the tank be filled?

$\Rightarrow$ Radius (

$r_1$ ) of circular end of pipe

$= \dfrac{20}{200} = 0.1$ m

$\Rightarrow$ Area of cross-section

$= \pi\times r_{1}^2 = \pi\times(0.1)^2=0.01\pi$ sq. m

$\Rightarrow$ Speed of water

$= 3$ kilometer per hour

$=$

$\dfrac{3000}{60} = 50$ meter per minute.

$\Rightarrow$ Volume of water that flows in

$1$ minute from pipe

$=$

$50 \times 0.01\pi=0.5\pi$ cu. m

$\Rightarrow$ From figure 2, Volume of water that flows in

$t$ minutes from pipe

$=$

$t \times 0.5\pi$ cu. m

$\Rightarrow$ Radius (

$r_2$ ) of circular end of cylindrical tank

$=$

$\dfrac{10}{2}= 5$ m

$\Rightarrow$ Depth (

$h_2$ ) of cylindrical tank

$= 2$ m

$\Rightarrow$ Let the tank be filled completely in

$t$ minutes.

$\Rightarrow$ The volume of water filled in tank in

$t$ minutes is equal to the volume of water flowed in

$t$ minutes from the pipe.

$\Rightarrow$ Volume of water that flows in

$t$ minutes from pipe

$=$ Volume of water in tank

Therefore,

$t \times 0.5\pi = \pi r_{2}^2 \times h_2$

$\Rightarrow t \times 0.5 = 5^2 \times 2$

$\Rightarrow t = \dfrac{25\times 2}{0.5}$

$\Rightarrow t = 100$

Therefore, the cylindrical tank will be filled in

$100$ minute.

The front compound wall of a house is decorated by wooden spheres of a diameter $21\ cm$ , placed on small supports as shown in the figure. Eight such spheres are used for this purpose and are to be painted silver. Each support is a cylinder of radius $1.5\ cm$ and height $7\ cm$ and is to be painted black. Find the cost of paint required if silver paint costs $25\ paise\ per\ cm^{2}$ .

$r=\cfrac{21}{2}=10.5$ cm, SA

$=4\pi {r}^{2}$

$=4\times \cfrac{22}{7}\times {(10.5)}^{2}$

$=1386$ cm

$^{2}$
Cylindrical support,

${r}_{1}$ of circular end

$=1.5$ cm,

$h=7$ cm
Curved Surface Area

$=2\pi rh=2\times \cfrac{22}{7}\times 1.5\times 7=66$ cm

$^{2}$
Area of circular end

$=\pi {r}^{2}=\cfrac{22}{7}\times {(1.5)}^{2}=7.07$ cm

$^{2}$
Area to be painted silver

$=8\times (1386-7.07)$ cm

$^{2}$

$=8\times 1378.9=11031.44$ cm

$^{2}$
Cost of silver color

$=$ Rs.

$0.25\times 11031.4$

$=$ Rs.

$2757.8$
Area to be painted black

$=8\times 66=578$ cm

$^{2}$
Cost of painting black color

$=1528\times 0.05$

$=$ Rs.

$26.40$
Total cost

$=$ Rs.

$2757.8+26.4$

$=$ Rs.

$2784.2$

A vessel is in the form of an inverted cone. Its height is $8\ cm$ and the radius of its top, which is open, is $5\ cm$ . It is filled with water up to the brim. When lead shots, each of which is a sphere of radius $0.5\ cm$ are dropped into the vessel, $\dfrac {1}{4}$ of the water flows out. Find the number of lead shots dropped in the vessel.

Given:
Height (

$h$ ) of conical vessel

$= 8$ cm
Radius (

$r_1$ ) of conical vessel

$= 5$ cm
Radius (

$r_2$ ) of lead shots

$= 0.5$ cm
Let

$x$ number of lead shots were dropped in the vessel.

Volume of the cone

$=\dfrac{1}{3} \pi (r_1)^3$

Volume of water spilled = Volume of dropped lead shots

According to the question

$\dfrac{1}{4}\times \dfrac{1}{3}\pi r_{1}^{2}h=x\times \dfrac{4}{3} \times \pi r_{2}^{3}$

$\Longrightarrow r_{1}^{2}h= x\times 16 r_{2}^{3}$

$\Longrightarrow 5^2 \times 8 =x \times 16 \times (0.5)^3$

$\Longrightarrow 200 = x \times 2$

$\Longrightarrow x = \dfrac{200}{2}$

$\Rightarrow$

$x = 100.$

Therefore, the number of lead shots dropped in the vessel is

$100$ .

Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is $3\ cm$ and its length is $12\ cm$ . If each cone has a height of $2\ cm$ , find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

$\textbf{Step 1: Finding the radius and height of cone and cylinder}$

$\text{Radius of the conical and cylindrical part }(r)=1.5\;cm$

$\text{Height of the cylindrical part }(H)=8\;cm$

$\text{Height of the conical part }(h)=2\;cm$

$\textbf{Step 2: Finding the volume of air in the model}$

$\text{Volume of air in the model=Volume of cylinder +}2\times\text{Volume of cone}$

$\text{Volume of cylinder}=\pi r^2H$

$=\dfrac{22}{7}\times(1.5)^2\times8$

$\text{Volume of cone}=\dfrac{1}{3}\pi r^2h$

$=\dfrac{1}{3}\times\dfrac{22}{7}\times(1.5)^2\times2$

$\text{Volume of air}=\dfrac{22}{7}\times(1.5)^2\times8+2\times\dfrac{1}{3}\times\dfrac{22}{7}\times(1.5)^2\times2\approx66\;cm^3$

$\textbf{Hence , the volume of air in the model is 66 }\mathbf{cm^3}$

A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are $2.1$ m and $4$ m respectively, and the slant height of the top is $2.8$ m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs $500$ per $m^2$ . (Note that the base of the tent will not be covered with canvas.) (Use $\pi = \dfrac{22}{7}$ ).

Given:
Height (h) of the cylindrical part =

$2.1$ m
Diameter of the cylindrical part =

$4$ m
Radius of the cylindrical part =

$2$ m
Slant height (l) of conical part =

$2.8$ m

Area of canvas used = CSA of conical part + CSA of cylindrical part

$=\pi rl + 2\pi r h$

$=\pi\times 2 \times 2.8 + 2\pi \times 2 \times 2.1$

$=2\pi[2.8 + 4.2]$

$= 2\times \dfrac{22}{7}\times 7$

$= 44 m^2$

Cost of

$1$ m

$^2$ canvas

$= 500$ rupees
Cost of

$44$ m

$^2$ canvas

$= 44 \times 500 = 22000$ rupees.

Therefore, it will cost

$22000$ rupees for making such a tent.

A gulab jamun contains sugar syrup up to about 30 $\%$ of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.

${\textbf{Step 1: Find out the volume}}{\textbf{.}}$

${\text{Given }}h = 5cm$

${\text{Length of cylinder }}h = 5 - 2.8 = 2.2$

${\text{Volume of gulab jamun = volume of cylinder + volume of 2 hemisphere}}$

$= 3\pi {r^2}h + \dfrac{4}{3}\pi {r^3}$

$= \dfrac{{22}}{7} \times 1.4 \times 1.4 \times \dfrac{{\left( {3 \times 2.2 + 4 \times 1.4} \right)}}{3}$

$= \dfrac{{6.16\left( {12.2} \right)}}{3}$

$= \dfrac{{75.152}}{3}$

$= 25.05cm$

${\text{There are 45 gulab jamuns}}{\text{.}}$

${\textbf{Step 2: Find the volume of 45 gulab jamuns}}{\textbf{.}}$

${\text{Volume = 45}} \times {\text{25}}{\text{.05}}$

$= 1127.2830{\text{ percentage of syrup}}$

$= 1127.28 \times \dfrac{{30}}{{100}}$

$= 338.184cm$

${\textbf{Hence, there are }}\mathbf{338.184}{\textbf{ cm of syrup be found in }}\mathbf{45}{\textbf{ gulab jamuns}}{\textbf{.}}$

A toy is in the form of a cone of radius $3.5$ cm mounted on a hemisphere of same radius. The total height of the toy is $15.5$ cm. Find the total surface area of the toy rounded off to the nearest integer.

As per the question we can draw the diagram.

Given:
The radius of the conical part and the hemispherical part is same (i.e.,

$3.5\text{ cm}$ )
Height of the hemispherical part,

$r= 3.5 = \dfrac{7}{2}$
Height of conical part,

$h = 15.5 - 3.5 = 12\text{ cm}$

Slant height (

$l$ ) of conical part

$l=\sqrt{r^2 + h^2}$

$\sqrt{\left( \dfrac{7}{2}\right)^2 + 12^2}= \sqrt{\dfrac{49}{4} + 144}=\sqrt{\dfrac{49+576}{4}}$

$=\sqrt{\dfrac{625}{4}}=\dfrac{25}{2}$

Total surface area of toy = CSA of conical part + CSA of hemispherical part

$=\pi rl +2\pi r^2$

$=\dfrac{22}{7}\times\dfrac{7}{2}\times \dfrac{25}{2} + 2 \times \dfrac{22}{7}\times \dfrac{7}{2}\times \dfrac{7}{2}$

$=214.7$

Water in canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Consider an area of cross-section of canal as ABCD

Area of cross-section

$= 6 \times 1.5 = 9$ m

$^2$

Speed of water

$= 10$ km/h =

$\dfrac{10000}{60}$ meter/minute.

Volume of water that flows in

$1$ minute from canal =

$9\times \dfrac{10000}{60}= 1500 \space\ m^2$

Volume of water that flows in

$30$ minutes from canal =

$30\times 1500$ =

$45000$

$m^3$

From figure 2: Assume the irrigated area be A. Volume of water irrigating the required area will be equal to the volume of water that flowed in

$30$ minutes from the canal.

Volume of water flowing in 30 minutes from canal = Volume of water irrigating the required area

$\Rightarrow$

$45000 =\dfrac{A\times 8}{100}$

$\Rightarrow$

$A = 562500$

$m^2$

So, area irrigated in

$30$ minutes is

$562500$

$m^2$ .

Derive the formula for the volume of the frustum of a cone.

Let

$ABC$ be a cone. A frustum

$DECB$ is cut by a plane parallel to its base. Let

$r_1$ and

$r_2$ be the radii of the ends of the frustum of the cone and h be the height of the frustum of the cone.

$\angle ABG$ and

$\angle ADF$ ,

$DF\parallel BG$
Therefore,

$\angle ABG \sim \angle ADF$

$\dfrac{DF}{BG}=\dfrac{AF}{AG}=\dfrac{AD}{AB}$

$\Rightarrow \dfrac{r_2}{r_1}=\dfrac{h_1 - h}{h_1}=\dfrac{l_1 - l}{1_1}$

$\Rightarrow \dfrac{r_2}{r_1}= 1-\dfrac{h}{h_1}=1-\dfrac{l}{l_1}$

$\Rightarrow 1-\dfrac{h}{h_1}=\dfrac{r_2}{r_1}$

$\Rightarrow \dfrac{h}{h_1}=1-\dfrac{r_2}{r_1}=\dfrac{r_1 - r_2}{r_1}$

$\Rightarrow \dfrac{h_1}{h}=\dfrac{r_1}{r_1 - r_2}$

$\Rightarrow h_1 = \dfrac{r_1 h}{r_1 - r_2}$

Volume of frustum of cone = Volume of cone ABC- Volume of cone ADE

$=\dfrac{1}{3}\pi r_{1}^2 h_{1}-\dfrac{1}{3}\pi r_{2}^2 (h_{1}-h)$

$=\dfrac{\pi}{3}[r_{1}^2 h_1 - r_{2}^2(h_1 - h)]$

$=\dfrac{\pi}{3}\left[r_{1}^2\left (\dfrac{r_1 h}{r_1 - r_2}\right) - r_{2}^2\left(\dfrac{r_1 h}{r_1 - r_2}- h\right)\right]$

$=\dfrac{pi}{3}\left[ \left(\dfrac{r_{1}^3 h}{r_1 - r_2}\right) - r_{2}^2 \left(hr_{1}-hr_{1}+hr_{2} \right)\right]$

$=\dfrac{\pi}{3}\left[\dfrac{r_{1}^3 h}{r_1 - r_2} - \dfrac{r_{2}^3 h}{r_1 - r_2}\right]$

$=\dfrac{\pi}{3}h \left[\dfrac{r_{1}^3-r_{2}^3}{r_1 - r_2}\right]$

$=\dfrac{\pi}{3}h\left[\dfrac{(r_1 -r_2)(r_{1}^2)+r_{2}^2+r_{1}r_{2}}{r_{1}-r_{2}}\right]$

$=\dfrac{1}{3}\pi h[r_{1}^2+r_{2}^2+r_{1} r_{2}]$

A solid is in the form of a hemisphere surmounted by a right circular cone. If height of the cone is $12\text{ cm}$ and diameter of the base is $10\text{ cm}$ . Calculate the surface area of the toy.

Given, Radius,

$r$ of cone

$= 5\text{ cm}$
Height,

$h$ of cone

$= 12\text{ cm}$

slant height,

$l = \sqrt{h^2+r^2}$

$l = \sqrt{12^2+5^2}$

$l=\sqrt{144+25}$

$l=\sqrt{169}$

$l = 13\text{ cm}$

Surface area of the toy = Curve surface area of cone + curved surface area of hemisphere

$=\pi r l + 2\pi r^2$

$=\pi r(l+2r)$

$=3.14 \times 5(13+2\times 5)$

$= 361.1 \text{ cm}^{2}$

A right-angled triangle whose sides are $3\ cm$ and $4\ cm$ (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose a value of $\pi$ as found appropriate.)

The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse is shown in the figure.

Hypotenuse, AC =

$\sqrt{3^2+4^2}$

$=\sqrt{9+16}=\sqrt{25}=5$ cm

Area of

$\Delta ABC = \dfrac{1}{2}\times AB \times AC$

$\Rightarrow \dfrac{1}{2}\times AC\times DB = \dfrac{1}{2}\times 4\times 3$

$\Rightarrow\dfrac{1}{2}\times 5\times DB = 6$

So, DB =

$\dfrac{12}{5}=2.4$ cm

The volume of double cone = Volume of cone 1 + Volume of cone 2

$=\dfrac{1}{3}\pi r^2 h_1 +\dfrac{1}{3}\pi r^2 h_2$

$=\frac{1}{3}\pi r^2[h_1 + h_2]=\dfrac{1}{3}\pi r^2[DA+DC]$

$=\dfrac{1}{3}\times 3.14\times 2.4^2\times 5$

$= 30.14\space\ cm^3$

The surface area of double cone = Surface area of cone 1 + Surface area of cone 2

$= \pi r l_1 + \pi r l_2$

$=\pi r[4+3]=3.14\times2.4\times 7$

$= 52.75$ cm

$^2$

An oil funnel made of tin sheet consists of a $10\ cm$ long cylindrical portion attached to a frustum of a cone. If the total height is $22\ cm$ , diameter of the cylindrical portion is $8\ cm$ and the diameter of the top of the funnel is $18\ cm$ , find the area of the tin sheet required to make the funnel .

Given:
Radius

$(R)$ of upper circular end of frustum part =

$\dfrac{18}{2}=9$ cm
Radius

$(r)$ of lower circular end of frustum part = Radius of circular end of cylindrical part =

$\dfrac{8}{2}=4$ cm

Height

$(h_1)$ of frustum part

$= 22 - 10 = 12$ cm
Height

$(h_2)$ of cylindrical part

$= 10$ cm

Slant height (l) of frustum part =

$\sqrt{(R-r)^2+h_{1}^2}=\sqrt{(9-4)^2+ 12^2}=\sqrt{25+144}$

$=\sqrt{169}=13$ cm

Area of tin sheet required = CSA of frustum part + CSA of cylindrical part

$=\pi(R+r)l+2\pi r h_2$

$=\dfrac{22}{7}\times (9+4)\times 13+2\times \dfrac{22}{7}\times 4\times 10$

$=\dfrac{22}{7}[169 +80]=\dfrac{22\times249}{7}$

$=782\dfrac{4}{7}\space\$ cm

$^2$

From a solid cylinder of height $7$ cm and base diameter $12$ cm, a conical cavity of same height and same base diameter is hollowed out. Find the total surface area of the remaining solid. $\left [ Use \ \pi = \dfrac{22}{7} \right ].$

Given:

Height

$(h)$ of cylindrical part

$=$ height

$(h)$ of the conical part

$= 7$ cm

Diameter of the cylindrical part

$= 12$ cm

Therefore, Radius

$(r)$ of the cylindrical part

$=\dfrac{12}{2}=6$ cm

So, Radius of the conical part

$= 6$ cm

Slant height

$(l)$ of conical part =

$\sqrt{r^2+h^2}$ cm

$=\sqrt{6^2+7^2}$

$=\sqrt{36+49}$

$=\sqrt{85}$

$\approx 9.22$ cm.

The total surface area of the remaining solid

$=$ CSA of the cylindrical part

$+$ CSA of the conical part

$+$ Base area of the circular part

$=2\pi r h + \pi r l+\pi r^2$

$=2\times \dfrac{22}{7}\times 6\times 7+\dfrac{22}{7}\times 6\times 9.22+\dfrac{22}{7}\times 6\times 6$

$=264+173.86+113.14$

$=551$

$cm^2$

A bucket is in the form of a frustum of a cone and it can hold $28.49\ liters$ of water. If the radii of its circular ends are $28\ cm$ and $21\ cm$ , find the height of the bucket. $\left [ Use \ \pi = \dfrac{22}{7} \right ].$

Let the height of the bucket be

$h$ cm.

Suppose

$r_1$ and

$r_2$ be the radii of the circular ends of the bucket.

Given,

$r_1 = 28$ cm and

$r_2 = 21$ cm

Capacity of bucket

$= 28.49$ litres

Therefore, Volume of the bucket

$= 28.49 \times 1000$ [1 litre = 1000

$cm^3$ ]

$\Rightarrow \dfrac{1} {3}\pi h(r_1^2+r_2^2+r_1r_2)=28.49 \times 1000$

$\Rightarrow \dfrac{1}{3}\times \dfrac{22}{7}\times h\times[28^2+21^2+28\times 21]=28490$

$\Rightarrow \dfrac{22}{21}\times h\times 1813=28490$

$\Rightarrow h=\dfrac{28490\times 21}{22\times 1813}$

$\Rightarrow h=15$ cm

Therefore, the height of the bucket is

$15$ cm.

Derive the formula for the curved surface area and total surface area of the frustum of a cone.

Let

$ABC$ be a cone. A frustum

$DECB$ is cut by a plane parallel to its base.

Let

$r_1$ and

$r_2$ be the radii of the ends of the frustum of the cone and

$h$ be the height of the frustum of the cone.

$\angle ABG$ and

$\angle ADF$ ,

$DF\parallel BG$

Therefore,

$\angle ABG \sim \angle ADF$

$\dfrac{DF}{BG}=\dfrac{AF}{AG}=\dfrac{AD}{AB}$

$\dfrac{r_2}{r_1}=\dfrac{h_1 - h}{h_1}=\dfrac{l_1 - l}{1_1}$

$\dfrac{r_2}{r_1}= 1-\dfrac{h}{h_1}=1-\dfrac{l}{l_1}$

$1-\dfrac{l}{l_1}=\dfrac{r_2}{r_1}$

$1-\dfrac{l}{l_1}=1-\dfrac{r_2}{r_1}=\dfrac{r_1 - r_2}{r_1}$

$\dfrac{l_1}{l}=\dfrac{r_1}{r_1 - r_2}$

$l_1 = \dfrac{r_1 l}{r_1 - r_2}$

CSA of frustum

$DECB =$ CSA of cone

$ABC -$ CSA cone

$ADE$

$=\pi r_{1}l_{1}-\pi r_{2}(l_{1}-l)$

$=\pi r_{1}\left(\dfrac{r_1}{r_1 - r_2} \right)-\pi r_{2}\left(\dfrac{r_1}{r_1 - r_2} - l\right)$

$\dfrac{\pi r_{1}^2l}{r_{1}-r_{2}}-\pi r_{2}\left (\dfrac{r_1 l- r_1 l+r_2 l}{r_1 - r_2}\right)$

$=\dfrac{\pi r_1^2 l}{r_1 - r_2}-\dfrac{\pi r_2^2 l}{r_1 - r_2}$

$=\pi l[\dfrac{r_{1}^2 - r_{2}^2}{r_1 - r_2}]$

CSA of frustum =

$\pi (r_1 + r_2)l$ .

Total surface area of frustum = CSA of frustum + Area of upper circular end + Area of lower circular end.

$=\pi (r_1 + r_2)l+\pi r_{2}^2+\pi r_{1}^2$

$=\pi[(r_1 + r_2)l+r_{1}^2+r_{2}^2]$

A bucket opens at the top, and made up of a metal sheet is in the form of a frustum of a cone. The depth of the bucket is $24\ cm$ and the diameters of its upper and lower circular ends are $30\ cm$ and $10\ cm$ respectively. Find the cost of metal sheet used in it at the rate of $Rs\ 10$ per $100\ cm^2$ . $\displaystyle \left[ Use\ \pi = 3.14 \right]$ .Insert answer in nearest integer.

Diameter of upper end of bucket =

$30 cm$

Diameter of lower end of bucket =

$10 cm$

Then, radius of upper end of bucket

$r_{1}=\cfrac{30}{2}=15$ cm

And, radius of lower end of bucket

$r_{2}=\cfrac{10}{2}=5$ cm

Height of bucket =

$24\ cm$

Slant height of bucket =

$\sqrt{(r_{2}-r_{1})^{2}+h^{2}}$

$=\sqrt{(15-5)^{2}+(24)^{2}}=\sqrt{(10)^{2}+(24)^{2}}$

$=\sqrt{100+256}=\sqrt{676}=26\ cm$

Area of metal sheet used to make bucket

$=\pi (r_{1}+r_{2})l+\pi( r_{2})^{2}=\pi (15+5)26+\pi (5)^{2}$

$=520\pi +25\pi =545\pi\ cm^{2}$

Given cost of

$100$ sq cm metal sheet =Rs.

$10$

Cost of

$545\pi\ cm^{2}$ metal sheet

$=\dfrac{545\times 3.14\times 10}{100}=171.13\ Rs.$

Cost of metal sheet used for making bucket

$= \ Rs. 171.13$

A cistern, internally measuring $150$ cm x $120$ cm x $110$ cm, has $129600$ cm $^3$ of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being $22.5$ cm x $7.5$ cm x $6.5$ cm

Volume of cistern

$= 150 \times 120 \times 110$

$= 1980000$ cm

$^3$

Volume to be filled in cistern

$= \dfrac{1980000}{ 129600}$

$= 1850400$ cm

$^3$

Let

$x$ numbers of porous bricks were placed in the cistern.
Volume of

$x$ bricks =

$x \times 22.5 \times 7.5 \times 6.5$

$= 1096.875x$

As each brick absorbs one-seventeenth of its volume, therefore, volume absorbed by these bricks

$=\dfrac{x}{17}(1096.875)$

$\Rightarrow 1850400+\dfrac{x}{17}(1096.875)=(1096.875)x$

$\Rightarrow 1850400=\dfrac{16x}{17}(1096.875)$

$x = 1792.41$

Therefore,

$1792$ bricks were placed in the cistern.

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to $7$ cm and the height of the cone is equal to its diameter. Find the volume of the solid. $\left [ Use \ \pi = \dfrac{22}{7} \right ].$

Given: radius

$= 7$ cm, height

$= 14$ cm

Volume of hemisphere =

$\dfrac{2}{3}\pi r^3$

$=\dfrac{2}{3}\times \dfrac{22}{7}\times 7^3$

$=718.66$

$cm^2$

Volume of cone

$=$

$\dfrac{1}{3}\pi r^2h$

$=\dfrac{1}{3}\times\dfrac{22}{7}\times 7^2\times 14$

$=718.66$

$cm^3$

Total volume

$=$ Volume of hemisphere

$+$ Volume of cone

$=718.66 + 718.66=1437.32$

$cm^3$

A funnel of tin sheet consists of a cylindrical portion, $10\ cm$ long, attached to a frustum of a cone. If a diameter of the top and bottom of the frustum is $18\ cm$ and $8\ cm$ respectively and the slant height of the frustum of cone is $13\ cm$ , find the surface area of the tin required to make the funnel.

We know

$\dfrac { EC }{ EB } =\dfrac { OD }{ KA }$ (Similar triangle)

$\dfrac { x }{ x+13 } =\dfrac { 4 }{ 9 }$

$\Rightarrow 9x=4x+52$

$\Rightarrow 5x=52$

$\Rightarrow x=10.4$

Surface area of cone

$=\pi r\left( r+l \right)$

where

$r$ is radius and

$l$ is slant height

So, surface area of conical part

$=$ Surface area of bigger cone

$-$ Surface area of small imaginary cone

$=\pi \times 9\times \left( 9+(13+10.4) \right) -\pi \times 9\times (4+10.4)$

$=\pi \times 234$

Surface area of cylindrical part

$=2\pi rh$

$=2\times \pi \times 4\times 10$

$=80\pi$

So, total surface area

$=234\pi +80\pi$

$=314\pi$ sq. units

From the top a conical shaped jaggery of base diameter $10$ cm and height $10$ cm, a small cone is cut off by slicing parallel to the base, $4$ cm from the vertex. Find the surface area and volume of the frustum so obtained.

Diameter of the cone ABO

$=10$ cm

$\therefore$ radius

$r_1=\displaystyle\frac{10}{2}=5$ cm
Height of the cone ABO

$=h_1=10$ cm
Height of the cone

$A^|B^|O=h_2=4$ cm
First, let us find the other required data.
Radius of cone

$A^|B^|O=r_2$
Let the Slant height of cones

$ABO$ and

$A^|B^|O,$ be

$\ l_1$ and

$l_2$ respectively.

$\therefore \displaystyle\frac{r_1}{r_2}=\frac{h_1}{h_2}, \frac{5}{r_2}=\frac{10}{4}$

$\therefore r_2=\displaystyle\frac{5\times 4}{10}=2$ cm

$l^2_1=h^2_1+r^2_1=10^2+5^2=100+25=125$

$\therefore l_1=\sqrt{125}=5\sqrt{5}$ cm

$l^2_2=h^2_2+r^2_2=4^2+2^2=16+4=20$

$\therefore l_2=\sqrt{20}=2\sqrt{5}$ cm

(i) Lateral surface area of cone ABO

$=\pi r_1l_1=\pi \times 5\times 5\sqrt{5}=25\sqrt{5}\pi$

$\therefore$ LSA of cone ABO

$=25\sqrt{5}\pi cm^2$

Lateral surface area of cone

$A^|B^|O=\pi r_2l_2=\pi\times 2\times 2\sqrt 5=4\sqrt 5\pi$

$\therefore$ LSA of cone

$A^|B^|O=4\sqrt 5\pi cm^2$

$\therefore$ Lateral surface area of frustum

$=$ LSA of cone ABO

$-$ LSA of come

$A^|B^|O$

$=25\sqrt{5}\pi -4\sqrt{5}\pi =21\sqrt{5}\pi=21\times \sqrt{5}\times \displaystyle\frac{22}{7}=66\sqrt{5}cm^2$

$\therefore$ LSA of frustum

$=66\sqrt{5}$ sq. cm. or

$147.84cm^2$

(ii) Total surface area of cone ABO

$=\pi r_1(r_1+l_1)$

$=\pi\times 5\times (5+5\sqrt{5})=\pi\times 5\times 5(1+\sqrt{5})$

$\therefore$ TSA of cone ABO

$=25(1+\sqrt{5})\pi \ cm^2$

Lateral surface area of cone

$A^|B^|O=\pi r_2l_2=\pi \times 2\times 2\sqrt{5}=4\sqrt{5}\pi$

$\therefore$ LSA of cone

$A^|B^|O=4\sqrt{5}\pi cm^2$

The total surface area of frustum

$=$ TSA of cone ABO

$-$ LSA of cone

$A^|B^|O+\pi r^2_2$

$=25(1+\sqrt{5})\pi - 4\sqrt{5}\pi +4\pi$

$=\pi (25+25\sqrt{5}-4\sqrt{5}+4)$

$=\pi(29+21\sqrt{5})$

$=\displaystyle\frac{22}{7}\times 75.95$

$\therefore$ TSA of frustum

$=238.70cm^2$

(iii)Volume of cone ABO

$=\displaystyle\frac{1}{3}\pi r^2_1h_1=\frac{1}{3}\times \pi\times 5\times 5\times 10$

$\therefore$ Volume of cone ABO

$=\displaystyle\frac{250}{3}\pi cm^3$

Volume of cone

$A^|B^|O=\displaystyle\frac{1}{3}\pi r^2_2h_2=\frac{1}{3}\times \pi\times 2\times 2\times 4$

$\therefore$ Volume of cone

$A^|B^|O=\displaystyle\frac{16}{3}\pi cm^3$

$\therefore$ Volume of the frustum

$=$ Volume of cone ABO

$-$ Volume of cone

$A^|B^|O$

$=\displaystyle\frac{250}{3}\pi -\frac{16}{3}\pi =\frac{234}{3}\pi$

$\therefore$ Volume of the frustum

$=78\pi cm^3$ or

$245.14cm^3$

We can also directly find the lateral surface area, total surface area, and volume of the frustum by using the formulae.

A circus tent is made if canvas and is in the form of a right circular cylinder and a right circular cone above it. The diameter and height of the cylindrical part of the tent are $126$ m and $5$ m respectively. The total height of the tent is $21$ m. Find the total cost of the canvas used to make the tent when the cost per $m^2$ of the canvas is Rs. $15$ .

Cylindrical part:

$r=\displaystyle\frac{126}{2}=63m, h=21-5=16m$

$l=\sqrt{r^2+h^2}=\sqrt{63^2+16^2}$

$=\sqrt{3969+256}=\sqrt{4225}$

$\therefore l=65m$
Total canvas used

$=$ CSA of cylindrical part

$+$ CSA of conical part

$=2\pi rh+\pi rl=2\times \displaystyle\frac{22}{7}\times 63\times 5m^2+\frac{22}{7}\times 63\times 65m^2$

$=1980m^2+12870m^2=14850m^2$

$\therefore$ Total canvas used

$=14, 850m^2$
Cost of canvas at the rate of Rs.

$16$ per

$m^2=14, 850\times Rs.15=Rs. 2,22,750$

$\therefore$ Cost of canvas

$=Rs. 2,22,750$ .

A solid is in the form of a cone mounted on a right circular cylinder, both having same radii of their bases. Base of the cone is placed on the top base of the cylinder. If the radius of the base and height of the cone be $7$ cm and $10$ cm respectively, and the total height of the solid be $30$ cm, find the volume of the solid.

Cylindrical part:

$r=7cm, h=30-10=20cm$
Conical part:

$r=7cm, h=10cm$
Volume of the solid

$=$ Volume of cylinder

$+$ Volume of cone

$=\displaystyle\pi r^2h+\frac{1}{3}\pi r^2h$

$=\displaystyle\frac{22}{7}\times 7\times 7\times 20cm^2+\frac{1}{3}\times \frac{22}{7}\times 7\times 7\times 10$

$=3080cm^3+\displaystyle\frac{1540}{3}cm^3$

$=3080+513.33cm^3$
Volume of the solid

$=3593.33cm^3$ .

A metallic right circular cone $20\ cm$ hight and whose vertical angle is $60^{\circ}$ is cut into two puts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter $\dfrac {1}{16}\ cm$ , find the length of the wire.

1159078_879070_ans_18abc0594f5f41e08fb3bc5845dd0493.jpg

A bucket made up of a metal sheet is in the form of a frustum of a cone of height $16cm$ with diameters of its lower and upper ends as $16cm$ and $40cm$ respectively. Find the volume of the bucket. Also, find the cost of the bucket if the cost of metal sheet used is Rs. $20$ per $100{cm}^{2}$ . (Use $\pi=3.14$ )

Given,

Hight of frustum of cone

$h=16\ cm$

Diameter of lower circular end

$d=16\ cm$

Diameter of upper circular end

$D=40\ cm$

Hence,

Radius of lower circular end

$r=\dfrac{d}{2}=8\ cm$

Radius of upper circular end

$R=\dfrac{D}{2}=20\ cm$

Now,

Volume of frustum of cone

$V=\dfrac{\pi}{3}h(R^{2}+r^{2}+R\times r)$

$=\dfrac{\pi}{3}\times 16\times(20^{2}+8^2+20\times 8)\ cm^3$

$=\dfrac{\pi}{3}\times 16\times(400+64+160)\ cm^3$

$=\dfrac{3.14}{3}\times 16\times624\ cm^3$

$=\dfrac{31349.76}{3}\ cm^3$

Volume of frustum of cone

$V=10449.92\ cm^3$

$\text{Surface area of bucket (S) = curved surface area of a frustum of cone + area of lower circular end}$

$S= \pi (R+r)\sqrt{(R-r)^{2}+h^{2}}+\pi r^{2}$

$= \bigg((R+r)\sqrt{(R-r)^{2}+h^{2}}+r^{2}\bigg)\pi$

$= \bigg((20+8)\sqrt{(20-8)^{2}+16^{2}}+8^{2}\bigg)\pi \ cm^{2}$

$= \bigg((28)\sqrt{(12)^{2}+256}+64\bigg)\pi\ cm^{2}$

$= \bigg((28)\sqrt{(144+256)}+64\bigg)\pi\ cm^{2}$

$= \bigg((28)\sqrt{(400}+64\bigg)\pi\ cm^{2}$

$= \bigg(28\times 20+64\bigg)\pi\ cm^{2}$

$= \bigg(560+64\bigg)\times 3.14\ cm^{2}$

$= 624\times 3.14\ cm^{2}$

$= 1959.36\ cm^{2}$

$\text{Total surface area of meatal sheet used}=1959.36 \ cm^{2}$

Cost of

$100\ cm^2$ metal sheet

$=20\ Rs.$

Cost of

$1\ cm^2$ metal sheet

$=\dfrac{20}{100}\ Rs=0.2\ Rs..$

Cost of

$1959.36\ cm^2$ metal sheet

$=0.2\times 1959.36\ Rs.=391.872\ Rs.$

Hence, cost of metal sheet used to make bucket

$391.872 \ Rs.$

A frustrum of a right circular cone has a diameter of base $20\text cm$ , of top $12cm$ , and height $3cm$ . Find the area of its whole surface and volume.

${ D }_{ 1 }=20cm,\quad { R }_{ 1 }=10cm$

${ D }_{ 2 }=12cm,\quad { R }_{ 2 }=6cm$

$h=3cm$

Volume of frustum

$=\dfrac { 1 }{ 3 } \pi h\left( { R }_{ 1 }^{ 2 }+{ R }_{ 2 }^{ 2 }+{ R }_{ 1 }{ R }_{ 2 } \right) =\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times 3\left[ { 10 }^{ 2 }+{ 6 }^{ 2 }+10\times 6 \right]$

$=\dfrac { 22 }{ 7 } \times 196=616{ cm }^{ 3 }$

Let slant height, be

$l$ . Then,

$l=\sqrt { { \left( { R }_{ 1 }-{ R }_{ 2 } \right) }^{ 2 }+{ h }^{ 2 } } =\sqrt { { \left( 10-6 \right) }^{ 2 }+{ 3 }^{ 2 } } =\sqrt { 16+9 } =5cm$

TSA of frustum of cone

$=\pi \left( { R }_{ 1 }+{ R }_{ 2 } \right) l+\pi \left( { R }_{ 1 }^{ 2 }+{ R }_{ 2 }^{ 2 } \right)$

$=\dfrac { 22 }{ 7 } \left[ 10+6 \right] \times 5+\dfrac { 22 }{ 7 } \left( { 10 }^{ 2 }+{ 6 }^{ 2 } \right)$

$=\dfrac { 22 }{ 7 } \left[ 80+1360 \right] =\dfrac { 22 }{ 7 } \times 216=678.85{ cm }^{ 2 }$

A vessel is a hollow cylinder fitted with a hemispherical bottom of the same base. The depth of the cylinder is $\cfrac{14}{3}m$ and the diameter of hemisphere is $3.5m$ . Calculate the volume and the internal surface of the solid.

$Volume=\dfrac{2}{3}\pi r^3+\pi r^2h$

$V=\dfrac{2}{3}\pi\times 1.75^3+\pi \times 1.75^2\times \dfrac{14}{3}$

$V=56.123m^3$

$Surface\ area=2\pi r^2+2\pi r h$

$A=2\pi\times1.75^2+2\pi\times1.75\times \dfrac{14}{3}$

$A=70.55m^2$

The perimeters of the ends of a frustum of a right circular cone are $44\ cm$ and $33\ cm.$ If the height of the frustum is $16\ cm,$ find its volume, the slant surface, and the total surface.

Let

$R_1$ and

$R_2$ be the radius of the ends of the frustum of the cone.

So,

$2\pi { R }_{ 1 }=44cm$

$\Rightarrow 2\times\dfrac{22}{7}\times{ R }_{ 1 }=44$

$\Rightarrow R_1=\dfrac { 44\times 7 }{ 44 }$

$\Rightarrow R_1 =7\ cm$

$2\pi { R }_{ 2 }=33\ cm$

$\Rightarrow 2\times\dfrac{22}{7}\times{ R }_{ 2 }=33$

$\Rightarrow R_2=\dfrac { 33\times 7 }{ 44 }$

$\Rightarrow R_2 =5.25\ cm$

Now,

$\begin{aligned}{}{\rm{Volume}} &= \frac{1}{3}\pi h\left[ {R_1^2 + R_2^2 + {R_1}{R_2}} \right]\\& = \frac{1}{3} \times \frac{{22}}{7} \times 16\left[ {{7^2} + {{5.25}^2} + 7 \times 5.25} \right]\\ &= 1899.33c{m^3}\end{aligned}$

Slant height of the frustum of the cone will be,

$\begin{aligned}{}l &= \sqrt {{{\left( {{R_1} - {R_2}} \right)}^2} + {h^2}} \\ &= \sqrt {{{\left( {7 - 5.25} \right)}^2} + {{16}^2}} \\ &= \sqrt {259.0625} \\& = 16.10\ cm\end{aligned}$

$\begin{aligned}{}CSA &= \pi \left( {{R_1} + {R_2}} \right)l\\ &= \frac{{22}}{7}\left( {7 + 5.25} \right) \times 16.10\\& = 619.85{\rm{c}}{m^2}\end{aligned}$

$\begin{aligned}{}TSA &= CSA + \pi \left( {R_1^2 + R_2^2} \right)\\ &= 619.85 + \frac{{22}}{7}\left( {{7^2} + {{5.25}^2}} \right)\\& = 860.475\;c{m^2}\end{aligned}$

A reservoir in the form of the frustum of a right circular cone contains $44\times {10}^{7}$ litres of water which fills it completely. The radii of the bottom and top of the reservoir are $50$ metres and $100$ metres respectively. Find the depths of water and the lateral surface area of the reservoir. (Take $\pi=22/7$ )

Volume of frustum cone

$=44\times { 10 }^{ 7 }litre=44\times { 10 }^{ 4}{ m }^{ 3 }$

$\Rightarrow \dfrac { 1 }{ 3 } \pi h\left( { R }_{ 1 }^{ 2 }+{ R }_{ 2 }^{ 2 }+{ R }_{ 1 }{ R }_{ 2 } \right) =44\times { 10 }^{ 4 }\Rightarrow \dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times h=\dfrac { 44\times { 10 }^{ 4 }\times 3\times 7 }{ 22\times 17500 } =24m$

$\Rightarrow \dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times h\left( 10000+2500+5000 \right) =44\times { 10 }^{ 4 }\Rightarrow h=\dfrac { 44\times { 10 }^{ 4 }\times 3\times 7 }{ 22\times 17500 } =24m$

slant height

$\left( l \right) =\sqrt { { \left( { R }_{ 1 }-{ R }_{ 2 } \right) }^{ 2 }+{ h }^{ 2 } } =\sqrt { { \left( 100-50 \right) }^{ 2 }+{ \left( 24 \right) }^{ 2 } } =\sqrt { 2500+576 } =55.46m$

$CSA=\pi \left( { R }_{ 1 }+{ R }_{ 2 } \right) l=\dfrac { 22 }{ 7 } \left( 100+50 \right) \times 55.46=26145.43{ m }^{ 2 }$

A cylindrical vessel of diameter $14\text{ cm}$ and height $42 \text{ cm}$ is fixed symmetrically inside a similar vessel of diameter $16\text{ cm}$ and height $42\text{ cm}$ . The total space between the two vessels is filled with cork dust for heat insulation purposes. How many cubic centimeters of cork dust will be required?

The volume of cork dust

$=$ Volume of outer cylinder

$-$ volume of the inner cylinder

$\implies V=\pi (r_2^2-r_1^2)h$

$\implies V=\pi(8^2-7^2)42=1980\text{ cm}^3$

The volume of cork dust will be

$1980\text{ cm}^3$

The height of a cone is $20\ cm$ . A small cone is cut off from the top by a plane parallel to the base. If its volume be $\dfrac{1}{125}$ of the volume of the original cone, determine at what height above the base the section is made.

$\triangle ABC\sim \triangle ADE$ [ By AA similarity]

$\therefore \dfrac { AB }{ AD } =\dfrac { BC }{ DE }$

$\Rightarrow \dfrac { h }{ 20 } =\dfrac { { r }_{ 1 } }{ { r }_{ 2 } }$

${ V }_{ ABC }=\dfrac { 1 }{ 125 } { V }_{ ADE }$

$\Rightarrow \dfrac { 1 }{ 3 } \pi { r }_{ 1 }^{ 2 }h=\dfrac { 1 }{ 3 } \pi { r }_{ 2 }^{ 2 }\times 20\times \dfrac { 1 }{ 125 }$

$\Rightarrow { \left( \dfrac { { r }_{ 1 } }{ { r }_{ 2 } } \right) }^{ 2 }=\dfrac { 20 }{ 125h }$

$\Rightarrow { \left( \dfrac { h }{ 20 } \right) }^{ 2 }=\dfrac { 20 }{ 125h }$

$\Rightarrow { h }^{ 3 }=\dfrac { 20\times 20\times 20 }{ 125 } =64$

$\Rightarrow \quad h=\sqrt [ 3 ]{ 64 } =4cm$

Therefore section is made

$(20-4=16cm)$ above the base.

A metallic right circular cone $20\text{ cm}$ high and whose vertical angle is ${90}^{\circ}$ is cut into two parts at the middle point of its axis by a plane parallel to the base. If the frustum so obtained be drawn into a wire of diameter $\dfrac {1}{16}\text{ cm}$ , find the length of the wire.

Volume of cylindrical wire

$=$ Volume of frustum

Volume of frustum ABDC

$V=\dfrac { \pi h }{ 3 } \left[ { R }_{ 1 }^{ 2 }+{ R }_{ 2 }^{ 2 }+{ R }_{ 1 }{ R }_{ 2 } \right]$

$h=20cm$ , cone is cut at the middle

$\therefore \quad OQ=QP=10cm$

$\angle QOB={ 45 }^{ 0 }$ , In

$\triangle QOB,\quad { R }_{ 1 }=PD\quad \& \quad { R }_{ 2 }=QB$

$\tan{ 45 }^{ 0 }=\dfrac { QB }{ OQ } \Rightarrow OQ=QB\Rightarrow { R }_{ 2 }=10cm$

$\triangle OPD,\quad tan{ 45 }^{ 0 }=\dfrac { OP }{ PD } \Rightarrow PD=OP\Rightarrow { R }_{ 1 }=20cm$

$\therefore \quad V=\dfrac { \pi \times 10 }{ 3 } \left[ { \left( 20 \right) }^{ 2 }+{ \left( 10 \right) }^{ 2 }+20\times 10 \right] =\dfrac { 10\pi }{ 3 } \left[ 400+100+200 \right]$

$V=\dfrac { 7000\pi }{ 3 }$

Volume of cylindrical wire

$=\pi { r }^{ 2 }h=\pi { \left( \dfrac { 1 }{ 32 } \right) }^{ 2 }\times h$

$\Rightarrow \pi \times { \left( \dfrac { 1 }{ 32 } \right) }^{ 2 }\times h=\dfrac { 7000\pi }{ 3 } \Rightarrow h=\dfrac { 7000\times 32\times 32 }{ 3 } =2389333.33cm$

$\Rightarrow h=23893.33m$

$\therefore$ Length of wire

$=23893.33m$

A milk container is made of metal sheet in the shape of frustum of a cone whose volume is $10459\cfrac{3}{7}\ {cm}^{3}$ . The radii of its lower and upper circular ends are 8 cm and 20 cm respectively. Find the cost of metal sheet used in making the container at the rate of Rs. 1.40 per ${cm}^{3}$ . (Use $\pi=\frac{22}{7}$ )

The above figure can be constructed using the given information.

Given that, the volume of the bucket

$V=10459\dfrac{3}{7} \ cm^{3}$

$\Rightarrow V=\dfrac{73216}{7}$

The radii of bottom circular end,

$R=20 \ cm$

The radii of top end,

$r=8 \ cm$

We know that, the volume of frustum of cone

$V=\dfrac{\pi}{3}h(R^{2}+r^{2}+R\times r)$

Hence, height of frustum of cone will be:

$h=\dfrac{3V}{\pi(R^{2}+r^{2}+R\times r)}\ cm$

$\therefore \ h=\dfrac{3\times\dfrac{73216}{7}}{\dfrac{22}{7}\times (20^{2}+8^{2}+20\times 8)}\ cm$

$\Rightarrow h=\dfrac{3\times 73216}{22\times (400+64+160)} \ cm$

$\Rightarrow h=\dfrac{219648}{22\times 624}\ cm$

$\Rightarrow h=\dfrac{219648}{13728}\ cm$

$\therefore \ h=16 \ cm$

Now, for calculating the area of metal sheet used, we have to calculate total surface area of frustum of the cone.

$\text{Total surface area of = curved surface area + area of bottom circular end}$

$= \pi (R+r)\sqrt{(R-r)^{2}+h^{2}}+\pi R^{2}$

$= \bigg((R+r)\sqrt{(R-r)^{2}+h^{2}}+R^{2}\bigg)\pi$

$= \bigg((20+8)\sqrt{(20-8)^{2}+16^{2}}+20^{2}\bigg)\pi \ cm^{2}$

$= \bigg((28)\sqrt{(12)^{2}+256}+400\bigg)\pi\ cm^{2}$

$= \bigg((28)\sqrt{(144+256)}+400\bigg)\pi\ cm^{2}$

$= \bigg((28)\sqrt{(400}+400\bigg)\pi\ cm^{2}$

$= \bigg(28\times 20+400\bigg)\pi\ cm^{2}$

$= \bigg(560+400\bigg)\times \dfrac{22}{7}\ cm^{2}$

$= 960\times \dfrac{22}{7}\ cm^{2}$

$= \dfrac{21120}{7}\ cm^{2}$

$\text{Total surface area of meatal sheet used}=\dfrac{21120}{7} \ cm^{2}$

Now, cost of

$1 \ cm^2$ metal sheet

$=Rs.\ 1.40$

$\therefore \$ cost of

$\dfrac{21120}{7} \ cm^2$ metal sheet

$=Rs.\ 1.40\times \dfrac{21120}{7}=Rs.\ 4224$

Hence, the cost of making the container is

$Rs.\ 4224$ .

A solid consisting of a right circular cone of height $120cm$ and radius $60cm$ standing on a hemisphere of radius $60cm$ is placed upright in a right circular cylinder full of water such that it touches the bottoms. Find the volume of water left in the cylinder, if the radius of the cylinder is $60cm$ and its height is $180cm$ .

Water left in cylinder=Volume of cylinder-volume of hemisphere-volume of cone

Water left

$=\pi r^2 h-\dfrac{2}{3}\pi r^3-\dfrac{1}{3}\pi r^2 h$

$Water\ left=\pi \times 60^2\times 180-\dfrac{2}{3}\pi\times 60^3-\dfrac{1}{3}\pi\times 60^2\times 120$

$V_{remaining}=1130973.355cm^3$

If the volumes of two cones are in the ratio of 1:4 and their diameters are in the ratio of 4:5, then find the ratio of their heights.

1669814_1714163_ans_d1f2731ad9984eba88792e1bada95c24.jpg

A hemispherical bowl of internal diameter 30 cm contains some liquid.This liquid is to be poured into cylindrical bottles of diameter 5 cm and height 6 cm each. Find the number of bottles necessary to empty the bowl.

1677513_1714239_ans_1f746278cd754b46b9559eb8ecdc100f.png

Write whether the following statements are true or false. Justify your answer:
(i) The volume of a sphere is equal to two-third of the volume of a cylinder whose height and diameter are equal to the diameter of the sphere.
(ii) The volume of the largest right circular cone that can be fitted in a cube whose edge is $2 r$ equals the volume of a hemisphere of radius $r$ .
(iii) A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is $1:2:3$ .

1976191_1784321_ans_63c1660831e24115937e2762aeb08ffd.jpg

The given figure shows a solid trophy made of shining glass. If one cubic centimetre of glass costs $Rs \ 0.75$ , find the cost of the glass for making the trophy.

1975262_1784364_ans_8a835ab27d604dbaacc8c9befb26049b.jpg

A conical tent is $10$ m high and the radius of its base is $24$ m. Find the slant height of the tent. If the cost of $1$ $m^2$ canvas is Rs. $70$ , find the cost of the canvas required to make the tent.

We have,

$r=24$ m and

$h=10$ m

$\therefore \ slant \ height = l=\sqrt{r^2+h^2}=\sqrt{576+100}=26$ m

moreover,

$C.S.A.=\pi \times r \times l=\left(\dfrac{22}{7}\times 24\times 26\right)$

$\therefore$ Cost of canvas required

$=$ Rs.

$\left\{\left(\dfrac{22}{7}\times 24\times 26\right)\times 70\right\}=$ Rs.

$137280$ .

In a figure (i) given below , Calculate the area of the shaded region correct to two decimal places .Take $\pi =3.142$

1976492_1812551_ans_b9a8b2f8d9084c5d975447169ff40cdb.jpeg

The diagram shows a toy.
The shape of the toy is a cone, with radius $4\ cm$ and height $9\ cm$ , on top of a hemisphere with radius $4\ cm$ .
Calculate the volume of the toy.
Give your answer correct to the nearest cubic centimeter.

[The volume, V of a cone with radius $r$ and height $h$ is $V = \dfrac{1}{3} \pi r^{2} h$ ]

[The volume, V of a sphere with radius $r$ is $V = \dfrac{4}{3} \pi r^{3}$ ].

Given that, a toy is in the shape of a cone on top of a hemisphere.

To find out: The volume of the toy.

$1)$ Volume of the cone -

Radius of cone,

$r=4\ cm$

Height of the cone,

$h=9\ cm$

We know that, the volume of a cone is given by,

${ V }_{ 1 }=\dfrac { 1 }{ 3 } \pi { r }^{ 2 }h$

$\therefore \ { V }_{ 1 }=\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times { \left( 4 \right) }^{ 2 }\times 9$

$\therefore \ { V }_{ 1 }=\dfrac { 1 }{ 3 } \times \dfrac { 22 }{ 7 } \times 16\times 9$

$\therefore \ { V }_{ 1 }= \dfrac { 22 }{ 7 } \times 16\times 3$

$\therefore \ { V }_{ 1 }=\dfrac { 1056 }{ 7 }$

$\therefore \ { V }_{ 1 }=150.857\ { cm }^{ 3 }$

$2)$ Volume of the hemisphere -

Radius of hemisphere,

$r=4\ cm$

We know that, the volume of a hemisphere is given by,

${ V }_{ 2 }=\dfrac { 2 }{ 3 } \pi { r }^{ 3 }$

$\therefore \ { V }_{ 2 }=\dfrac { 2 }{ 3 } \times \dfrac { 22 }{ 7 } \times { \left( 4 \right) }^{ 3 }$

$\therefore\ { V }_{ 2 }=\dfrac { 2 }{ 3 } \times \dfrac { 22 }{ 7 } \times 64$

$\therefore \ { V }_{ 2 }=\dfrac { 2816 }{ 21 }$

$\therefore { V }_{ 2 }=143.095\ { cm }^{ 3 }$

Now, the total volume of the toy

$=$ volume of the cone

$+$ volume of the hemisphere

$\therefore \ V={ V }_{ 1 }+{ V }_{ 2 }$

$\therefore \ V=150.857+134.095$

$\therefore\ V=284.952\ { cm }^{ 3 }\approx 285\ cm^3$

Hence, the required volume of the toy is

$285\ { cm }^{ 3 }$ .

A cone of radius 10 cm is divided two parts by a plane parallel to its base at the middle of its height. Find the ratio of volumes of two parts.

2044018_1558375_ans_d4279063fc3d490fbfb2476da3866330.jpg

If a square is inscribed in a circle, what is the ratio of the areas of the circle and the square?

$\textbf{Step 1: Creating equations}$

$\text{If we closely observe the given picture we can see that }$

$\text{diameter of circle}=\text{diagonal of square}$

$\text{Area of circle}=\pi \dfrac{d^2}4$

$\text{Area of square}=\dfrac {d^2}4$

$\dfrac{\text{Area of circle}}{\text{Area of square}}=\dfrac{\pi \dfrac{d^2}{4}}{\dfrac{d^2}{4}}$

$\implies \dfrac{\text{Area of circle}}{\text{Area of square}}=\pi$

$\textbf{Hence,The ratio of the area of the circle and square is }\mathbf{\pi}$

Surface Areas And Volumes - Class 10 Maths - Extra Questions

Consider a frustum of cone with larger and smaller radius equal to r1 and r2r_1 \ and\ r_2, respectively. Then, the total surface area of frustum of a cone is πl(r1+r2)+π(r21+r22)\pi l(r_1+r_2)+\pi (r_{1}^{2}+r_{2}^{2}). if correct enter 1 else enter 0

State True(1) or False(0).The volume of the frustum of a cone is 13πh[r21r22−r1r2],\frac{1}{3}\pi h\left [ r_{1}^{2}r_{2}^{2}-r_{1}r_{2} \right ], where h is verticalheight of the frustum and r1,r2r_1, r_2 are the radii of the ends.

Consider a frustum of the cone with larger and smaller radius equal to r1 and r2r_1 \ and\ r_2, respectively. Then, the curved surface area of a frustum of a cone is πl(r1+r2)\pi l (r_1+r_2). If true enter 1 else 0.

A spherical ball of radius 3 cm3\ cm is melted and recast into three spherical balls. The radii of the two of the balls are 1.5 cm1.5\ cm and 2 cm2\ cm respectively. Find the radius of the third ball.

A fraction clutch is in the form of a frustum of a cone, the diameter of the ends being 32 cm32\ cm and 20 cm20\ cm and length 8 cm8\ cm. Find its bearing surface and volume.[Take π=3.14\pi=3.14]

The decorative block shown in Fig. 13.7 is made of two solids - a cube and a hemisphere. The base of the block is a cube with edge 5 cm. and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block. (Takeπ=227)\left(Take \pi =\dfrac { 22 }{ 7 } \right)

How many spherical bullets each of 5cm5cm in diameter can be cast from a rectangular block of metal 11dm×1m×5dm11dm\times 1m\times 5dm?

A 16 m16\text{ m} deep well with diameter 3.5 m3.5\text{ m} is dug up and the earth from it is spread evenly to form a platform 27.5 m27.5\text{ m} by 7 m7\text{ m}. Find the height of the platform.

A well of diameter 2 m2\ m is dug 14 m14\ m deep. The earth taken out of it is spread evenly all around it to form an embankment of height 40 cm40\ cm. Find the width of the embankment.

Rain water, which falls on a flat rectangular surface of length 6m6m and breadth 4m4m is transferred into a cylindrical vessel of internal radius 20cm20cm. What will be the height of water in the cylindrical vessel if a rainfall of 1cm1cm has fallen? (Use π=227\pi =\cfrac { 22 }{ 7 } )

Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 9cm9cm.

5050 circular plates each of diameter 14cm14cm and thickness 0.5cm0.5cm are placed one above the other to form a right circular cylinder. Find its total surface area.

A cylindrical tub of radius 12cm12cm contains water to a depth of 20cm20cm. A spherical form ball of radius 9cm9cm is dropped into the tub and thus the level of water is raised by hcmhcm. What is the value of hh?

A conical flask is full of water. The flask has base-radius rr and height hh. The water is poured into a cylindrical flask of base-radius mrmr. Find the height of water in the cylindrical flask.

A hollow sphere of internal and external radii 2 cm2\ cm and 4 cm4\ cm respectively is melted into a cone of base radius 4 cm4\ cm. Find the height and slant height of the cone.

A hollow sphere of internal and external diameters 4cm4cm and 8cm8cm respectively is melted into a cone of base diameter 8cm8cm. Calculate the height of the cone.

2525 circular plates, each of radius 10.5cm10.5cm and thickness 1.6cm1.6cm, are placed one above the other to form a solid circular cylinder. Find the curved surface area and the volume of the cylinder so formed.

If hh, ss, VV be the height, curved surface area and volume of a cone respectively, then (3πVh3+9V2−s2h2)=?\left( {3\pi V{h^3} + 9{V^2} - {s^2}{h^2}} \right) = ?

The largest sphere is to be carved out of a right circular cylinder of radius 7cm7cm and height 14cm14cm. Find the volume of the sphere.

A vessel in the shape of a cuboid contains some water. If three identical spheres are immersed in the water, the level of water is increased by 2cm2cm. If the area of the base of the cuboid is 160cm2160{cm}^{2} and its height 12cm12cm, determine the radius of any of the spheres.

The diameter of a metallic sphere is equal to 9cm9cm. It is melted and drawn into a long wire of diameter 2mm2mm having a uniform cross-section. Find the length of the wire.

A copper sphere of radius 3 cm3\ cm is melted and recast into a right circular cone of height 3 cm3\ cm. Find the radius of the base of the cone.

A self help group wants to manufacture jokes caps (conical shapes) of 3cm3 \text cm radius and 4cm4cm height. If the available colour paper sheets is 1000cm21000 \text cm^{2}, then how many caps can be manufactured from that paper sheet?

A spherical ball of radius 3\ cm3\ cm is melted and recast into three spherical balls. The radii of two of the balls are 1.5\ cm1.5\ cm and 2\ cm2\ cm. Find the diameter of the third ball.

Metal spheres, each of radius 2cm2cm, are packed into a rectangular box of internal dimension 16cm\times 8cm\times 8cm16cm\times 8cm\times 8cm when 1616 spheres are packed the box is filled with preservative liquid. Find the volume of this liquid. (Use \pi =\dfrac{669}{213}\pi =\dfrac{669}{213})

A washing tub in the shape of a frustum of a cone has height 21\ cm21\ cm. The radii of the circular top and bottom are 20\ cm20\ cm and 15\ cm15\ cm respectively. What is the capacity of the tub ? (\pi=\dfrac{22}{7})(\pi=\dfrac{22}{7})

The radius of a solid sphere is 77 cm. If the solid sphere is melt to form small spheres of radius 3.53.5 cm. Then how many small spheres can be formed?

Given : ED = 14 cm ED = 14 cm

The slant height of a cone with hemispherical base is 55 cm. If the total surface area of the article is 103.62 cm^2103.62 cm^2, find its height.

A rectangular reservoir is 120 m long and 75 m wide. At what speed per hour must water flow into it through a square pipe of 20 cm wide so that the water rises by 2.4 m in 18 hours ?

Threee cubes of the same metals, whose edges are 6,8,10 cm are melted and formed into a single cube. find the diagonal of these single cube.

From a solid cylinder whose height is 2.4cm2.4cm and diameter 1.4cm1.4cm a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest {cm}^{2}{cm}^{2}

Find the volume of the largest circular cone that can be cut out of a cube whose edge is 9\ cm. \left (\pi = \dfrac {22}{7}\right )9\ cm. \left (\pi = \dfrac {22}{7}\right ).

A hemispherical bowl of internal radius 9 cm is full of liquid. The liquid is to be filled into cylindrical shaped small bottles, each of diameter 3 cm and height 4 cm. How many bottles are needed to empty the bowl ?

A cylindrical container with diameter of base 42\ cm42\ cm contains sufficient water to submerge a rectangular solid of iron with dimensions 22\ cm\times 14\ cm \times 10.5\ cm22\ cm\times 14\ cm \times 10.5\ cm. Find the rise in level of the water when the solid is sub-merged.

The height of a cone is 40\ cm40\ cm A Small cone is cut off at the top by a plane parallel to its base. If its volume be \dfrac {1}{64}\dfrac {1}{64} of the volume of the given cone at what height above the base is the section cut?

How many silver coins, 1.75cm1.75cm in diameter and of thickness 2mm2mm, must be melted to form a cuboid of dimensions 5.5cm\times 10cm\times 3.5cm5.5cm\times 10cm\times 3.5cm?

In figure, the object is made of two solids a cube and a hemisphere. The base of the figure is a cube with edges of 5 cm and the hemisphere fixed on the top has a diameter of 4.2 cm . Find the total surface area of the object (take pi = 22/7 ) .

The radii of circular ends of a bucket of height 24 cm24 cm. Find the area of its curved surface.

The slant height of the frustum of a cone is 5\ cm5\ cm. If the difference between the radii of its two circular ends is 4cm4cm. Calculate the height of the frustum.

The slant height of the frustum of a cone is 4cm4cm. If the perimeters of its circular bases be 18cm18cm and 6cm6cm, find the curved surface area of the frustum and also find the cost of painting its total surface at the rate of Rs. 12.5012.50 per 100{cm}^{2}100{cm}^{2}

Fill in the blanks:A rectangular pyramid has____ faces.

If the total surface area of a solid hemisphere is 46c{m^2}46c{m^2}, find its volume

What length of tarpaulin 3 m wide will be required to make a conical tent of height 8 m and base radius 6 m ? Assume that the extra length of material will be required for stitching margins and wastage in cutting is approximately 20 cm.

A sphere is inscribed in a cylinder than find the ratio of volume of sphere to volume of cylinder.

A bucket is in the form of a frustum of a cone of height is 30\ cm30\ cm with radii of its lower and upper ends as 10\ cm10\ cm and 20\ cm20\ cm respectively. Find the capacity of the bucket.

Write the formula for finding volume of a frustum of a cone.

The silant height and base diametors of a conical to mb we 25m and 14 m qupertiuity. Find the cost of washisg its surface surface at the rate of RS 210 per 100m^{ 2 }m^{ 2 }

A metallic bucket, open at the top, of height 24cm24cm is in the form of the frustum of a cone, the radii of whose lower and upper circular ends are 7cm7cm and 14cm14cm respectively. Findthe volume of water which can completely fill the bucket;the area of the metal sheet used to make the bucket.

A sphere of diameter 18\text{ cm}18\text{ cm} is dropped into a cylindrical vessel of diameter 36\text{ cm},36\text{ cm}, partly filled with water. If the sphere is completely submerged, then the increase in water level (\text{in cm}\text{in cm}) will be

In the above figure, area of a square ABCDABCD is SS and area of each shaded portion is PP and area of inner circle is QQ. Then the area of remaining part of a square is

Find the ratio of the volumes of a cone and of a cylinder whose base diameter and heights are equal.

How many planks of size 2\ cm\times 25\ cm\times 8\ cm2\ cm\times 25\ cm\times 8\ cm can be prepared from a wooden block 5\ m5\ m long. 70\ cm70\ cm broad and 32\ cm32\ cm thick, assuming that there is no wastage?

Fill in the blanks to make the correct statements true.A cube of side 4 cm is painted on all its sides. If it is sliced in 1 cubic cm cubes, then number of such cubes that will have exactly two of their faces painted is ____________.

Fill in the blanks the statements are true.The surface area of a cylinder which exactly fits in a cube of side b is ________.

A wooden cylindrical pole is 7\ m7\ m high and its base radius is 10\ cm10\ cm. Find its weight if the wood weights 225\ kg225\ kg per cubic meter.

What figure is formed if only the height of a cube is increased or decreased?

Fill in the blanks to make the statement true.The volume of a cylinder which exactly first in a cube of side a is ________.

How many envelopes of size 25 cm \times\times 15 cm can be made from a rectangular sheet of size 4 m \times\times 1.2 m?

A solid sphere of radius 6cm6cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 55 cm and its height is 3232 cm, then find the thickness of the cylinder.

A solid right circular cone of radius 4cm and height 7cm is melted to a sphere. Find the radius of sphere.

A bucket is in the form of a frustum of a cone and holds 28.49028.490 litres of water. The radii of the top and bottom are 28\ cm28\ cm and 21\ cm21\ cm respectively. Find the height of the bucket.

State True(1) or False(0).The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the Fig. 12.7 is \pi \cfrac{r^{2}}{3}(3h-2r)\pi \cfrac{r^{2}}{3}(3h-2r)

Two identical solid hemispheres of equal base radius rr cm are stuck together along their bases. The total surface area of the combination is k \pi r^2.k \pi r^2. Find kk.

An ice cream cone full of ice cream having radius 55 cm and height 1010 cm as shown. Calculate the volume of ice cream (to the nearest integer, in cm^3^3), provided that its \dfrac{1}{6}\dfrac{1}{6}th part is left unfilled with ice cream. Insert answer in nearest integer.

The curved surface area of the frustum of a cone is 180180 sq. cm and the circumference of its circular bases are 1818 cm and 66 cm respectively. Find the slant height of the frustum of a cone (in cm).

A bucket is 40cm40cm in diameter at the top and 28cm28cm in diameter at the bottom. Find the capacity of the bucket in litres, if it is 21cm21cm deep. Also find the cost of tin sheet used in making the bucket, if the cost of tin is Rs.\space 1.50Rs.\space 1.50 per sq dm.

From a cylinder whose height is 8 cm and radius is 6 cm, a conical cavity of height 8 cm and base radius 6 cm is hollowed out. Find the volume of the remaining solid. \displaystyle \left ( \pi =\frac{22}{7} \right )\displaystyle \left ( \pi =\frac{22}{7} \right ).

Three cubes whose edges are 33 cm, 44 cm and 55 cm, respectively are melted to from a single cube. Find the surface area of the new cube.

Three solid spheres of gold whose radii are 11 cm, 66 cm and 88 cm, respectively are melted into a single solid sphere. Find the radius of the sphere.

A spherical ball of radius 3\ cm3\ cm is melted and recast into three spherical balls. The radii of two of the balls are 1.5\ cm1.5\ cm and 2\ cm.2\ cm. Find the radius of the third ball.

If a sphere of radius 6 \ cm6 \ cm is melted and drawn into a wire of radius 0.02\ cm0.02\ cm then what is the length of the wire ?

A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14\ cm14\ cm and the total height of the vessel is 13\ cm13\ cm. Find the inner surface area of the vessel.

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1\ cm1\ cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of \pi\pi.

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is 10\ cm10\ cm, and its base is of radius 3.5\ cm3.5\ cm, find the total surface of the article.

A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 1515 cm by 1010 cm by 3.53.5 cm. The radius of each of the depressions is 0.50.5 cm and the depth is 1.41.4 cm. Find the volume of wood in the entire stand .

A cubical block of side 7\ cm7\ cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

The frustum cone curved surface area is 210 m^2210 m^2. The upper and lower circle area of a cone is 2424 and 15 m^215 m^2. Find the total surface area.

Consider a frustum of cone with larger and smaller radius equal to $r_1 \ and\ r_2$ , respectively. Then, the total surface area of frustum of a cone is $\pi l(r_1+r_2)+\pi (r_{1}^{2}+r_{2}^{2})$ . if correct enter 1 else enter 0

State True(1) or False(0).

The volume of the frustum of a cone is $\frac{1}{3}\pi h\left [ r_{1}^{2}r_{2}^{2}-r_{1}r_{2} \right ],$ where h is vertical
height of the frustum and $r_1, r_2$ are the radii of the ends.

surface area of a frustum of a cone is $\pi l (r_1+r_2)$ . If true enter 1 else 0.

A spherical ball of radius $3\ cm$ is melted and recast into three spherical balls. The radii of the two of the balls are $1.5\ cm$ and $2\ cm$ respectively. Find the radius of the third ball.

A fraction clutch is in the form of a frustum of a cone, the diameter of the ends being $32\ cm$ and $20\ cm$ and length $8\ cm$ . Find its bearing surface and volume.[Take $\pi=3.14$ ]

The decorative block shown in Fig. 13.7 is made of two solids - a cube and a hemisphere. The base of the block is a cube with edge 5 cm. and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block. $\left(Take \pi =\dfrac { 22 }{ 7 } \right)$

How many spherical bullets each of $5cm$ in diameter can be cast from a rectangular block of metal $11dm\times 1m\times 5dm$ ?

A $16\text{ m}$ deep well with diameter $3.5\text{ m}$ is dug up and the earth from it is spread evenly to form a platform $27.5\text{ m}$ by $7\text{ m}$ . Find the height of the platform.

A well of diameter $2\ m$ is dug $14\ m$ deep. The earth taken out of it is spread evenly all around it to form an embankment of height $40\ cm$ . Find the width of the embankment.

Rain water, which falls on a flat rectangular surface of length $6m$ and breadth $4m$ is transferred into a cylindrical vessel of internal radius $20cm$ . What will be the height of water in the cylindrical vessel if a rainfall of $1cm$ has fallen? (Use $\pi =\cfrac { 22 }{ 7 }$ )

Find the volume of the largest right circular cone that can be cut out of a cube whose edge is $9cm$ .

$50$ circular plates each of diameter $14cm$ and thickness $0.5cm$ are placed one above the other to form a right circular cylinder. Find its total surface area.

A cylindrical tub of radius $12cm$ contains water to a depth of $20cm$ . A spherical form ball of radius $9cm$ is dropped into the tub and thus the level of water is raised by $hcm$ . What is the value of $h$ ?

A conical flask is full of water. The flask has base-radius $r$ and height $h$ . The water is poured into a cylindrical flask of base-radius $mr$ . Find the height of water in the cylindrical flask.

A hollow sphere of internal and external radii $2\ cm$ and $4\ cm$ respectively is melted into a cone of base radius $4\ cm$ . Find the height and slant height of the cone.

A hollow sphere of internal and external diameters $4cm$ and $8cm$ respectively is melted into a cone of base diameter $8cm$ . Calculate the height of the cone.

$25$ circular plates, each of radius $10.5cm$ and thickness $1.6cm$ , are placed one above the other to form a solid circular cylinder. Find the curved surface area and the volume of the cylinder so formed.

If $h$ , $s$ , $V$ be the height, curved surface area and volume of a cone respectively, then $\left( {3\pi V{h^3} + 9{V^2} - {s^2}{h^2}} \right) = ?$

The largest sphere is to be carved out of a right circular cylinder of radius $7cm$ and height $14cm$ . Find the volume of the sphere.

A vessel in the shape of a cuboid contains some water. If three identical spheres are immersed in the water, the level of water is increased by $2cm$ . If the area of the base of the cuboid is $160{cm}^{2}$ and its height $12cm$ , determine the radius of any of the spheres.

The diameter of a metallic sphere is equal to $9cm$ . It is melted and drawn into a long wire of diameter $2mm$ having a uniform cross-section. Find the length of the wire.

A copper sphere of radius $3\ cm$ is melted and recast into a right circular cone of height $3\ cm$ . Find the radius of the base of the cone.

A self help group wants to manufacture jokes caps (conical shapes) of $3 \text cm$ radius and $4cm$ height. If the available colour paper sheets is $1000 \text cm^{2}$ , then how many caps can be manufactured from that paper sheet?

A spherical ball of radius $3\ cm$ is melted and recast into three spherical balls. The radii of two of the balls are $1.5\ cm$ and $2\ cm$ . Find the diameter of the third ball.

Metal spheres, each of radius $2cm$ , are packed into a rectangular box of internal dimension $16cm\times 8cm\times 8cm$ when $16$ spheres are packed the box is filled with preservative liquid. Find the volume of this liquid. (Use $\pi =\dfrac{669}{213}$ )

A washing tub in the shape of a frustum of a cone has height $21\ cm$ . The radii of the circular top and bottom are $20\ cm$ and $15\ cm$ respectively. What is the capacity of the tub ? $(\pi=\dfrac{22}{7})$

The radius of a solid sphere is $7$ cm. If the solid sphere is melt to form small spheres of radius $3.5$ cm. Then how many small spheres can be formed?

Given : $ED = 14 cm$

The slant height of a cone with hemispherical base is $5$ cm. If the total surface area of the article is $103.62 cm^2$ , find its height.

From a solid cylinder whose height is $2.4cm$ and diameter $1.4cm$ a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest ${cm}^{2}$

Find the volume of the largest circular cone that can be cut out of a cube whose edge is $9\ cm. \left (\pi = \dfrac {22}{7}\right )$ .

A cylindrical container with diameter of base $42\ cm$ contains sufficient water to submerge a rectangular solid of iron with dimensions $22\ cm\times 14\ cm \times 10.5\ cm$ . Find the rise in level of the water when the solid is sub-merged.

The height of a cone is $40\ cm$ A Small cone is cut off at the top by a plane parallel to its base. If its volume be $\dfrac {1}{64}$ of the volume of the given cone at what height above the base is the section cut?

How many silver coins, $1.75cm$ in diameter and of thickness $2mm$ , must be melted to form a cuboid of dimensions $5.5cm\times 10cm\times 3.5cm$ ?

The radii of circular ends of a bucket of height 24 cm. Find the area of its curved surface.

The slant height of the frustum of a cone is $5\ cm$ . If the difference between the radii of its two circular ends is $4cm$ . Calculate the height of the frustum.

The slant height of the frustum of a cone is $4cm$ . If the perimeters of its circular bases be $18cm$ and $6cm$ , find the curved surface area of the frustum and also find the cost of painting its total surface at the rate of Rs. $12.50$ per $100{cm}^{2}$

Fill in the blanks:
A rectangular pyramid has____ faces.

If the total surface area of a solid hemisphere is $46c{m^2}$ , find its volume

A bucket is in the form of a frustum of a cone of height is $30\ cm$ with radii of its lower and upper ends as $10\ cm$ and $20\ cm$ respectively. Find the capacity of the bucket.

The silant height and base diametors of a conical to mb we 25m and 14 m qupertiuity. Find the cost of washisg its surface surface at the rate of RS 210 per 100 $m^{ 2 }$

A metallic bucket, open at the top, of height $24cm$ is in the form of the frustum of a cone, the radii of whose lower and upper circular ends are $7cm$ and $14cm$ respectively. Find
the volume of water which can completely fill the bucket;
the area of the metal sheet used to make the bucket.

A sphere of diameter $18\text{ cm}$ is dropped into a cylindrical vessel of diameter $36\text{ cm},$ partly filled with water. If the sphere is completely submerged, then the increase in water level ( $\text{in cm}$ ) will be

In the above figure, area of a square $ABCD$ is $S$ and area of each shaded portion is $P$ and area of inner circle is $Q$ . Then the area of remaining part of a square is

How many planks of size $2\ cm\times 25\ cm\times 8\ cm$ can be prepared from a wooden block $5\ m$ long. $70\ cm$ broad and $32\ cm$ thick, assuming that there is no wastage?

Fill in the blanks to make the correct statements true.
A cube of side 4 cm is painted on all its sides. If it is sliced in 1 cubic cm cubes, then number of such cubes that will have exactly two of their faces painted is ____________.

Fill in the blanks the statements are true.
The surface area of a cylinder which exactly fits in a cube of side b is ________.

A wooden cylindrical pole is $7\ m$ high and its base radius is $10\ cm$ . Find its weight if the wood weights $225\ kg$ per cubic meter.

Fill in the blanks to make the statement true.
The volume of a cylinder which exactly first in a cube of side a is ________.

How many envelopes of size 25 cm $\times$ 15 cm can be made from a rectangular sheet of size 4 m $\times$ 1.2 m?

A solid sphere of radius $6cm$ is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is $5$ cm and its height is $32$ cm, then find the thickness of the cylinder.

A bucket is in the form of a frustum of a cone and holds $28.490$ litres of water. The radii of the top

and bottom are $28\ cm$ and $21\ cm$ respectively. Find the height of the bucket.

State True(1) or False(0).

The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the Fig. 12.7 is
$\pi \cfrac{r^{2}}{3}(3h-2r)$

Two identical solid hemispheres of equal base radius $r$ cm are stuck together along their bases. The total surface area of the combination is $k \pi r^2.$ Find $k$ .

An ice cream cone full of ice cream having radius $5$ cm and height $10$ cm as shown. Calculate the volume of ice cream (to the nearest integer, in cm $^3$ ), provided that its $\dfrac{1}{6}$ th part is left unfilled with ice cream. Insert answer in nearest integer.

The curved surface area of the frustum of a cone is $180$ sq. cm and the circumference of its circular bases are $18$ cm and $6$ cm respectively. Find the slant height of the frustum of a cone (in cm).

A bucket is $40cm$ in diameter at the top and $28cm$ in diameter at the bottom. Find the capacity of the bucket in litres, if it is $21cm$ deep. Also find the cost of tin sheet used in making the bucket, if the cost of tin is $Rs.\space 1.50$ per sq dm.

From a cylinder whose height is 8 cm and radius is 6 cm, a conical cavity of height 8 cm and base radius 6 cm is hollowed out. Find the volume of the remaining solid. $\displaystyle \left ( \pi =\frac{22}{7} \right )$ .

Three cubes whose edges are $3$ cm, $4$ cm and $5$ cm, respectively are melted to from a single cube. Find the surface area of the new cube.

Three solid spheres of gold whose radii are $1$ cm, $6$ cm and $8$ cm, respectively are melted into a single solid sphere. Find the radius of the sphere.

A spherical ball of radius $3\ cm$ is melted and recast into three spherical balls. The radii of two of the balls are $1.5\ cm$ and $2\ cm.$ Find the radius of the third ball.

If a sphere of radius $6 \ cm$ is melted and drawn into a wire of radius $0.02\ cm$ then what is the length of the wire ?

A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is $14\ cm$ and the total height of the vessel is $13\ cm$ . Find the inner surface area of the vessel.

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to $1\ cm$ and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$ .

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is $10\ cm$ , and its base is of radius $3.5\ cm$ , find the total surface of the article.

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are $15$ cm by $10$ cm by $3.5$ cm. The radius of each of the depressions is $0.5$ cm and the depth is $1.4$ cm. Find the volume of wood in the entire stand .

A cubical block of side $7\ cm$ is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

The frustum cone curved surface area is $210 m^2$ . The upper and lower circle area of a cone is $24$ and $15 m^2$ . Find the total surface area.

Find the volume of a frustum cone, whose base and upper area of a circle is $40cm^2$ and $90cm^2$ . (Insert answer in nearest integer).

The total surface area of frustum cone is $1200 ft^2$ . The radius of a top cone is $3$ and the bottom cone is $6$ ft. What is the curved surface area of a frustum of cone?

The radii of the circular ends of a frustum of height 4 cm are 7 cm and 4 cm respectively. Find the total surface area. (Use $\pi = 3$ ).

Find the volume of the frustum cone whose base and top radius is $8$ cm and $4$ cm respectively. The height of the cone is $5$ cm. (Use $\pi = \dfrac{22}{7}$ ).
(Insert answer in nearest integer).

The base and top radius of a cone is $2.1$ m and $0.8$ m respectively. The height of the cone is $24$ m. What is the volume of frustum of a cone? (Use $\pi = 3.14$ ). Insert answer in nearest integer,

A cone has base radius cm is cut from the top. Calculate the volume of the frustum. Insert answer in terms of $\pi$ .

The slant height of the frustum of a cone is $4$ cm, and the perimeter of its circular bases are $18$ cm and $6$ cm respectively. Find the curved surface area and total surface area of the frustum.

The radii of two circular ends of a frustum shaped dust bin are $15$ cm and $8$ cm. If its depth is $63$ cm, find volume of the dust bin.

From the top of a cone of base radius $12$ cm and height $20$ cm, a small cone of base radius $3$ cm is to be cut off. How far down the vertex is the cut to be made? Find the volume of the frustum so obtained.

A flower vase is in the form of a frustum of a cone. The perimeter of the ends are $44$ cm and $8.4\pi$ cm. If the depth is $14$ cm, find how much water it can hold?

A vessel is in the form of a frustum of a cone. Its radius at one end is $8\ cm$ and the height is $14\ cm$ . If its volume is $\displaystyle\dfrac{5676}{3}\ cm^3$ , find the radius of the other end.

A petrol tank is in the shape of a cylinder with hemispheres of same radius attached to both ends. If the total length of the tank is $6$ m and the radius is $1$ m, what is the capacity of the tank in litres.