Class 8 Maths - Mensuration

A cuboidal vessel is $12\ \text{cm}$ long and $8\ \text{cm}$ wide. How high should it be made so that it may hold $480$ cubic centimetre of a liquid?

Given length vessel,

$L = 12\ \text{cm}$

Width of vessel,

$B = 8\ \text{cm}$

Let, height of the vessel be

$'H'$ .

Given, volume

$V= 480\ \text{cm}^{3}$

Volume of cuboid

$= LBH$

$LBH=480\ \text{cm}^3$

$h=\dfrac{480}{12\times 8}\ \text{cm}=5\ \text{cm}$

Hence, the vessel should be

$5\ \text{cm}$ high.

Find the side of a cube, whose total surface area is 216 $cm^{2}$ ?

Let the side of a cube be

$a\ cm$

As we know that

total surface area of a cube with side

$a$ is

$6 a^2$

Given total surface area of a cube is

$216\ cm^2$

$\implies 6 a^2=216$

$\implies a^2=36$

$\implies a=6$

So side of a cube is

$6\ cm$

Find the volume of a cuboid of length $4\ cm$ ,breadth $3\ cm$ and height $1\ cm$

Volume of given cuboid

$=$ length

$\times$ Breadth

$\times$ Height

$=4\times 3\times 1$ cu.cm

$=12$ cu.cm

Find the capacity of a closed rectangular cistern whose length is $8$ m,breadth $6$ m and depth $2.5$ .

Here

$l=8$ m;

$b=6$ m;

$h=2.5$ m

Capacity of the closed rectangular cistern

$=$ Volume of the cistern

$=l\times b\times h$

$\therefore$ Capacity of the cistern

$=8\times 6\times 2.5{m}^{3}=120{m}^{3}$

Find the area of trapezium whose parallel sides are $24\ cm$ and $20\ cm$ and the distance between them is $15\ cm$ .

We know that the length of the parallel sides is

$24cm$ and

$20cm$ and height

$(h)=15cm$

By using the formula

$Area\ of\ trapezium$

$=1/2 \times (sum\ of\ parallel\ side )\times height$ we get,

$=1/2 \times (24+20) \times 15$

$=33\ cm^2$

The parallel sides of a trapezium are $8.4$ cm and $12.3$ cm. If its height is $7.2$ cm; then the value of its area is $74.52\,\,cm^2$
If true then enter $1$ and if false then enter $0$

Area of a trapezium

$= \frac { 1 }{ 2 } \times (Sum\quad of\quad parallel\quad sides)\times height = \frac { 1 }{ 2 } \times (8.4 + 12.3) \times 7.2 = 74.52 sq cm$

If a solid cuboid of metal has dimensions 24 cm, 18 cm and 4 cm, then its volume 1728 cm $^3$ .
If true enter 1 else 0.

Volume of a cuboid of length l, breadth b and height h

$= l \times b \times h$
So, the volume of the cuboid

$= 24 \times 18 \times 4 = 1728 {cm}^{3}$

How much length of the iron sheet $11$ cm wide is required for making an open cylinder $15$ cm high and $7$ cm as base radius $\displaystyle \left ( \pi =22/7 \right )$ ?

Surface of an open cylinder

$=$ area of the iron sheet

Let

$A$ be the surface area of open cylinder

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

$\displaystyle =\frac{22}{7}\times 7^{2}+2\times \frac{22}{7}\times 7\times 15$

$\displaystyle =\left ( 154+660 \right )cm^{2}$

$\displaystyle =814cm^{2}$

Sheet is of the rectangular shape whose area is given as length times breadth.

$\displaystyle \therefore$ Length of the sheet

$\displaystyle =\left ( 814\div 11 \right )cm$

$\displaystyle =74cm$

A cuboid is $3$ cm high, $4$ cm wide and $5$ cm long. Its volume is ____ (in $\text{cm}^3$ ).

Height of a cuboid

$(h)=3\,cm$

Breadth of a cuboid

$(b)=4\,cm.$

Length of a cuboid

$(l)=5\,cm.$

$\Rightarrow$ Volume of cuboid

$=l\times b\times h$

$=5\,cm\times 4\,cm\times 3\,cm$

$=60\,cm^3$

The parallel sides of trapezium are $9cm$ and $7cm$ long and the distance between them is $6cm$ . Find the area of the trapezium

Area of trapezium

$=\dfrac { 1 }{ 2 } \left( sum\quad of\quad parrallel\quad sides \right) \times h$

$=\dfrac { 1 }{ 2 } \left( 9+7 \right) \times 6=48{ cm }^{ 2 }$

Find the area of a trapezium whose parallel sides are $6cm$ and $10cm$ long and the distance between them is $4cm$ .

Area of a trapezium is given by

$=\cfrac { 1 }{ 2 } \times (Sum \,of\, parallel\, sides)\times height$

$\therefore\,$ Area

$=\cfrac { 1 }{ 2 } \times (6+10)\times 4$

$=32$ sq.cm

Find the total surface area of a cube with side $5$ $cm$ .

Given, the side of a cube is

$5$

$cm$ .
Total surface area of cube

$= 6 \times$

$($ side

$)^{2}$

$= 6 \times 5 \times 5$

$= 150$

$\therefore$ The total surface area of cube is

$150\ {cm}^{2}$ .

The radius of the base of a right circular cylinder is $5$ $cm$ and height is $21$ $cm$ , find the curved surface area.

The radius of the base of a right circular cylinder is

$5$

$cm$ and height is

$21$

$cm$ .
Here,

$r = 5$

$cm$ and

$h = 21$

$cm$ .
Curved surface area of cylinder

$= 2 \pi rh$

$=2\times \dfrac { 22 }{ 7 } \times 5\times 21$

$=44\times 15$

$=660\ { cm }^{ 2 }$

$\therefore$ The curved surface area of cylinder is

$660\ { cm }^{ 2 }$ .

A rectangular sheet of paper 44 cm $\times$ 18 cm is rolled along the length to form a cylinder. Assuming that the cylinder is solid (Completely filled), find its total surface area and the radius.

Given a rectangular sheet of paper

$44 cm \times 18 cm$ is rolled along the length to form a cylinder. And cylinder is solid

Then height of cylinder =18 cm

And Circumference base of cylinder =44 cm

$\therefore 2\pi r=44$

$\rightarrow \frac{44\times 7}{2\times 22}$

$\rightarrow 7 cm$

So radius of cylinder = 7 cm

Then Total surface area of cylinder =

$2\pi r(r+h)=2\times \frac{22}{7}\times 7(7+18)=2\times 22\times 25=1100 cm^{2}$

Find the total surface area of a cube with side $7$ $cm$ .

Given, the side of a cube is

$7$

$cm$
Total surface area of cube

$= 6 \times ($ side

$)^{2}$

$=6 \times 7 \times 7$

$= 294$

$\therefore$ the total surface area of cube is

$294\ {cm}^{2}$ .

A water tank is 6 m in length, 3 m in breadth, and 4 m in height. How much liter water can be stored in it?

Give: Length of tank

$=l=6$ m
Breadth of tank

$= b=3$ m
Height of tank

$=h=4$ m

Volume of cuboidal tank

$=l \times b \times h$

$=6$ m

$\times\ 3$ m

$\times\ 4$ m

$= 72\ m^3$

We know,

$1\ m^3=1000$ liter

Hence Volume of tank

$=72\ m^3$

$=(72 \times 1000)$ litre

$=72,000$ litre

$\therefore 72,000$ liter water can be stored in the water tank.

The length, breadth, and height of a tank are $3 \text{ m},\;2 \text{ m}$ and $6 \text{ m}$ respectively. What is the volume of the tank in $\text{m}^3$ ?

Given:-

Length of the tank,

$l=3 \text{ m}$

The breadth of the tank,

$b=2 \text{ m}$

Height of the tank,

$h=6 \text{ m}$

Volume of cuboid is equal to product of length, breadth and height. So,

$\text{Volume of tank}= l\times b\times h$

$= \left( {3 \times 2 \times 6} \right)$

$= 36{\text{ m}^3}$

Hence, the volume of the tank is equal to

$36{\text{ m}^3}.$

Salma bought a tank measuring 3 m X 3 m X 2 m for storing water.How many litres of water can be stored in it ?(in L)

Salma bought a tank measuring 3m X 3m X 2m .

$\therefore$ volume of the tank is

$( 3m \times 3m \times 2 m ) \\ = 18 m^3 \, \, [\therefore \, V = a b c \, \textrm{unit}]$

We know

$1 m^3 = 1000 \, \textrm{litres}.$

$\therefore (18 \times 1000 ) = 18000$ litres of water can be stored in it.

The radius and height of a right circular cylinder are $4$ and $14$ cm respectively. Find the curved surface area of a cylinder.

The curved surface area of a cylinder is given by the formula,

$2\pi \times r \times h$ , where

$r$ is the radius of the cylinder and

$h$ is the height.

Substituting the given values in the formula, we get

Curved Surface Area =

$2\times\dfrac{22}{7}\times4\times14$

$2 \times 22 \times 4 \times 2$

$352\, cm^2$

Find the area of a trapezium whose parallel side are $24cm$ and $20cm$ B the distance them is $15cm$

Area of trapezium =

$\dfrac{a+b}{2}*h$ where

$a$ and

$b$ are the length of the two parallel sides and

$h$ is the distance between them.

$\therefore Area = \dfrac{24+20}{2}*15=330cm^2$

The perimeter of a trapezium is $52cm$ and its non-parallel sides are each equal to $10cm$ and its altitude is $8cm$ its Area $=?$

Perimeter

$=52cm$

$a+b+10+10=52$

$a+b=32cm$

$h=8cm$

Area

$=\dfrac{1}{2}(a+b)\times h$

Area

$=\dfrac{1}{2}\times 32\times 8=128cm^{2}$

The are of a trapezium id $1080cm^2$ If the length of its parallel sides are $55.6cm$ and $34.4cm$ find the distance between them.

Area of trapezium

$=\dfrac 12\times h(a+b)$

$1080=\dfrac 12\times (55.6+34.4)\times h$

$1080=\dfrac 12 \times 90\times h$

$h=\dfrac{1080\times 2}{90}$

$h=\dfrac {2160}{90}$

$h=24cm$

Find the volume in $cm^3$

Let length

$(l)=18cm,$ Breadth

$(b)=15cm,$ Height

$(h)=10.5cm$

Volume

$=lbh=18\times 15\times 10.5=2835{ cm }^{ 3 }$

Two cylinders of same radii $14\ cm$ , and heights $20\ cm$ , $25\ cm$ are placed one upon another then find the $T.S.A$ (Total Surface Area) of the shape formed.

Letting the cylinders to be hollow.

Now the newly made cylinder has radius

$(r)=14\ cm$ and height

$(h)=20+25=45\ cm$ .

Now total surface are of newly formed hollow cylinder is

$2\pi r h\ cm$

$^2=88\times45=3960\ cm$ ^2$$.

If the ratio between the curved surface area and the total surface area of a right circular cylinder is $3:4$ , find the ratio of height to the radius of the cylinder.

$\dfrac{C.S.A}{T.S.A}=\dfrac{2\pi rh}{2\pi r(r+h)}=\dfrac{3}{4}$

$\Rightarrow \dfrac{h}{r+h}=\dfrac{3}{4}$

$\Rightarrow 4h=3r+3h$

$\Rightarrow h=3r$

$\Rightarrow\dfrac{h}{r}=3$

A $4 cm$ cube is cut into $1 cm$ cubes. Calculate the total surface area of all the small cubes.

Vol. of 4cm cube

$= (4)^3 cm^3$

$=64 cm^3$

Vol. of 1cm cube

$= (1)^3 cm^3$

$=1 cm^3$

Total no. of 1cm cubes

$= \dfrac {Vol. of \ 4 \ cm \ cube }{Vol. \ of \ 1 \ cm \ cube}$

$= \dfrac {64}{1}$

$=64$

Surface area of 1 cube

$=6(side)^2 cm^2$

Surface area of 1 cube

$=6(1)^2 cm^2$

Surface area of 1 cube

$=6 cm^2$

Total surface area of all small cubes

$=64 \times 6cm^2$

$=384 cm^2$

Find the lateral or curved surface area of a closed cylindrical petrol storage tank that is $4.2\ m$ in diameter and $4.5\ m$ high.

Closed cylindrical pipe is in the form of a cylinder

$\therefore$ Curved surface area of a cylinder

$= 2\pi rh$

$=\pi dh$ where

$d=2r$

$=\dfrac{22}{7}\times 4.2\times 4.5$ where

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

$=59.4 m^{2}$

Find the sum of length of the parallel sides of a trapezium whose area is $4.2 \ m^2$ and whose height is $280 \ cm$ .

Area of trapezium

$=\dfrac{a+b}{2}\times h$

Where,

$a$ and

$b$ are parallel sides and

$h$ is perpendicular height.

So,

$(a+b)$ is sum of length of bases of a trapezium.

Now,

Area of trapezium

$=\dfrac{a+b}{2}\times h$

$\Rightarrow$

$4.2\times 100^2=\dfrac{a+b}{2}\times 280$ {

$1m = 100cm$ }

$\Rightarrow$

$\dfrac{8.4}{280} \times 10000=a+b$

$\Rightarrow$

$a+b=300\,cm$

The lengths of the parallel sides of a trapezium are $30$ centimetres, $10$ centimetres and the distance between them is $20$ centimetres. What is its area?

Area of a trapezium

$=\dfrac{1}{2}\times$ Sum of parallel sides

$\times$ distance between them.

$=\dfrac{1}{2}\times\left(30+10\right)\times 20$

$=\dfrac{1}{2}\times 40\times 20$

$=400$ sq.cms

Find the volume of cuboid , if its length , breadth and height are 6 cm , 5 cm and 3 cm respectively .

Volume of cuboid =

$lbh = 6 \times 5 \times 3 = 90$ cm

$^{3}$

A rectangular sheet of dimensions $25\ cm\times 7\ cm$ is rotated about its longer side. Find the whole surface area of the solid thus generated.

When a rectangular sheet is rotated about its length, we get a cylinder

where the height of cylinder

$=7\ cm$

and the circumference of the base of the cylinder

$=25\ cm$

Thus curved surface area

$=2\pi rh$

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

$=25\times7$

$=175\ cm^2$

If the areas of three adjacent faces of a cuboid are $X, Y$ and $Z$ respectively, then find the volume of a cuboid.

Let the sides of cuboid be

$a,b,c$

Volume of cuboid

$abc$

$X=ab,Y=bc,Z=ca$

Consider

$XYZ=(ab)(bc)(ac)$

$XYZ=(abc)^2$

$\implies abc=\sqrt{XYZ}$

So the Volume of Cuboid is

$\sqrt{XYZ}$

Find the area of a trapezium whose parallel sides are $24cm$ and $20cm$ and the distance between them is $15cm$

We know,

$\text{Area of trapezium}=\dfrac {1}{2} × \text {Sum of parallel sides} × \text{Distance between them}.$

Area of trapezium =

$\dfrac{h}{2}(a+b)$ , where

$a$ and

$b$ are base and h is height.

Here,

$a=24$ cm,

$b=20$ cm and

$h=15$ cm

Therefore, Area=

$\dfrac{15}{2}(24+20)=330cm^2$

The length of an edge of a cube is $l$ . Find the formula for the sum of lengths of all the edges of the cube.

According to question,

Edge length of a cube

$=l$

Hence, cube has

$12$ edges.

So,sum of length of edges

$=12l$

Hence formula for the sum of edge length

$=(12\times$ one edge length)

Find the area of a trapezium whose parallel sides are $11$ m and $25$ m and the non parallel sides are $15$ m and $13$ m.

Trapezium

$ABCD$ is divide into parallelogram

$AECD$ and triangle

$CEB$ .

Now in

$\triangle{CEB}$ , we have

$EB = 25 - 11 = 14 \; cm$

Using Heron’s theorem, we will first evaluate the semi-perimeter of

$\triangle{CEB}$ and then evaluate its area.

$S = \cfrac{BE + EC+ BC}{2}$

$\Rightarrow S = \cfrac{13 + 14 + 15}{2} = 21 \; cm$

Therefore,

area of

$\triangle{CEB} = \sqrt{S \left( S - BE \right) \left( S - EC \right) \left( S - BC \right)}$

$\Rightarrow$ Area of

$\triangle{CEB} = \sqrt{21 \left( 21 - 13 \right) \left( 21 - 14 \right) \left( 21 - 15 \right)} = 84 \; {cm}^{2}$

As we know that,

Area of triangle

$= \cfrac{1}{2} \times$ Base

$\times$ height

Therefore,

Area of

$\triangle{CEB} = \cfrac{1}{2} \times 14 \times BF$

$\Rightarrow BF = \cfrac{84 \times 2}{14} = 12 \; cm$

Again, as we know that,

Area of trapezium

$=$ Sum of parallel sides

$\times$ height

Therefore,

Area of trapezium

$ABCD = \cfrac{1}{2} \times \left( 11 + 25 \right) \times 12 = 216 \; {cm}^{2}$

Thus the area of trapezium is

$216 \; {cm}^{2}$ .

1143384_1159143_ans_8e7bbdcc526f41b0a31251b96ac1e74b.png

Given equilateral $ABCD$ with right angles in $B$ and $D$ . We also know that $BC=1, CD=4$ and $DA=3\ units$ . What is the area of $ABCD$

Length of

$AE = 1$

$\therefore l(DE)=3-1=2$

$\therefore l(CE)=\sqrt{16+4}=\sqrt{20}$

$\therefore$ Area of

$\square ABCE = (1) (\sqrt{20})=\sqrt{20}\, units^{2}$

$=2\sqrt{5}\,units^{2}$

$\therefore$ Area of

$\Delta CDE=\dfrac{1}{2}\times (4)(2)$

$= 4 \,units^{2}$

$\therefore$ Area of

$\square ABCD=(4+2\sqrt{5})\,units^{2}$

1067099_1159514_ans_6b5bc390117148ef92785224080ba895.png

A cuboid has volume four times the volume of a cube whose each edge is $7\ \mathrm{m}$ . Find the volume of the cuboid.

Volume of cube with side

$7\ \mathrm{m}=(7)^3\ \mathrm{m}^3= 343\ \mathrm{m}^3$

According to the statement:

Volume of cuboid

$= 4\times$ (Volume of cube)

Volume of cuboid

$= (4\times 343)\ \mathrm{m}^{3}$

$= 1372\ \mathrm{m}^{3}$

$ABCD$ is a trapezium in which $AB || CD$ and $AB=3cm ,DC=5cm$ . Find the of area if height is $4cm$

The area of trapezium is

$\dfrac 12\times(a+b)\times h\\\dfrac 12\times (3+5)\times 4\\2\times (8)=16cm^2$

Find the ratio between TSA of a cylinder to CSA given that its height and radius are $7.5$ and $3.5$ cm.

Given radius of the cylinder

$(r)=3.5$

$cm$

height of the cylinder

$(h)=7.5$

$cm$

curved surface area(CSA)

$=2\pi rh\\ =2\pi \times 3.5\times 7.5\\ =52.5\pi \quad cm$

total surface area(TSA)

$=2\pi rh+2\pi { r }^{ 2 }\\ =2\pi \times 3.5\times 7.5+2\pi \times { 3.5 }^{ 2 }\\ =52.5\pi +24.5\pi \\ =77\pi \quad cm$

Ratio

$=\dfrac { TSA }{ CSA } \\ =\dfrac { 77\pi }{ 52.5\pi } \\ =1.47$

Find the total surface area of cubes having the following sides.
$5\ cm$

Given side of cube

$=5\,cm$ .

then Total surface area of cube

$= 6a^2$

then

$TSA = 6 \times (5)^2 =6\times 25$

TSA of cube

$=150 \, cm^2$

Find the total surface area of cubes having the following sides.
$7.2\ m$

Given TSA of cube

$=6a^2$ .

Given side of cube

$=7.2\, m$

then

$TSA = 6 \times (7.2)^2$

$TSA = 311.04\, cm^2$

The volume of a solid cylinder is $448 \pi \ cm^3$ and height $7 \ cm$ . Find its lateral surface area and total surface area.

Let the radius of the base and height of the cylinder be r cm and h cm respectively. Then,

$h=7 \ cm$

Now, Volume =

$448\pi \ cm^3$

$\pi r^2h=448 \pi$

$r^2=\dfrac{448}{7}=64$

$r=8 \ cm$

Therefore,

Lateral surface area

$=2\pi rh =2\times \dfrac{22}{7}\times 8 \times 7$

Lateral surface area

$=352 \ cm^2$

Total surface area

$=2\pi r(r+h)=2\times \dfrac{22}{7} \times 8(7+8)$

Total surface area

$=\dfrac{5280}{7}$

Total surface area

$=754.28 \ cm^2$

The diameter of a cylindrical tin is $80$ $cm$ and height is $1.5$ $m$ . How many square metre will be its curved surface area? $(\pi = 3.14 )$

curved surface area

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

$=2\times 3.14\times 0.8\times 1.5$

$=7.536$ metre square

A cuboid of dimensions $1$ cm $\times 2$ cm $\times 5$ cm, then how many small cubes of dimension $0.75$ cm can be cut.

Cubes that can be cut

$=\cfrac{1\times 2\times 5}{0.75\times 0.75\times 0.75}=23.7$

So,

$23$ cubes.

The area of a rectangular stage is $750$ sq.m if its height is $1.5$ m find its volume.

Area of rectangular stage

$= 750$ sq.m

Height of stage

$= 1.5$ m

Volume of stage

$=$ area

$\times$ height

$= 750 \times 1.5 = 1125$ cubic metres

A small indoor green house (her barium) is made entirely of glass panes (including house) held together with tape.It is $30$ cm long, $25$ cm wide and $25$ cm high. Find the volume of the house.

Given house dimensions

$30$ cm,

$25$ cm,

$25$ cm

It is in the form of cuboid.

Volume of cuboid =

$l \times b \times h$

Volume of cuboid =

$30\times 25 \times 25 = 18750$ cm

$^3$ .

If the dimensions of a room are $4$ m $\times 3.5$ m $\times 3$ m. What is the volume of air contained in the room?

Volume of air contained in the room

$=$ Volume of Room

$=4\times 3.5\times 3\ m^3$

$=42 m^3$

Find the volume of the container of measurement 5.5 m , 4 m ,3 m.

1338064_1277318_ans_1a23233becc4448a9adcd120feced740.PNG

Lateral surface area of a cube whose edge is $3.5cm$ is

Given the side of the cube

$=3.5 cm$

lateral surface of the cube

$4 \times sid{e^2}$

$= = 4 \times {\left( {3.5} \right)^2}$

$= 4 \times 12.5$

$= 49\,c{m^2}$

we get, lateral surface area is

$49\,c{m^2}.$

Find the surface area and lateral surface area of a cube whose edge is $9$ m.

1349751_1298887_ans_900489cbbde24ac2ba02b4c1ecac0a28.PNG

Find the volume of a cuboid which is $15\ \text{m}$ long, $12\ \text{m}$ wide and $4.5\ \text{m}$ high.

$L = 15\ \text{m}$

$B = 12\ \text{m}$

$H = 4.5\ \text{m}$

Volume

$= V = L\times B \times H$

$= 15\times 12\times 4.5\ \text{m}^3 = 810\ \text{m}^3$

Find the volume of a cube if its total surface area is $294\ sq\ cm$ .

Total surface area of cube

$= 6 {a}^{2} = 294 \; {cm}^{2} \quad \left( \text{Given} \right)$

Here

$a$ is the edge length of cube

Therefore,

$6 {a}^{2} = 294\\$

${a}^{2} = \cfrac{294}{6}\\$

$\Rightarrow a = \sqrt{49} = 7 \; cm$

$\therefore$ Volume of cube

$= {a}^{3} \\ = {\left( 7 \right)}^{3} = 343 \; {cm}^{3}$

Find the total surface area of cubes having the following sides.
1) 3 cm
2) 5 cm
3) 5.5 cm

TSA of cube =

$6a^{2}$

1) 3 cm

TSA =

$6(3)^{2} = 54$ sq. cm

2) 5 cm

TSA =

$6(5)^{2} = 150$ sq. cm

3) 5.5 cm

TSA =

$6(5.5)^{2} = 181.5$ sq. cm

The edges of a cuboid are in the ration 3:2:1 and it's surface area is $88\,c{m^2}$ . Find the volume of a cuboid.

Let the edge of a cuboid be

$3x, 2x, x$ .

We know that,

$SA=2(l\times b + b\times h+ h\times l)$

Since,

$2(3x\times 2x+2x\times x+x\times 3x)=88$

$6x^2+2x^2+3x^2=44$

$11x^2=44$

$x^2=4$

$x=2\ cm$

So, the edges are

$6\ cm, 4\ cm, 2\ cm$

We know that the volume of the cuboid

$=l\times b\times h$

Therefore,

Volume

$=6\times 4\times 2$

Volume

$=48\ cm^3$

Hence, this is the answer.

Total surface area of cylinder is $60\pi\ cm^ {2}$ and its height $7\ cm$ . Find the curved surface area of cylinder.

As we know that,

Total surface area of cylinder

$= 2 \pi r h + 2 \pi {r}^{2} = 60 \pi \; {cm}^{2} \quad \left( \text{Given} \right)$

Therefore,

$2 \pi {r}^{2} + 2 \pi \cdot r \cdot 7 = 60 \pi\\$

${r}^{2} + 7r = 30\\$

${r}^{2} + 7r - 30 = 0\\$

${r}^{2} + 10r - 3r - 20 = 0\\$

$\left( r + 10 \right) \left( r - 3 \right) = 0\\$

$\because r \ne -ve\\$

$\therefore r = 3 \; cm$

Therefore,

Curved surface area of cylinder

$= 2 \pi r h \\ = 2 \times \pi \times 3 \times 7 = 42 \pi \; {cm}^{2}$

The adjoining figure shows the cross-section of a concrete wall to be constructed. It is 1.9 m wide at the bottom, 90 cm wide at the top, and 5 m high. If its length is 20 m, find:
(i) the cross-sectional area;
.

Cross sectional area = Area of trapezium =

$\dfrac{1}{2}\times{sum \ of \ parallel \ sides}\times{height}$

$=\dfrac{1}{2}\times{(0.9+1.9)}\times{5}$

$=7 m^2$ {

$100cm = 1m$ }

Calculate the area of the isosceles trapezium drawn above .

Area of Trapezium =

$\dfrac{1}{2}\times$ Sum of Parallel sides

$\times$ Height

$=\dfrac{1}{2}\times(7+3)(4)=20cm^{2}$

$ABCD$ is a trapezium with $AB \parallel CD$ . $AD = BC= 5 cm, AB =12 cm$ and $CD =7 cm$ , find the area of trapezium $ABCD$ .

In trapezium

$ABCD,$

$AB|| CD$ ,

$AD=BC$

Let

$DE$ and

$CF$ be the perpendiculars on

$AB$ from

$D$ and

$C.$

Then as

$DC=EF,EF=7$ cm, therefore

$AE=FB\\=\dfrac{12-7}{2}\\=\dfrac{5}{2}\\=2.5 cm$

Then in triangle

$ADE$ by pythagoras theorem,

$AD^2=AE^2+DE^2$

$5^2=(2.5)^2+DE^2$

$DE^2=18.75$

$DE=4.34 cm$

Therefore height

$=4.34$ cm

Hence area of trapezium,

$=\dfrac{1}{2}\times{Sum \ of\ parallel \ sides}\times{H}$

$=\dfrac{1}{2}\times{(7+12)}\times{4.34}$

$=41.23\ cm^2$

1298745_1371803_ans_ec67e640a9bc4c31a9559a9cd19d1b13.png

If total surface area of a cube is $150\ cm^{2}$ , find the edge.

Given: Total surface area of a cube

$= 150$ cm

$²$

Let

$a$ be the side of a cube

Total surface area of a cube

$= 6a²$

$\Rightarrow 150 = 6a²$

$\Rightarrow a²= \dfrac{150}{6}$

$\Rightarrow a²= 25$

$\therefore\ \ a= 5$

Find the area of the following trapeziums

Given parallel sides of trapezium are 10cm,5cm

Height of trapezium = 6 cm

Area of trapezium with parallel sides a,b and height h = (a+b/2)*h

Area of trapezium = (10+5/2)*6 = (15/2) * 6 = 45 sq.cm

Find the area of the following trapeziums

Given parallel sides of trapezium are

$7$ cm,

$9$ cm

Height of trapezium =

$3$ cm

Area of trapezium with parallel sides

$a$ ,

$b$ and height

$h$ =

$\dfrac{(a+b)}{2}\times h$

Area of trapezium

$= \dfrac{(7+9)}{2}\times 3 = 8 \times 3 =24$ sq.cm.

Find the area of a rhombus whose diagonals are $12,16$ $cm$

The length of diagonals are

$12,16cm$

The area of Rhombus is

$\dfrac 12 (12)(16)=96cm^2$

The volume of cuboid with length breadth and height as $3\ cm,4\ cm,5\ cm$ respectively is

Length of Cuboid

$l =3\ cm$

Breadth of Cuboid

$b =4\ cm$

Height of Cuboid

$h = 5\ cm$

Volume of Cuboid V

$=l \times b \times h \\=3\times 4\times 5\\=60\ cm^3$

The volume of a tank, which is cuboid in shape, is 7.2 m $^3$ . The dimensions of its base are 1.2 m $\times$ 0.6 m. Find the depth of the tank.

The volume of tank is

$7.2m^3$

The dimensions of base is

$1.2m\times 0.6m$

Depth is given by

$\dfrac{7.2}{1.2\times 0.6}\\\dfrac{7.2}{0.72}=10m$

If side of a cube is double then how many times its total surface area will increase?

Let

$a$ be the side of the cube.

Then its total surface area will be

$S_1=6a^2$ .

Now if the sides are doubled then the length of new sides be

$2a$ .

And the total surface area will be

$6\times(2a)^2=4\times 6a^2=4\times S_1$ .

So surface area will be four times the original surface area.

Find the volume of the cuboid whose dimensions are given below.
Length Breadth Height

8.2 m 5.3m 2.6m

Volume of cuboid

$= lbh$

$= 8.2\times 5.3\times 2.6 =112.996 {m}^{3}$

The diagonal of a cube is $16 \sqrt{3}$ cm. Find its surface area and volume.

1319122_1485128_ans_2eb423f4d9374062a90a83823207bd24.jpeg

Find the surface area of a cube whose edge is 14 cm.

We know that the

$TSA$ of a cube

$=6\times side^2 = 6\times 14\times 14 = 1176cm^2$

Find the curved surface area and total surface area of a right circular cylinder of height $15$ cm and whose base radius is $7$ cm.

Radius of cylinder

$( r ) = 7 cm$
Height of cylinder

$( h ) = 15 cm$

Curved Surface Area

$=2 \pi rh$

$=2\times (22/7) \times 7 \times 15$

$= 660cm^2$

Total Surface Area

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

$= 2 \times (22/7) \times 7 ( 15 + 7 )$

$= 2 \times 22 \times 22$

$= 968 cm^2$

Radius of the base of a cylinder is 20 cm and its height is 13 cm. Find its curved surface area of and total surface area.

Radius of the cylinder,

$r=20\ cm$
Height of the cylinder,

$h=13\ cm$

$\therefore$ Curved surface area of the cylinder

$=2\pi rh =2\times 3.14 \times 20 \times 13$

$=1632.80\ cm^2$
Also,
Total surface area of the cylinder

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

$=2\times 3.14 \times 20\times (20+13)$

$=2 \times 3.14 \times 20\times 33=4144.80\ cm^2$
Thus, the curved surface area and the total surface area of the cylinder is

$1632.80\ cm^2$ and

$4144.80\ cm^2$ , respectively.

Find the volume, curved surface area and total surface area of a cylinder whose dimensions are, radius of the base $=7\ cm$ and height $=50\ cm$

We know that volume of the cylinder

$=\pi r^2h$

Here

$r=7\ cm\ h=50\ cm$

So,

$V=22/7\times 7\times 7\times 50$

$V=22\times 7\times 50$

$V=7700\ cm^3$

Also we know that curved surface area of cylinder

$=2\pi rh$

So,

Curved surface area

$=2\times 22/7 \times 7\times 50$

Curved surface area

$=2200\ cm^2$

We know that total surface area of cylinder

$=2\pi (r+h)$

So,

Total surface area

$=2\times 22/7 \times 7(7+50)$
Total surface area

$=2580\ cm^2$

Find the volume, curved surface area and total surface area of a cylinder whose dimensions are radius of the base $=14\ dm$ and height $=15\ m$

We know that volume of the cylinder

$=\pi r^2h$

Here

$r=14\ dm=1.4\ m,\ h=15\ m$

So,

$V=22/7\times 1.4\times 1.4\times 15$

$V=92.4\ m^3$

Also we know that curved surface area of cylinder

$=2\pi rh$

So,

Curved surface area

$=2\times 22/7 \times 1.4\times 15$

Curved surface area

$=132\ m^2$

We know that total surface area of cylinder

$=2\pi (r+h)$

So,

Total surface area

$=2\times 22/7 \times 1.4(1.4+15)$
Total surface area

$=144.32\ m^2$

Find the area of the shaded portion in the following figures.

Area of shaded portion = Area of trapezium - area of rectangle - area of circle

$\Rightarrow$ Area of trapezium =

$\frac{1}{2} \times(sum parallel side) \times height$
=

$\frac{1}{2} \times (120+160) \times 100$
=

$\frac{1}{2} \times 280\times 100$ =

$28000/2$ =

$14000 cm^{2}$

Fill in the blanks statements are true.
The curved surface area of a cylinder is reduced by _________ peer cent if the height is half of the original height.

50%

The area of a trapezium is $475 cm^2$ and the height is 19 cm, then the lengths of smaller side of two parallel sides if one side is 4 cm greater than the other is

$Let\quad the\quad length\quad of\quad one\quad parallel\quad side\quad of\quad trapezium\quad is\quad a.\\ So\quad length\quad of\quad other\quad parallel\quad side\quad =\quad a\quad +\quad 4\\ Given\quad the\quad area\quad of\quad trapezium\quad =\quad 475\quad \& \quad the\quad height\quad =\quad 19\\ Now\quad area\quad of\quad a\quad trapezium\quad =\quad \frac { 1 }{ 2 } \quad \times \quad height\quad \times \quad \left( sum\quad of\quad the\quad parallel\quad sides \right) \\ \Rightarrow \quad \frac { 1 }{ 2 } \quad \times \quad 19\quad \times \quad \left[ a\quad +\quad \left( a\quad +\quad 4 \right) \right] \quad =\quad 475\\ \Rightarrow \quad 2a\quad +\quad 4\quad =\quad \frac { 475\quad \times \quad 2 }{ 19 } \\ \Rightarrow \quad 2a\quad +\quad 4\quad =\quad 25\quad \times \quad 2\\ \Rightarrow \quad 2a\quad +\quad 4\quad =\quad 50\\ \Rightarrow \quad 2a\quad =\quad 50\quad -\quad 4\\ \Rightarrow \quad 2a\quad =\quad 46\\ \Rightarrow \quad a\quad =\quad \frac { 46 }{ 2 } \quad =\quad 23\\ \therefore \quad Length\quad of\quad smaller\quad parallel\quad side\quad of\quad trapezium\quad =\quad 23\\ \& \quad Length\quad of\quad other\quad parallel\quad side\quad of\quad trapezium\quad =\quad 23\quad +\quad 4\quad =\quad 27$

The following figure shows the cross-section ABCD of a swimming pool which is trapezium in shape. If the width DC of the swimming pool is $64$ m, depth (AD) at the shallow end is $80$ m and depth (BC) at the deepest end is $24$ m, find its area of the cross-section( $m^2$ ).

Area of a trapezium

$=\cfrac{1}{2} \times (\text{sum of the parallel sides}) \times height$

Area

$=\cfrac{1}{2} \times (80+24) \times 64$

$=\cfrac{1}{2} \times 104 \times 64$

$= 3328 m^2$

The area of a trapezium is $279$ sq. cm and the distance between its two parallel sides is $18$ cm. If one of its parallel sides is longer than the other side by $5$ cm, find the value of smallest lengths of its parallel sides.

Let the lengths of the parallel sides of the trapezium be $a\ cm$ and $a + 5 \ cm$

Area of a trapezium $= \dfrac { 1 }{ 2 } \times (Sum\quad of\quad parallel\quad sides)\times height$

$\Rightarrow \dfrac { 1 }{ 2 } \times (a + a + 5) \times 18 cm = 279 {cm}^{2}$

$\Rightarrow \dfrac { 1 }{ 2 } \times (2a + 5) \times 18 = 279$

$2a+5=31$

$\Rightarrow a = 13$

Hence, the lengths of the parallel sides of the trapezium are $a = 13 cm$ and $a + 5 = 18 cm$

Thus length of smallest parallel side is $13$ cm

The parallel sides of a trapezium are in the ratio 3 :If the distance between the parallel sides is 9 dm and its area is $126\,\,dm^2$ , find the lengths of smallest its parallel sides.

Let the lengths of the parallel sides of the trapezium be

$3a$

$dm$ and

$4a$

$dm$

Area of a trapezium $= \frac { 1 }{ 2 } \times (Sum\quad of\quad parallel\quad sides)\times height$

$=> \frac { 1 }{ 2 } \times (3a + 4a) \times 9 dm = 126 {dm}^{2}$

$=>\frac { 1 }{ 2 } \times 7a \times 9 = 126$

$=> a = 4$

Hence, the lengths of the parallel sides of the trapezium are $3a = 12 dm$ and $4a = 16 dm$

The two parallel sides and the distance between them are in the ratio 3 : 4 :If the area of the trapezium is $175 \,\,cm^2$ ; find its height.

Let the lengths of the parallel sides of the trapezium be

$3a cm$ and

$4a cm$ . Let the height of the trapezium or the distance between them be

$2a cm$ .

Area of a trapezium $= \frac { 1 }{ 2 } \times (Sum\quad of\quad parallel\quad sides)\times height$

$=> \frac { 1 }{ 2 } \times (3a + 4a) \times 2a = 175 {cm}^{2}$

$=>\frac { 1 }{ 2 } \times 7a \times 2a = 175$

${a}^{2} = 25$

$=> a = 5$

Hence, the height of the trapezium $= 2a = 10 cm$

If $A, B, C$ denote the area of the three continuous faces of a rectangular solid cuboid and $V$ be its volume, then prove that $\displaystyle V=\left ( ABC \right )^{1/2}$ .

Let

$l,~b,~h$ be edges of the rectangular solid cuboid so that

$A=lb,~ B=bh,$ and

$C=lh$
Now

$\displaystyle ABC=lb\times bh\times lh=l^{2}b^{2}h^{2}$
or

$lbh=\left ( ABC \right )^\left(\frac{1}{2}\right)$
Thus volume

$V=\left ( ABC \right )^\left(\frac{1}{2}\right)$
Hence proved

The total surface of a rectangular block is $846$ $\displaystyle \text{cm}^{2}$ . Find the volume, if the dimensions are in proportion $5 : 4 : 3$ .

Let the dimensions be

$5x,4x,$ and

$3x$ .
Total surface area

$=2(lb+bh+hl)$

$\displaystyle =2\left ( 5x\times 4x+4x\times 3x+3x\times 5x \right )cm^{2}$

$\displaystyle =94x^{2}cm^{2}$

Now,

$\displaystyle 94x^{2}=846cm^{2}\left ( given \right )$

$\displaystyle \Rightarrow x^{2}=9$ or

$\displaystyle x=3cm$

$\displaystyle \therefore$ Volume of the block

$\displaystyle =5x\times 4x\times 3x$

$\displaystyle =60x^{3} =1620cm^3$

The three coterminous edges of a rectangular solid are $36\text{ cm}, \ 75 \text{ cm}$ and $80 \text{ cm}$ respectively, find the edge of a cube which will be of the same capacity.

Let the edge of the cube having same capacity be

$a$ .

Volume of the cube

$=$ Volume of the cuboid

$\left ( a\right )^{3}= 36\times 75\times 80$

$=\left ( 3\times 3\times 4\times 5\times 5\times 3\times 4\times 4\times 5 \right )$

$=\left ( 3\times 4\times 5 \right )^{3}$

$a=\left ( 3\times 4\times 5 \right )$

$=60\text{ cm}$

Find the area of a trapezium whose parallel sides are of length 10 cm and 6 cm respectively and the perpendicular distance between the two parallel sides is 4.5 cm. ( in $cm^2$ )

Area of trapezium =

$\displaystyle \frac{1}{2}$ (Sum of parallel sides

$\times$ Distance between them)

$\therefore$ Area =

$\displaystyle \frac{1}{2} (10 + 6) \times 4.5 cm^2$
=

$\displaystyle \frac{1}{2} \times 16 \times 4.5 cm^2$
= 36

$cm^2$
Hence, the area of the given trapezium is 36

$cm^2$

The shape of the top surface of a table is trapezium. Find its area if it's parallel sides are $1\ m$ and $1.2\ m$ and perpendicular distance between them is $0.8\ m$ .

Given:

$a=1m , b=1.2m , h=0.8m$

Area of trapezium

$=\dfrac12\times\text{Sum of Parallel Sides }\times\text{ Distance between parallel sides}$

$= \dfrac{1}{2}\times (a+b)\times h$

$= \dfrac{1}{2}\times (1+1.2)\times0.8 =0.88$ sq. metres

Sohan wants to buy a trapezium shaped field. Its sides along the river is parallel to and twice the sides along the road. If the area of this field is $10500$ sq metre and the perpendicular distance between the two parallel sides is $100$ m , find the length of the side along the river.

Let the length of the sides of the field along the road be

$x m$ .

Then, length of its sides along the river

$=2x\ m$ .

Distance between them

$= 100\ m$ .

Area of trapezium

$=\dfrac {1}{2} × \text {Sum of parallel sides} × \text{Distance between them}.$

$10500= \cfrac{\text{2x}+\text{x}}{2}\times 100$

$105=\cfrac{3\times\text{x}}{2}$

Road side

$x=70\: m$

River side

$= 2\times$ Road side

$=2x=2\times70=140\: m$

The area of the trapezium is $34$ sq cm and the length of one of the parallel side is $10$ cm and its height is $4$ cm. Find the length of the parallel side.

Area of Trapezium

$=\dfrac12 \times\text{Sum of Parallel Sides }\times\text{Distance between parallel sides}$

$\Rightarrow 34 = \dfrac12 \times(10+x)\times 4$

$\Rightarrow x =7$

Length of other parallel side of trapezium is

$7$ cm

Calculate the surface area(in $cm^2$ ) of a solid cylinder with diameter $14$ cm and height $10$ cm. $\left ( \pi = \dfrac{22}{7} \right)$

$S.A.=2\pi r^2+2\pi rh$

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

$=2\times \dfrac{22}{7}\times 7(7+10)$

$=44\times 17$

$={748}\ cm^2$

In a trapezoid of area $20$ , the two bases measure $4$ and $6$ . What is the height of the trapezoid?

Area of trapezoid is

$\dfrac{1}{2} \times h(6+4) = 5h = 20$ (where

$h$ is height of trapezoid)
We get

$h = \dfrac{20}{5} = 4$

Circular discs of $5 mm$ thickness, are placed one above the other to form a cylinder of curved surface area $462 cm^2$ . Find the number of discs, if the radius is $3.5 cm$ .

Thickness of each disc

$= 5 mm = \dfrac{5}{10} cm = 0.5 cm$
Radius of disc

$= 3.5 cm$ .
Curved surface area of cylinder

$= 462 cm^2$

$\therefore 2 \pi rh = 462$ ....... (i)
Let the no of discs be

$x$

$\therefore$ Height of cylinder,

$\ h =$ Thickness of disc

$\times$ no of discs

$= 0.5 x$ .......(ii)

From (i) and (ii) we get

$2 \times \dfrac{22}{7} \times 3.5 \times 0.5 x = 462$

$\therefore x = \dfrac{462 \times 7}{2 \times 22 \times 3.5 \times 0.5} = 42 \ discs$

The length of a cold storage is double its breadth. Its height is 3 metres. The area of its four walls ( including doors ) is 108 m $^2$ . Find its volume.

Given,

$l=2b$ and

$h=3 \ m$

Now, Area of four walls =

$108 \ m^3$

$2(l+b)h=108$

$18b=108$

$b=6 \ m$

Therefore,

$l=2b=12 \ m$

Hence, volume of the cold storage =

$lbh=12 \times 6 \times 3 =216 \ m^3$

A tent is in the form of a cylinder surmounted by a cone having its diameter of the base equal to 24 m. The height of cylinder is 11 m and the vertex of the cone is 5 m above the cylinder. Find the cost of making the tent, if the rate of canvas is Rs. 10 per m $^2$ .

Diametre of base of cylinder = diametre of cone = 24 m

$\therefore$ Radius of base = 12 m
Height of cylindedr = 11 m = h

$_1$
Height of Cone = 5 m = h

$_2$
Let slant height of cone be

$l$

$l = GD = \sqrt{r^2 + h^2} = \sqrt{12^2 + 5^2} = 13 m$
Area of canvas required = CSA of cylinder + CSA of cone

$= 2 \pi r h_1 + \pi rl$

$= \pi r (2h_1 + l)$

$= \dfrac{22}{7} \times 12 (2 \times 11 + 13) m^2$

$= \dfrac{22 \times 12}{7} \times 35 m^2$

$= 22 \times 60 m^2$

$= 1320 m^2$
Rate of canvas = Rs. 10 per m

$^2$

$\therefore$ Cost of canvas = Rate

$\times$ area of canvas

$=Rs. 10 \times 1320$

$= Rs. 13,200$

Find the area of trapezium $ABCD$ as given in the figure in which $ADCE$ is a rectangle.

Given :

BE= 3 cm , EA = 3cm and AD = 8cm

Area of trapezium (ABCD)

$=\dfrac{1}{2}\times$ sum of parallel sides

$\times$ height.

$\Rightarrow$ Area of trapezium ABCD

$=\dfrac{1}{2} \times ( AB + CD )\times AD$

$\Rightarrow$ Area of trapezium ABCD

$=\dfrac{1}{2}\times (6+3)\times 8$

$\Rightarrow$ Area of trapezium ABCD

$=\dfrac{1}{2} \times 9 \times 8$

$\Rightarrow$ Area of trapezium ABCD

$=36$

$m^{2}$

The radius of the base of a right circular cylinder is $10$ $cm$ and height is $28$ $cm$ , find the curved surface area.

The radius of the base of a right circular cylinder is

$10$

$cm$ and height is

$28$

$cm$ .
Here,

$r=10$ cm and

$h=28$ cm.
Curved surface area of cylinder

$= 2 \pi rh$

$= 2 \times \dfrac{22}{7} \times 10 \times 28$

$= 44 \times 40$

$= 1760 {cm}^{2}$ .

$\therefore$ The curved surface area of cylinder is

$1760\ {cm}^{2}$ .

Curved surface area and circumference of the base of a solid right circular cylinder are $4400$ sq. cm and $110$ cm respectively. Find its height and diameter.

It is given that C.S.A. of the cylinder

$=4400$ sq. cm.

We know that C.S.A of a right circular Cylinder

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

$\Rightarrow 2 \pi r h = 4400$

$\Rightarrow (110) h = 4400$

$\Rightarrow h = \dfrac{4400}{110}$

$= 40$ cm
Height

$= 40$ cm
Circumference of the base of the cylinder is

$110$ cm

$\Rightarrow 2\pi r=110$

$\Rightarrow 2\times \dfrac {22}{7}\times r=110$

$\Rightarrow r=\dfrac {35}{2}$

$\Rightarrow r=17.5$ cm

Diameter

$2 r = 2 \left( \dfrac{35}{2} \right ) = 35$ cm

Write the formula to find the total surface area of a right circular cylinder.

Solution:

Total surface area of a right circular cylinder

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

$=2\pi r(h+r)$ square units ;

where,

$h$ is height of the cylinder and

$r$ is radius of the base of the cylinder.

Construct trapezium PQRS with the following measurement. Find also its area.
$\overline{PQ}$ is parallel to $\overline{SR}$ , PQ = 7.5 cm, PS = 6.5 cm, $\angle$ QPS = $85^o$ and $\angle$ PQR = $45^o$

ar PQRS =

$\dfrac { 1 }{ 2 } \left( PQ+SQ \right) \times QT$

$\dfrac { 1 }{ 2 } \left( 7.5+0.5 \right) \times 6.5$

${ 26m }^{ 2 }$

Construct trapezium PQRS with the following measurement. Find also its area.
$\overline{PQ}$ is parallel to $\overline{SR}$ , PQ = 6.5 cm, QR = 7 cm, $\angle$ PQR = $85^o$ PS = 9 cm.

Area of PQRS =

$\dfrac { 1 }{ 2 } \left( PQ+SR \right) \times QR$

$\dfrac { 1 }{ 2 } \left( 6.5+12 \right) \times 7$

$64.75{ cm }^{ 2 }$

Find the volume of cuboid whose dimensions are given by 11 m, 10 m and 7 m.

Given that

$l=11 m, b= 10 m, h = 7 m$
volume

$=lbh$

$=11\times 10\times 7$

$=770 cu. m.$

The dimensions of a cuboid in cm are $40\times 18\times 10$ . Find its volume.

Fact: volume of a cuboid whose dimensions are

$l\times b\times h$ is

$lbh$

Hence volume of cuboid, whose dimensions are

$40\times 18\times 10$ is

$=40\cdot 18\cdot 10=7200$ cm

$^3$

The length , breadth and height of a cuboid are 4 cm , 3 cm and 3 cm respectively, find the volume of a cuboid.

Volume of cuboid

$= l \times b \times h$

$= 4 cm \times 3 cm \times 3 cm$

$=36 m^3$ (cubic metre)

The sea water gets filled up in the salt pan 30 m long, 10 m wide and 10 cm deep. How many litres of sea water is filled in this salt-pan?

We have,

length of salt-pen

$=30m$

Breadth of salt-pen

$=10m$

and, Height of salt-pan

$=10cm$

Now,

Capacity of salt-pen (in lit)

$=$ volume of salt-pen

$= 30 \times \frac{{10}}{{100}} \times 10$

$[\because 100cm=1m]$

$=30$

$m^3$

$1$

$m^3=1000$ L

$=30,000$ litres.

If a cuboidal tank is 20 m in length, 1 m in breadth and 2 m in height, find the volume of the tank in $m^3$

We have given from the question

length of cuboidal tank

$=20m$

breadth of cubodial tank

$=1m$

and,

height of cubodial tank

$2m$

now,

Volume of tank

$= length \times breadth \times height$

$= \left( {20 \times 1 \times 2} \right){m^3}$

$= 40{m^3}$

Hence the volume of tank

$= 40{m^3}$ .

A solid iron rectangular block of dimensions 4.4 m $\times$ 2.6 m $\times$ 1 m is cast into a hollow cylindrical pipe of internal radius 30 cm and thickness 5 cm. Find the length of the pipe.

Given that:

Internal radius

$r = 30$ cm
Thickness

$w = 5$ cm
External radius

$R = 30 + 5 = 35$ cm

Let the length of pipe be

$h$ .

Here,

The volume of cuboid

$=$ Volume of a hollow cylinder

$440 × 260 ×100 = π(R^2– r^2)h$

$⇒ 440 × 260 × 100 = π(35^2 – 30^2)h$

$⇒ 440 × 260 × 100 = π(325)h$

$\Rightarrow h=11204$

The length of the pipe

$h = 11204\ cm = 112.04\ m$

Find the area of a trapezium whose parallel sides are $10$ cm ad $12$ cm long and the distance between them is $4$ cm?

Area of trapezium

$=\cfrac{1}{2}\times$ sum of parallel sides

$\times$ distance between them

$=\cfrac{1}{2}\times (10+12)\times 4=44cm^2$

1169133_879189_ans_297567dd4f1041e2b007bf8333466cda.PNG

If radii of two cylinders are in the ratio $4 : 3$ and their heights are in the ratio $5 : 6$ , find the ratio of their curved surfaces.

$\cfrac { { S }_{ 1 } }{ { S }_{ 2 } } =\cfrac { 2\pi { r }_{ 1 }{ h }_{ 1 } }{ 2\pi { r }_{ 2 }{ h }_{ 2 } } =\cfrac { 4\times 5 }{ 3\times 6 } =\cfrac { 10 }{ 9 } =10:9$

The parallel sides of a trapezium are 25 cm and 13 cm.Its non parallel sides are 10cm each.Find the area of trapezium.

$A=A(I)+A(II)+A(III)$

$=\dfrac { 1 }{ 2 } \times 6\times 8+8\times 13+\dfrac { 1 }{ 2 } \times 6\times 8$

$=24+104+24$

$=152$ sq units.

1222105_890952_ans_604e62a0f3ce47a69d5af9f450158920.png

The difference between two parallel sides of a trapezium is 4 cm.The perpendicular distance between them is 19 cm.Find the area of the trapezium and find the length of parallel sides.

Let the two parallel sides of the trapezium be

$a$ cm and

$b$ cm.

Given that difference between the two parallel sides is 4 and the perpendicular distance

$(h)$ is 19

$\Rightarrow a-b=4\quad \longrightarrow (1)$

We know that Area of Trapezium =

$\dfrac { 1 }{ 2 } \times \left( a+b \right) \times h$

Substituting the values of

$h$ we get,

$\dfrac { 1 }{ 2 } \times \left( a+b \right) \times 19=475$

$\Rightarrow \left( a+b \right) =\dfrac { 475\times 2 }{ 19 }$

$\Rightarrow a+b=50\quad \longrightarrow \left( 2 \right)$

Adding (1) and (2),

$a-b=4$

$\underline { a+b=50 }$

$-2b=-54$

$b=\dfrac { 54 }{ 2 } =27$

Putting

$b=27$ in

$(2)$ we get

$a+27=50$

$a=50-27=23$

Two parallel sides are

$27$ cm and

$23$ cm.

In $\triangle$ ABC, P divides the side AB such that AP:PB = 1 :Q is a point in AC such $PQ \parallel BC$ . Find the ratio of the area of $\triangle$ APQ and trapezium BPQC.

If a line is drawn parallel to one side of a triangle to intersect the

other two sides in distinct points, the other two sides are divided in the same ratio.

$PQ\parallel BC$

$\dfrac{AP}{PB}=\dfrac{AQ}{QC}$

$\angle AQP=\angle ACB$

$\angle APQ=\angle ABC$

So by

$AAA$

$\triangle AQP\sim \triangle ACB$

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Hence

$\dfrac{Area(APQ)}{Area(ABC)}=\dfrac{(AP)^{2}}{(AB)^{2}}$

$\dfrac{Area(APQ)}{Area(ABC)}=\dfrac{(AP)^{2}}{(AP+PB)^{2}}$

$\dfrac{Area(APQ)}{Area(ABC)}=\dfrac{(x)^{2}}{(3x)^{2}}$

$\dfrac{Area(APQ)}{Area(ABC)}=\dfrac{1}{9}$

Let

$Area(APQ)=k$

$Area(ABC)=9k$

$Area(BPQC)=Area(ABC)-Area(APQ)=9k-k=8k$

$\dfrac{Area(APQ)}{Area(BPQC)}=\dfrac{1}{8}$

$\therefore$ the ratio of the

$\triangle APQ$ and trapezium

$BPQC$

$=\dfrac{1}{8}$

Three cubes whose edges measure $3cm,4cm$ and $5cm$ respectively are used to form a single cube. Find its edge. Also, find the surface area of the new cube.

Let

$x$ cm be the edge of the new cube.

Then, the volume of the new cube

$=$ Sum of the volumes of three cubes

$\Rightarrow { x }^{ 3 }={ 3 }^{ 3 }+{ 4 }^{ 3 }+{ 5 }^{ 3 }=216\quad$

$\Rightarrow { x }^{ 3 }={ 6 }^{ 3 }\Rightarrow x=6cm\quad$

The edge of the new cube is

$6cm$ long.

Surface area of the new cube

$=6{ x }^{ 2 }=6\times { 6 }^{ 2 }{ cm }^{ 2 }=216{ cm }^{ 2 }\quad$

In the adjoining figure, $\text{ABCD}$ is a trapezium in which parallel sides are $\text{AB} =78 \text{ cm} ,\text{DC = 52 cm}$ and the non parallel sides are $\text{BC = 30 cm}$ and $\text{AD = 28 cm}$ . find the area of the trapezium.

Let us construct

$=DE~ || ~CB$

$\therefore DC=EB=52\text{ cm}$

$DE=CB=30\text{ cm}$

$AE=AB-EB$

$=78-52=26\text{ cm}$

For

$\Delta ADE,\quad s=\dfrac{28+30+26}2=42$

$\Delta DAE$ is a scalene triangle

$Area (\Delta DAE)=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{42(42-30)(42-28)(42-26)}$

$=\sqrt{42\times12\times14\times16}$

$=\sqrt{\sqrt{6\times7\times6\times2\times7\times2\times4\times4}}$

$=\sqrt{6\times6\times7\times7\times2\times2\times4\times4}$

$=6\times7\times2\times4$

$=336 \text{ cm}^2$

But we know

$Area(\Delta DAE)=\cfrac{1}{2}\times AE\times Height$

$\Rightarrow 336=\cfrac{1}{2}\times26\times h$

Area of trapezium

$ABCD=\cfrac{1}{2}\times h\times(sum\ of\ parallel\ sides)$

$=\cfrac{1}{2}\times\cfrac{336}{13}\times130=1680 \text{ cm}^2$

1099813_1027260_ans_c22cfcc77b764a44b5456900ad0bfb66.png

ABCD is a trapezium in which $AB\parallel DC$ $AB=78cm ,CD=52cm ,AD=28cm$ and $BC=30cm$ find the area of the trapezium.

In trapezium

$ABCD$ construct a line parallel to

$AD = 28\ cm$ meeting

$AB$ at point

$E$ . Than

$EB= AB-AE = 78-52=26\ cm$ .

$\triangle CEB$

$EC=28\ cm,\ CB=30\ cm,\ EB=26cm$

$S= \dfrac{(28+30+26)}{2}=42\ cm$

$Area\ of\ \triangle CEB=\sqrt{s (s-a)(s-b)(s-c)}$

$=\sqrt{42(42-28)(42-30)(42-26)}$

$=\sqrt{42\times14\times12\times16}$

$=336\ cm$

$Area\ of\ \triangle CBE= \dfrac{bh}{2}$

$\dfrac{bh}{2}=336$

$bh=(336)2$

$26(h)=(336)2$ ----- (

$EB=b=26$ )

$h=\dfrac{(336)2}{26}$

$h=\dfrac{(168)2}{13}\ cm$

Area of trapezium

$= \dfrac{(sum\ of\ \parallel\ sides)\ h}{2}$

$=\dfrac{((52+78)\times \frac{168\times2}{13}}{2}$

$=\dfrac{(130\times168\times2}{13\times2}$

$=1680\ sq.cm$

1161813_1025606_ans_30d9082d33534db4839303bddcbc4838.JPG

Praksit make a cuboid of sides 5cm, 2cm, 5cm. How many such cuboid will be needed to form a cube?

Volume of cuboid

$=l\times b\times h=5\times 2\times 5=50{ cm }^{ 3 }$

To make a cube, the volume of cuboid should be a perfect cube. Therefore, the volume of

$50{ cm }^{ 3 }$ should be multiplied by

$2\times 2\times 5$ to make a perfect cube. i.e.,

$50\times 2\times 2\times 5=1000$

To, that 20, cuboids are required to form a cube.

In given figure, AB||DC. Find the value of x.

Since the diagonals of trapezium divide each other proportionally.

$\therefore$

$\dfrac { AO }{ OC } =\dfrac { BO }{ OD }$

$\Rightarrow$

$\dfrac { 3x-19 }{ x-5 } =\dfrac { x-3 }{ 3 }$

$\Rightarrow$

$3(3x-19)=(x-5)(x-3)$

$\Rightarrow$

$9x-57={x}^{2}-8x+15$

$\Rightarrow$

${x}^{2}-17x+72=0$

$\Rightarrow$

$(x-8)(x-9)=0$

$\Rightarrow$

$x-8=0$ or

$x-9=0$

$\Rightarrow x=8$ or

$x=9$

Two parallel sides of a trapezium are $60$ cm and $77$ cm and the other two sides are $15$ cm and $26$ cm. Find the area of the trapezium.

Area

$=\cfrac { 1 }{ 2 }$ (sum of parallel sides)

$\times$ height

$=\cfrac { 1 }{ 2 } \left( 77+60 \right) \times 15$

$=\cfrac { 2055 }{ 2 } \\=1027.5\ cm^2$

959899_1028229_ans_9ea4d1de4d2a487496469e8d7232e25c.png

Find the area of a trapezium $ABCD$ whose parallel sides are $AB = 19\ cm,\ DC = 9\ cm$ and non parallel sides are $BC = 8\ cm$ and $DA = 6\ cm.$

1314989_1056498_ans_ac07ff2950174a0aa653f7dac891848e.jpg

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. (see in Fig.). The length of the entire capsule is $14\ mm$ and the diameter of the capsule is $5\ mm$ . Find the surface area.

Surface area of capsule

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

Daimeter of capsule

$=5mm$

Radius of cylinder

$=$ Radius of hemisphere

$=\dfrac { 5 }{ 2 } =2.5mm$

Height (h)

$=14-2.5-2.5=9mm$

$TSA=2\pi r\left( h+2r \right)$

$=2\times \dfrac { 22 }{ 7 } \times 2.5\left( 9+5 \right)$

$=220 { mm }^{ 2 }$

959915_1030497_ans_711c7e934aee411999d2e45016ecbbba.png

A closed metallic cylindrical box is $1.25\ m$ high and its base radius is $35\ cm$ . If the metal sheet costs $Rs.80$ per ${m}^{2}$ , then find the cost of the material used in the box ?

$total surface area = 2\pi r^2+2\pi rh\\=2\pi r(r+h)\\=2\times 22/7\times 0.35(0.35+1.25)\\=3.52 m^2\\\therefore cost = 801\times 3.52=281.6 Rs$

The dimension of metallic cuboid are $100cm \times 80cm \times 64cm$ . It is method and recast into a cube. find the surface area of the cube.

Let Volume of cube

$=a^3$

After melting the volume will be same

$=100*80*64=2^3*4^3*10^3$

Equating,

$a^3=2^3*4^3*10^3\Rightarrow a=80\ cm$

Surface Area of Cube will be

$=6*80=480\ cm^2$

There is a cylindrical platform of $2m$ radius $50 cm$ height in school playground. What will be the cost of white washing the curved surface of this platform at the rate of $Rs.5$ per $100 sq. cm(\pi = 3.14)$

Given

$\Rightarrow r=200cm$

$\Rightarrow h=50cm$

circumference (c)

$=2\pi r$

$=2\times 3.14\times 200$

$=1256cm$

cylinder surface area

$=c\times h =1256\times 50$

$=62,800 cm^2$

cost of whitewashing

$=628\times 100cm^2$

Given

$12.5$ per Rs-

$100sqcm$

$\Rightarrow 628\times 12.5=7,850$ Rs.

The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are $1$ m and $1.2$ m and perpendicular distance between them is $0.8$ m.

Area of trapezium

$=\dfrac {1}{2}$ (height) (sum of parallel sides)

$=\dfrac {1}{2}(0.8)(1+1-2)$

$=(0.4)(2-2)=0.88\ m^2$

Find the area of a trapezium is $1080\ \text{cm}$ . If the lengths of its parallel sides be $55\ \text{cm}$ and $35\ \text{cm}$ . Find the distance between them.

Area of trapezium,

$A= \dfrac{h(a+b)}{2}$ ,

where

$h$ is the height between the parallel lines,

$a$ and

$b$ are the lengths of the parallel sides respectively.

Given that:

$A=1080\ \text{cm}^2$ ,

$a=55\ \text{cm}$ ,

$b=35\ \text{cm}$

Substituting in above equation, we get:

$h=\dfrac{2 \times 1080}{55+35}\ \text{cm}=24\ \text{cm}$

Find the curved surface area and volume of the cylinder whose height is $21 cm$ and area of its one end is $154\,c{m^2}$

We have,

$h=21\ cm$

$\pi r^2=154$

$r=7\ cm$

We know that the curved surface area of the cylinder

$A_c=2\pi r h$

$A_c=2\times \dfrac{22}{7}\times 7\times 21$

$A_c=22\times 42$

$A_c=924\ cm^2$

We know that the volume of the cylinder

$V=\pi r^2 h$

$V=\dfrac{22}{7}\times 7^2\times 21$

$V=22\times 21\ times$

$V=3234\ cm^3$

Hence, this is the answer.

Find the curved surface area, total surface area, and volume of right angle circular cylinder having radius of the base $3 cm$ and height $7 cm$ .

We have,

$r=3\ cm, h=7\ cm$

We know that the volume of the cylinder

$V=\pi r^2 h$

$V=\dfrac{22}{7}\times 3^2\times 7$

$V=22\times 9$

$V=198\ cm^3$

We know that the curved surface area of the cylinder

$A_c=2\pi r h$

$A_c=2\times \dfrac{22}{7}\times 3\times 7$

$A_c=22\times 6$

$A_c=132\ cm^2$

We know that the total surface area of the cylinder

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

$A_T=2\times \dfrac{22}{7}\times 3^2 +2\times \dfrac{22}{7}\times 3\times 7$

$A_T=56.57+132$

$A_T=188.57\ cm^2$

Hence, this is the answer.

What are the possible expressions for the dimensions of the cuboids whose volume are given?
Volume: $12ky^2+8ky-20k$

here function of volume is in

$y$ &

$k$

so, here Function

$v(k, y)=12\ ky^2+8\ ky-20\ k$

$v(k, y)=4\ k (3y^2)+4k(2y)+4k(-5)$

$=4k(3y^2+2y-5)$

$=4k(3y^2+5y-3y-5)$

$=4k [(y)(3y+5)+(-1)(3y+5)]$

$=(4k)(y-1) (3y+5)$

so, the dimension of the cuboids will be

$4k, y-1$ &

$3y+5$

Ravi made a cuboid of plasticine of dimension $12 cm, 8 cm$ and $3 cm$ . How many times number of such cuboids will be needed to form a cube?

1323407_1079801_ans_12d378fb3e654c0eb995a8866a80ba8d.jpg

The area of a trapezium is $34$ $cm^2$ and the length of one of the parallel sides is $10 cm$ and its height is $4\ cm$ . Find the length of the other parallel side.

Area of trapezium

$=34cm^2$

$=>$ Area of trapezium

$=\dfrac{1}{2}\times (l_1+l_2)\times h$

$=>34=\dfrac{1}{2}\times (10+x)\times 4$

$=>34=\dfrac{1}{2}\times (40+4x)$

$=>34=20+2x$

$=>x=7$

The height and radius of a cylinder are $7.5$ cm and $3.5$ cm respectively. Find the ratio of its total surface area and curved surface area.

Given that

$Radius (r) = 3.5cm$

$Height(h) = 7.5cm$

Curved Surface Area of cylinder

$= 2πrh$

$= 2\times \dfrac{22}{7}\times 3.5\times 7.5$

Curved Surface Area of cylinder

$= 165 cm^2$

Total Surface Area of cylinder

$= 2πr(h+r)$

$= 2\times \dfrac{22}{7}\times 3.5\times (7.5+3.5)$

Total Surface Area of cylinder

$= 242 cm^2$

$\therefore$ The ratio of Total Surface Area and Curved Surface Area of cylinder

$= 242 : 165$

$= 22:15$

Hence, the required ratio is

$22:15$

Form a sheet of $8$ square meter, a closed cylindrical tank is formed which has one meter height and $140\ cm$ diameter. How much sheet will be left after making tank?

Sheet left = Area of sheet - surface are of cylinder

$=8m^2-(2\pi r^2+2\pi rh)$

$8m^2-2\pi r(r+h)$

$8-2\times \dfrac{22}{7}\times \dfrac{140}{2\times 100}\left(\dfrac{140}{2\times 100}+1\right)$

$=8-2\times \dfrac{22}{7}\times \dfrac{140}{2\times 100}\times 1.7$

$=8-7.48\ m^2$

$=0.52\ m^2$

The height of a circular cylinder is $20 cm$ and the diameter of its base is $14 cm$ . Find:
(i) its volume
(ii) its total surface area

Given :

Diameter

$= 14 cm \ \$ Radius

$=\dfrac{14}{2}= 7 \ cm$

Height

$(h)= 20 cm$

Volume

$= \pi r^{2}h=\pi (7)^{2}\times 20\ cm^3=980 \pi \ cm^{3}$

T.S.A

$=2\pi r(r+h)=2\pi \times 7(7+20)\ cm^2$

$=14\pi \times 27\ cm^2$

T.S.A

$=378\pi \ cm^{2}$

If the perimeter of a trapezium-shaped floor is $56$ m, its non-parallel sides are $12$ m each and the altitude is $10$ m, find the area that the floor covers.

The perimeter of floor is

$56m$

The length of non parallel sides is

$12m$

So the sum of lengths of parallel sides is

$(a+b)56-12-12=32m$

The height of trapezium=

$10m$

The area of floor =

$\dfrac 12(a+b)\times h$

$\dfrac 12(32)\times 10$

$=160m^2$

Find the area of the trapezium whose parallel sides are $6cm, 4cm$ and the height is $3cm$ .

Here parallel sides are

$a=6$ cm and

$b=4$ cm and height is

$h=3$ cm

Area of trapezium

$=\dfrac{1}{2}\left(a+b\right)\times h$

$=\dfrac{1}{2}\times \left(6+4\right)\times 3$

$=5\times 3=15cm^2$

$12$ cylindrical pillars of a building have to be cleaned.If the diameter of each pillar is $42$ cm and the height of each pillar is $5$ m, what will be the cost of cleaning these at the rate of Rs. $5$ per sq.m

Diameter of pillar

$=42$ cm

Radius

$=r=\dfrac{42}{2}=21$ cm

$=\dfrac{21}{100}$ m

Curved area of pillar

$=2\pi rh=2\times\dfrac{22}{7}\times\dfrac{21}{100} \times 5=6.6$ sq.m

Total area of

$12$ pillars

$=12\times 6.6$ sq.m

$=79.2$ sq.m

Cost of cleaning for

$1$ sq.m

$=$ Rs.

$5$

Cost of cleaning for

$79.2$ sq.m

$=$ Rs.

$5\times 79.2=$ Rs.

$396$

$\therefore$ Cost of cleaning for

$79.2$ sq.m

$Rs.396$

Two parallel sides of a tranpezium are of lengths $27cm$ and $19cm$ respectively and the distance between them is $14cm$ . Find the area of the trapezium.

Here parallel sides are

$a=27$ cm and

$b=19$ cm and height is

$h=14$ cm

Area of trapezium

$=\dfrac{1}{2}\left(a+b\right)\times h$

$=\dfrac{1}{2}\times \left(27+19\right)\times 14$

$=46\times 7=322$ sq.cm

A trapezium PQRS. Meassuremnt in centimeters are given as shown in the figure. Find the area of the trapezium.

From

$R$ draw

$RT$ perpendicular

$PQ$

$\therefore,\,SR = PT=20$ cm

$TQ=PT-PQ=29-20=9$ cm

$\triangle{TQR},\,\angle{T}={90}^{\circ}$

By using Pythagoras theorem we have

${RT}^{2}={RQ}^{2}-{TQ}^{2}={15}^{2}-{9}^{2}=225-81=144$

$\Rightarrow RT=12$ units

$SP=RT=12$ cm

$\therefore,\,PTRS$ is a rectangle

Area of trapezium

$PQRS=$ Area of rectangle

$PTRS+$ area of triangle

${TQR}$

$=PT\times RT+\dfrac{1}{2}\times TQ\times RT$

$=12\times 20+\dfrac{1}{2}\times 9\times 12$

$=240+54=294$ sq.cms

$\therefore$ Area of trapezium

$PQRS=294$ sq.cms

1334944_1114172_ans_e33f92bf532e44cf802e5d9c6da3ed91.png

A cuboidal water tank dimensions are 6 cm, 8 cm and 10 cm. Find the capacity of the tank.

We have,

$l=10\text{ cm}$

$b=8\text{ cm}$

$h=6\text{ cm}$

Capacity of the tank is nothing but the Volume of the tank i.e.,

$=l\times b\times h$

$=10\times 8\times 6$

$=480\text{ cm}^3$

The adjoining figure shows the cross-section of a concrete wall to be constructed. It is $1.9$ m wide at the bottom, $90$ cm wide at the top, and $5$ m high. If its length is $20$ m, find the cross-sectional area.

REF.Image

Area of crossection =

$A_{1}+A_{2}$

$A_{1}= 0.9\times 5$

$= 4.5 m^{2}$

$A_{2}=\dfrac{1}{2}(1)\times 5= 2.5 m^{2}$

Area of crossection =

$7m^{2}$

1237358_1136994_ans_46f8fec4d4814d7da19be3b265315213.jpg

The area of a trapezium is $540 cm^2$ . If the ratio of parallel sides is $7 : 5$ and the distance between them is $18 cm$ . find the length of parallel sides.

1298831_1143210_ans_b9d75751ba08459098d10cdae9d81897.jpg

The area of a trapezium is $300\ m^{2}$ . The perpendicular distance between the two parallel sides is $15\ m$ . If the difference of the parallel side is $16\ m$ , find the length of the parallel sides.

Area of trapezium is given by

$A=\frac{1}{2}(a+b)h$

If one side is

$a$ then the other side is

$a-16$ since the difference of parallel sides is

$16$

$\therefore$ A

$=\dfrac{1}{2}(a+a-16)\times15=300$

$\Rightarrow\frac{1}{2}(2a-16)\times15=300$

$\Rightarrow(2a-16)\times15=600$

$\Rightarrow(2a-16)=40$ [By dividing both side by 15]

$2a=56$

$a=28 \ m$

The other parallel side

$= a -16$

$a-16=28-16=12$

Therefore, the sides are

$28m$ and

$12m.$

A cubical block of side $7cm$ is surmounted by a hemisphere of the largest size. Find the surface area of the resultant solid.

Consider the problem

The hemisphere can occupy whole of the side of cube.

Therefore, Greatest diameter of hemisphere

$=side\;of\;cube$

$=7cm$

Here, base of hemisphere falls on cube, so that area should not from part of solid.

$Surface\,area\,of\,solid = Area\,of\,cube + curved\,surface\,area\,of\,hemisphere - base\,area\,of\,hemisphere$

Area of cube

$= 6{a^2}$

Here,

$a=side=7cm$

So,

$\begin{array}{l} =6{ \left( 7 \right) ^{ 2 } } \\ =6\times \left( { 7\times 7 } \right) \\ =6\times 49 \\ =294c{ m^{ 2 } } \end{array}$

And,

Curved surface area of hemisphere

$= 2\pi {r^2}$

Diameter of hemisphere

$=7cm$

Therefore, radius is

$\begin{array}{l} r=\dfrac { { diameter } }{ 2 } \\ =\dfrac { 7 }{ 2 } cm \end{array}$

So,

$\begin{array}{l} =2\times \dfrac { { 22 } }{ 7 } \times { \left( { \dfrac { 7 }{ 2 } } \right) ^{ 2 } } \\ =2\times \dfrac { { 22 } }{ 7 } \times \dfrac { 7 }{ 2 } \times \dfrac { 7 }{ 2 } \\ =77c{ m^{ 2 } } \end{array}$

And,

Base area of hemisphere

$= \pi {r^2}$

$\begin{array}{l} =\dfrac { { 22 } }{ 7 } \times { \left( { \dfrac { 7 }{ 2 } } \right) ^{ 2 } } \\ =\dfrac { { 22 } }{ 7 } \times \dfrac { 7 }{ 2 } \times \dfrac { 7 }{ 2 } \\ =\dfrac { { 77 } }{ 2 } c{ m^{ 2 } } \end{array}$

Now,

$Surface\,area\,of\,solid = Area\,of\,cube + curved\,surface\,area\,of\,hemisphere - base\,area\,of\,hemisphere$

$\begin{array}{l} =294+77-\dfrac { { 77 } }{ 2 } \\ =294+77-38.5 \\ =371-38.5 \\ =332.5c{ m^{ 2 } } \end{array}$

Hence, surface area of solid

$= 332.5c{m^2}$

Find the area of a trapezium ABCD whose parallel sides are AB = 19cm, DC = 9cm and non parallel sides are BC = 8cm and DA = 6cm.

Const- Draw a line

$CE$ perpendicular to

$AB$

So,

$DC = AE$

$FB = AB - AE$

$EB = 19 - 9$

$EB = 10\ cm$

$\Delta CEB$ is right angle triangle

$(BC)^2 = (CE)^2 + (EB)^2$

$(8)^2 = (CE)^2 + (10)^2$

$64 = (CE)^2 + 100$

$100 - 64 = (CE)^2$

$\sqrt{36} = CE = 6\ cm$

So, area of trapezium

$= \dfrac{1}{2} (\text{Sum of side}\parallel \text{side}) \times h$

$= \dfrac{1}{2} (9 +19) \times 6\\$

$= \dfrac{1}{2} \times 28 \times 6\\$

$= 14 \times 6$

$= 84 \ cm^2$

1494131_1151666_ans_2e905956ccc74d3fbafe56ffd96f3e98.png

A cylinderical tank has a capacity of $6160\ cu\ m.$ Find its depth if its radius is $14\ m.$ Calculate the cost of painting its curved outer surface at the rate of $Rs.\ 3$ per square metre. $\left(Take\ \pi =\dfrac{22}{7}\right)$

Let the depth of the tank be

$h$ .

Given:-

Volume of tank

$= 6160 \; {m}^{3}$

Radius of tank

$= 14 \; m$

As we know that, volume of tank is given as-

$V = \pi {r}^{2} h$

Therefore,

Volume of tank

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

$\Rightarrow \cfrac{22}{7} \times {\left( 14 \right)}^{2} h = 6160$

$\Rightarrow 616 h = 6160$

$\Rightarrow h = \cfrac{6160}{616} = 10 \; m$

Hence the depth of the tank is

$10 \; m$

Now again as we know that the curved surface area of cylinder is given as-

$A = 2 \pi rh$

Therefore,

curved surface area of tank

$= 2 \times \cfrac{22}{7} \times 14 \times 10 = 880 \; {m}^{2}$

Given that the cost of painting is

$Rs. 3 \text{ per } {m}^{2}$ .

Therefore,

Cost of painting the curved outer surface of tank

$= 3 \times 880 = Rs. 2640$

Hence the cost of painting the curved outer surface is

$Rs. 2640$ .

The radii of two cylinders are in the ratio $3:2$ and their heights are in the ratio $4:5.$ Calculate the ratio of their curved surface areas.

Let

${r}_{1}$ and

${r}_{2}$ be the radii of the cylinders respectively.

$\Rightarrow\,\dfrac{{r}_{1}}{{r}_{2}}=\dfrac{3}{2}$

Let

${h}_{1}$ and

${h}_{2}$ be the heights of the cylinders respectively.

$\Rightarrow\,\dfrac{{h}_{1}}{{h}_{2}}=\dfrac{4}{5}$

Ratio of Curved surface areas

$=\dfrac{2\pi{r}_{1}{h}_{1}}{2\pi{r}_{2}{h}_{2}}$

$=\dfrac{{r}_{1}{h}_{1}}{{r}_{2}{h}_{2}}$

$=\dfrac{{r}_{1}}{{r}_{2}}\times\dfrac{{h}_{1}}{{h}_{2}}$

$=\dfrac{3}{2}\times\dfrac{4}{5}$

$=\dfrac{12}{10}=\dfrac{6}{5}$

Hence,ratio of their curved surface areas is

$6:5$

Biswas made a box 20cm long, 15cm broad and 10cm high but without a lid. How much cardboard did he use for it?

Length of the box

$=20\ cm$

Breadth of the box

$=15\ cm$

Height of the box

$=10\ cm$

Total Surface area

$=2\left(lb+bh+lh\right)$

$=2\left(20\times 15+15\times 10+10\times 20\right)$

$=2\left(300+150+200\right)=2\times 650=1300\,sq.cm$

Area of the lid

$=length\times breadth=20\times 15=300\,sq.cm$

Area of the box without lid

$=1300-300=1000\,sq.cm$

The pillars of a temple are cylindrically shaped. Each pillars has a circular base of radius $20\ cm$ and height $10\ cm$ . How much concrete mixture would be required to build $14$ such pillars?

Considering it to a solid cylinder, for amount of concrete making are cylinder

$=$ volume of cylinder

$\therefore \$ Amount of concrete mixture required to fild

$14$ such pillars

$=14\times$ volume of a cylinder

$-----(1)$

If Volume of cylinder where

$=\pi r^2h$

$x=radius$

$h=height$

Volume

$=\pi \times x^2\times h$

$=\pi \times 20\times 20\times 10\ cm^3$

$=4000\pi \ cm^3$

$=4000\times 3.142\ cm^3$

$=12568\ cm^3$

Substituting it in equation

$(1)$

$=14\times 12568$

$=150816\ cm^3$

$\therefore \$ Amount of concrete mixture required to build

$14$ such pillars

$=150816\ cm^3$

What will happen to the volume of the cube, if its edge is (a) tripled (b) reduced to one-fourth ?

Volume of cube

$\propto\ (edge)^3$

Case 1: If edge is tripled.

$\dfrac{V_{old}}{V_{new}}=\dfrac{(edge_{old})^3}{(3\ edge_{old})^3}$

$V_{new}=27\ V_{old}$

Volume will increase by

$27$ times.

Case 2: If edge is reduced to one-fourth.

$\dfrac{V_{old}}{V_{new}}=\dfrac{\left(edge_{old}\right)^3}{\left(\dfrac{edge_{old}}{4}\right)^3}$

$V_{new}=\dfrac{1}{64}\ V_{old}$

Volume will be decreased by

$\dfrac{1}{64}$ times.

A close cylindrical tank of diameter $14m$ and height $5m$ is made from a sheet of metal. How much sheet of metal will be required.

Given: diameter

$=14m$

$\therefore \$ radius

$=\dfrac {14}{2}=7m$ height

$=5\ m$

$\therefore \$ sheet of metal required

$=$ Total surface area

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

$=2\pi \times 7 (7+5)=2\pi \times 7\times 12$

$=528\ sq.m$

The parallel sides of a trapezium are $23\ cm$ and $13\ cm$ . Its non-parallel sides are $10\ cm$ each. Find the area of trapezium.

Given

$AB=13\ cm, CD=23\ cm, AC=BD=10\ cm$ each

construction Draw

$AP\bot CD$ &

$BQ\bot CD$

Formulae:(i)Area of a trapezium

$=\dfrac{1}{2}\times 15\ cm$ of parallel sides)

$\times$ (perpendicular distance between then)

(ii) Pythagoras theorem

In quad

$ABPQ$

$AB\parallel PQ$

$AP\parallel BQ$ &

$AP=BQ$ {parallel lines} -

$(1)$

$\triangle APC$ &

$\triangle BQD$

$\angle APC=\angle BQD =90^{o}$ {construction}

$AC=BD=10\ cm$ = given

$AP=BQ$ {from

$(1)$ }

$\therefore$ By

$RHS$ rule of congruence,

$\triangle APC\cong \triangle BQD$

Now let

$PC=QD=x$

$\Rightarrow 2x+PQ=CD$

$2x=23-13=10\ cm\Rightarrow x=5\ cm$

$\triangle APC$ &

$\triangle BQD$

$AP$ &

$BQ=\sqrt{(10)^{2}-(5)^{2}}=5\sqrt{3}\ cm$ {Pythagoras Theorem}

Now area of Trapezium

$ABCD$

$=\dfrac{1}{2}\times (AB+CD)\times AP$

$=\dfrac{1}{2}\times (13+23)\times 5\sqrt{3}\ cm^{2}$

$=\dfrac{1}{2}\times 36\times 5\sqrt{3}\ cm^{2}$

$=90\sqrt{3}\ cm^{2}$

Find the area and the perimeter of the trapezium shown above.

REF.Image

Area

$=\dfrac{1}{2}$ (sum of parallel sides)

$\times$ distance between them

$=\dfrac{1}{2}\times (4+5)\times 2.5= 11.25dm^{2}$

Perimeter =

$4+2.5+3+5= 14.5 dm$

1141025_1187262_ans_917e0ebaa3504878a1eea10ce00aad24.png

Find the area of each trapezoid.

Area of a trapezoid

$=\dfrac{1}{2}\left(sum\,\, of\,\, the \,\, parallel \,\, sides \right)\times \left(perpendicular \,\, distance\right)$

$=\dfrac{1}{2}\times\left(6+16\right)\times 6$

$=22\times 3=66$ sq.cm

Find the volume of wood used to make a closed rectangular box of outer dimensions $60 cm\times 45 cm\times 32cm$ , the thickness of wood being 2.5 cm all around. Also find the capacity of the box.

REF.Image.

outer dimensions :

$60\times 45\times 32$

Thickness of wood

$= 2.5 cm$

$l = 60-2(2.5) = 55 cm$

$b = 45-2(2.5) = 45 cm$

$h = 32-2(2.5) = 27 cm$

$\therefore$ The capacity of the box =

$l\times b\times h = (55\times 40\times 27)cm^{3}$

$= 59400 \,cm^{3}$

Total volume of box

$= L\times B\times H$

$= (60\times 45\times 32)cm^{3}$

$= 86400\, cm^{3}$

$\therefore$ Volume of wood used

$=$ Total volume - Actual capacity

$= 86400-59400$

$= 27000\,cm^{3}$

1151118_1201654_ans_d760917b68b84ef4bc2125195c04157f.jpg

$ABCD$ is a trapezium in which $AB\parallel DC$ , AB=29cm, DC=20cm, and height=15cm.Find the area of trapezium.

Area of trapezium

$=\dfrac{1}{2}$ (Sum of parallel sides)

$\times$ height

$=\dfrac{1}{2}\times(29+20)\times15$

$=\dfrac{1}{2}\times49\times15$

$=367.5\ cm^2$

In a building there are $48$ cylindrical pillars.The radius of each pillar is $28$ cm and height is $4m$ .Find the total cost of pointing the curved surface area of all pillars at the rate of $Rs.8$ per $m^2$

$\because r=28cm=0.28m\\ h=4m$

curved surface area of a cylindrical pillars is given as;

$CSA=2\pi rh$

$=2.\dfrac { 22 }{ 7 } .0.28.4$

$=7.04{ m }^{ 2 }\\$

cost of painting the CSA of each pillar

$=8\times 7.04$

$=Rs56.32$

therefore total cost of painting 48 pillars

$=48\times 56.32$

$=Rs2703.36$

From the adjoining sketch, calculate
i) the length $AD$
ii) the area of trapezium $ABCD$
iii) the area of triangle $BCD$

Area of

$ABCD$

area of

$\triangle BCD$

$AD^2=(41)^2-(40)^2=(41+40)(41-40)$

$AD^2=81$

$AD=9$

area of

$ABCD=\dfrac {1}{2}|40+15|\times 9=\dfrac {1}{5}\times 55\times 9\dfrac {495}{2}$

$247.5$

area of

$\triangle BCD=$ area of

$ABCD-$ area of

$\triangle ADB$

$=247.5-\dfrac {1}{2}\times 9\times 20$

$=247.5-180=16.75=167.5$ unit

1353680_1204402_ans_088bf0c73a194222a7c095dac5296740.png

The TSA of cylinder is $231$ $\text { cm } ^ { 2 }$ if the CSA is $\dfrac {2}{3}^{rd}$ of the TSA , find radius and height.

$TSA = (2\pi r^2+2\pi r h ) \>and\>CSA = 2\pi r h \\now (\dfrac{2}{3})(2\pi r^2+2\pi r h )=2\pi r h \\(4\pi r^2+4\pi r h )=6\pi r h \\or\>4\pi r^2=2\pi r h\\2r=h\\now$

$=2\pi rh = (\dfrac{2}{3})231\\$

$=2\pi r \cdot2r = (\dfrac{2}{3})231\\$

$=r^2=(\dfrac{2\cdot 231}{3\cdot 4\pi })=(\dfrac{2\cdot 231 \cdot 7}{4\cdot 22\cdot 3})=(\dfrac{49}{4})\\\therefore r = (\dfrac{7}{2})cm\\and\>hence\>h =2r = 7cm$

Find the curved surface area and the total surface area of a right circular cylinder whose height is 15 cm and the diameter of the cross-section is 14 cm.

Diameter

$=14\ cm$

Radius

$=7\ cm$

Height

$=15\ cm$

So,

TSA of cylinder

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

$=2\times\dfrac{22}{7}\times7(7+15)$

$=2\times\dfrac{22}{7}\times7\times22$

$=968\ cm^2$

CSA of cylinder

$=2\pi rh$

$=2\times\dfrac{22}{7}\times7\times15$

$=660\ cm^2$

Length of the fence of a trapezium shaped field $ABCD$ is $120$ $m$ . If $BC = 48 m , CD = 17 m$ and $A D = 40 m$ , find the area of this field. Side $AB$ is perpendicular to the parallel sides $AD$ and $BC$ .

Perimeter

$=120\ m$ .

$48+17+40+x=120$

$x=15\ m$ .

Area

$= \dfrac{1}{2}$ (sum of parallel sides)

$\times$ height

$= \dfrac{1}{2} (48+40 ) \times 15$

$=660\ m^{2}$ .

1354548_1204661_ans_3e382750dcbe4ba2923fc627dd4c99f7.png

Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is $1.45m$ and its radius is $30cm$ . Find the total surface area of the bird-bath.

Curved surface area of cylinder

$=2\pi rh$

Here,

Height of the cylinder

$h=1.45m=145\ cm$

Radius

$r=30\ cm$

Curved surface area of cylinder

$=2\times\pi \times30\times145$

$=8700\pi\ cm^2$

Curved surface area of hemisphere

$=2\pi r^2$

$=2\times\pi\times30\times30$

$=1800\pi\ cm^2$

Now,

Total surface area of the bird-bath

$=8700\pi+1800\pi$

$=10500\pi$

$=10500\times\dfrac{22}{7}$

$=1500\times22$

$=33000\ cm^2$

$=33000\times\dfrac{1}{100}\times\dfrac{1}{100}$

$=3.3\ m^2$

Hence, surface area of the bird-bath

$=3.3\ m^2$

The mean length of the bases of a trapezium is $150\ cm$ . Find the area of the trapezium if its altitude is $80\ cm$ .

Area of trapezium

$= \dfrac{1}{2}$ (Sum of bases)

$\times$ Height

$= 150 \times 80$

$= 12000 \, cm^{2}$

Water is being poured at the rate of $36 m^3/sec$ in a cylindrical vessel of base radius $3$ meters. Find the rate at which water level is rising.

Volume of cylinder,

$V=\pi r^2h$

differentiating on both sides,

$\dfrac{dV}{dt}=\pi r^2\dfrac{dh}{dt}$

$36= \pi \times 3^2\times \dfrac{dh}{dt}$

$\dfrac{dh}{dt}=\dfrac{36}{9\pi}=\dfrac{4}{\pi}$

height of water level increasing at a rate of

$\dfrac{4}{\pi}$

Two parallel side of a trapezium are $60\ m$ and $77\ m$ and other sides are $25\ m$ and $26\ m$ . Find the area of the trapezium?

Draw a line

$BC$ parallel to

$AD$
Draw a perpendicular line

$BE$ on

$DF$

$ABCD$ is a parallelogram.

$BC=AD=25 \ cm$

$CD=AB=60 \ cm$

$CF=77-CD=17cm$

For

$Δ BCF$

$semiperimeter \ of \ triangle=\dfrac {25+26+17}{2}$

$=\dfrac{68}{2}=34$

By Heron's Formula of triangle

$=\sqrt{ s(s-a)(s-b)(s-c)}$

$=\sqrt{34(34-25)(34-26)(34-17})$

$=204cm²$

Also, area of

$ΔBCF=\dfrac{1}{2}\times base \times height$

$204 =\dfrac{1}{2} BE \times CF$

$204 =\dfrac{1}{2} BE \times 17$

$BE=408/17$

$BE=24 \ cm$

Area of Trapezium

$=\dfrac{1}{2}(AB+DF)\times BE$

$=1/2(60+77)\times 24$

Area

$=1644cm²$

1076753_1212727_ans_c2da7fdc7f3743f6b98bdb21a69192a6.png

In the following table information about a few cylinder is given copy and complete the table
Radius(cm) height(cm) Curved surface Area of the base Total surface
i) 7 9
ii) 14 10
iii) 21 5

Curved surface area of a cylinder

$=2\pi rh$

Base area

$= \pi r^2$

Total Surface area

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

	Radius(cm)	height(cm)	Curved surface Area	Area of the base	Total Surface Area
i)	7	9	396	154	704
ii)	14	10	880	616	2112
iii)	21	5	660	1386	3432

Find the volume of solid.

$l=12$

$b=10$

$h=12$

$\Rightarrow$ Volume of cuboid

$=\left( l\times b\times h\right) \left(unit\right)^3$

$=\left( 12\times 12\times 10\right) \left( unit\right)^3$

$=1440\left( unit \right)^3$

Hence, the answer is

$1440.$

1190703_1218649_ans_2de5e796f1f94fea854588998e47d1c5.png

The total surface area of a cube is $1350\ sq.m$ . Find its volume.

We have

$Tsa$ of Cube

$=6{a}^{2}$

$1350=6{a}^{2}$

$\cfrac{1350}{6}={a}^{2}$

$225={a}^{2}$

$\sqrt{225}=a$

$a=\sqrt{225}$

$A=15cm$

then volume of cube

$={a}^{3}=153$

$3375\,cm^2$

The diameter of a garden roller is 1.4 m and it is 2 m long. How much area will it cover in 5 revolutions ?

${\textbf{Step -1: Calculating the area covered in one rotation}}{\text{.}}$

${\text{To find how much area will be covered in one rotation,}}$

${\text{we have to find the curved surface area of cylindrical roller}}{\text{.}}$

${\text{Given that Diameter, D}} = 1.4{\text{ m}}{\text{.}}$

${\text{and length h}} = 2{\text{m}}{\text{.}}$

${\text{so, radius r}} = \dfrac{{\text{D}}}{2} = \dfrac{{1.4}}{2} = 0.7{\text{m}}{\text{.}}$

${\text{As per formula, curved surface area}} = 2\pi {\text{rh}}{\text{}}$

$= 2 \times \dfrac{{22}}{7} \times 0.7 \times 2$

$= 8.8{{\text{m}}^2}.$

${\textbf{Step -2: Calculating the area covered in 5 revolutions}}{\text{.}}$

${\text{So in 1 revolution total 8}}{\text{.8 }}{{\text{m}}^2}{\text{ area is covered}}{\text{.}}$

$\therefore {\text{ in 5 revolutions total area covered is 8}}{\text{.8}} \times {\text{5 }}{{\text{m}}^2}.$

$= 44{\text{ }}{{\text{m}}^2}.$

${\textbf{Hence, It will cover total 44 }}{{\textbf{m}}^2}{\textbf{ in 5 revolutions}}{\text{.}}$

A box is in the form of a cuboid whose external measures are $50$ cm $\times 40$ cm $\times 25$ cm. Then base, the side faces and back face are to be painted black. Find the area to the painted.

The dimensions of cuboid are

$50\times 40\times 25$

The Lateral surface area of cuboid is given by

$2h(l+b)$

Here

$l=50;b=40;h=25$

Lateral surface area is

$2\times 25(50+40)=50(90)=4500$

And the base area is given by

$lb=(40)(50)=2000$

So The total area is given by

$4500+2000=6500$

Find the surface area of a cubical box whose each side is $1\ m$ , assuming that the box has no lid.

A typical cube has

$6$ faces, cubical box without a lid will have five faces.

Each face is a square of side

$a$ and area of each face

$=a^2$

Surface area of a cubical box

$=5 \times a^{2}\\=5 \times 1^{2}$

$=5\ m^{2}$

The cross section of a canal is trapezium. If width of the upper side is $10$ m and that of base is $6$ m and cross section area of the canal is $72$ $m^2$ , find its depth.

Let depth

$=h$

cross section area

$=72$

$\cfrac{1}{2}h(10+6)=72$

$\cfrac{h}{2}\times 16=72$

$h\times 8=72$

$h=9$

$\therefore\quad depth =9m$

1197318_1240731_ans_7e4f6495a76742fd809f67b2ab6345b7.jpg

Top surface of a table is a trapezium in shape as shown in the figure alongside. Find its area if its parallel sides are $1.5 m$ and $2 m$ and the perpendicular distance between them is $1.2 m$ .

Area of trapezium

$=\left(\dfrac{a+b}{2}\right)\times h$

$h=1.2m$ ;

$a=2m$ ;

$b=1.5$ m

$\left(\dfrac{2+1.5}{2}\right)\times 1.2=2.1m^2$ .

The area of a trapezium is $156\ sq.\ cm$ . If one of the parallel sides is $9\ cm$ and the distance between them is $12\ cm$ , find the length of the other parallel side.

Given:

Area of trapezium

$= 156 \,cm^{2}$

Length of parallel side

$=9$ cm

Distance between parallel side

$=12$ cm.

Area of trapezium

$=$

$\dfrac{one \,parallel\, side + other \,side}{2}\times \$ DIstance between parallel sides

$\therefore A = \dfrac{a+b}{2}\times h$

$\therefore 156 = \dfrac{a+9}{2}\times 12$

$\therefore \dfrac{a+9}{2} = 13$

$\therefore a+9 = 26$

$\therefore a = 17 cm$

Area of a trapezium is $210$ $\text{m}^2$ . If perpendicular distance between the parallel sides is $15$ m and one of these parallel sides be $12$ m, then find its other parallel side.

One parallel side

$=12m$ , distance between parallel side

$=15m$

Area

$={ 210 }m^{ 2 }$

Let

$x$ be other parallel side

$\text{Area of trapezium}=\dfrac {1}{2} × \text {Sum of parallel sides} × \text{Distance between them}.$

Area

$=\dfrac { 1 }{ 2 } \times 15\times \left( 12+x \right) \\$

$210=\dfrac { 15 }{ 2 } \left( 12+x \right)$

$28=12+x$

$x=16m$ (Answer)

The two parallel sides of a trapezium are $19.8\ cm$ and $7.2\ cm$ . If the area of the trapezium is $135\ cm^{2}$ , find the distance between its parallel sides.

two parallel side of trapezium are

$19.8\ cm$ &

$7.2\ cm$

Area of trapezium

$=135 \,cm^{2}$

distance between parallel sides

$= (?)$

$\Rightarrow$ Area of trapezium

$=\dfrac{parallel\, side + other\, side}{2}\times h$

$A = \dfrac{a+b}{2}\times h$

$\therefore 135 = \dfrac{19.8+7.2}{2}\times h$

$\therefore h = \dfrac{135\times 2}{27}$

$\therefore h = \dfrac{270}{27}$

$\therefore h = 10 \,cm$

The volume of a cube is 343 $\mathrm { cm } ^ { 3 } .$ Find its surface area.

$let\>side\>of\>cube\>be\>x\>\\then\>x^3=343\\x^3=7^3\\\therefore\>x=7cm\\now\>surface\>area\>=6x^2\\=6\cdot49\\=294$

The area of a trapezium is $450\ sq. cm$ and the lengths of the parallel sides are $37\ cm$ and $23\ cm$ . Find the distance between them.

Area of trapezium

$= (1/2) \times$ (distance between parallel side)

$\times$ (sum of parallel side)

Let

$h$ be the distance between the parallel sides

$450 = (1/2) \times h \times (37+23)$

$450 = (1/2) \times h \times 60$

$h = \dfrac{450}{30}$

$h = 15$

$cm$

A teak wood log is cut first in the form of a cuboid of length 2.3 m, width 0.75 m and of a certain thickness. Its volume is 1.104 m $^3$ . How many rectangular planks of size 2.3 m $\times$ 0.75 m $\times$ 0.04 m can be cut from the cuboid ?

Number of rectangular planks

$=$ Volume of the cuboid / Volume of a plank

Number of rectangular planks

$=\dfrac{1.104}{2.3 \times 0.75 \times 0.04}$

$=\dfrac{1104}{69}$

$=16$

Find the quantity of oil that can be poured into a container of dimensions 13 cm, 8 cm, 11 cm.

1338059_1277356_ans_7b1ee2e6c65f4380ae153243287061c5.PNG

The curve surface area of a right circular cylinder of height $14\ cm$ is $88\ cm^{2}$ . Find the diameter of the base of the cylinder.

We have,

Curved surface area of cylinder $2\pi rh=88$

And height $h=14\,cm.$

Then,

$2\pi rh=88$

$2\pi r\times 14=88$

$r=\dfrac{88\times 7}{28\times 22}$

$r=1$

Then, Diameter

$d=2r$

$d=2\times 1=2\,cm.$

Hence, this is the answer.

In the adjacent figure the area of rapezium PQRS

$\dfrac { { { { Product } }\, \, { { of } }\, \, { { diagonal } } } }{ { { 2 } } } \\ { { = } }\dfrac { { { { 18\times 6 } }{ { .4 } } } }{ { { 2 } } } { { =57 } }{ { .6 } }\, \, { { c } }{ { { m } }^{ { 2 } } }$

Area of trapezium

$PQRS =$

$\begin{array}{l} area\, \, of\, \, \Delta QTR+area\, \, of\, \, PTRS \\ =\dfrac { 1 }{ 2 } \times 4\times 3+4\times 3 \\ =6+12=18\, \, c{ m^{ 2 } } \end{array}$

Hence, this is the answer.

In a hot water heating system, there is a cylindrical pipe of length $28\ m$ and diameter $5\ cm$ . Find the total radiating surface in the system.

Given,

Diameter of cylindrical pipe

$=5$

$cm$

$=5\times 10^{-2}m$

Length of cylindrical pipe

$=28$ m

Now, curved surface area of pipe

$=2\pi rh$

$=2\pi\times \dfrac{5}{2}\times 28\times 10^{-2}$

$=140\pi \times 10^{-2}m^2$

Total radiating surface of pipe

$=1.40\pi m^2$

1206723_1281066_ans_985e41c822ff4e12b2ef064e11c84946.JPG

A pit $5\ m$ long and $3.5\ m$ wide is dug to a certain depth. If the volume of earth taken out of it is $14\ m^{2}$ . What is the depth of the pit.

R.E.F image

$l = 5 m, w = 3.5 m$

volume of pit

$= 14 m ^{3}$

volume

$= l\times b\times h$

$14 = 5\times 3.5\times h$

$\therefore h = \dfrac{14}{5\times 3.5} = 0.8 m$

$\therefore$ height of the pit

$= 0.8 m$

1162325_1278401_ans_af1d0383a363446c93e4fe8a1770eb6e.jpg

The length of two parallel sides of a trapezium are $21.5\ cm$ and $23.5\ cm$ . If the distance between them is $14\ cm$ , find the area of the trapezium.

Area of a trapezium

$=\dfrac{1}{2}\times$ (Sum of parallel sides)

$\times$ distance between them.

$=\dfrac{1}{2}\times\left(21.5+23.5\right)\times 14$

$=45\times 7\\=315\ sq.cm$

$\therefore$ Area of a trapezium

$=315$ sq.cms

The area of a trapezium is $34\,c{m^2}$ and the length of the parallel sides is $10 cm$ and its height is $4 cm.$ Find the length of the other parallel side $.$

Let the required other side

$=x cm$

Area of the trapezium

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

$($ sum of parallel sides

$)$

$\times$

$($ Height

$)$

$34\,c{m^2} = \dfrac{1}{2} \times \left( {10 + x} \right) \times 4\,c{m^2}$

$or,\,\,\dfrac{{2 \times 34}}{4}\,c{m^2} = 10 + x$

$or,\,\,10 + x = 17$

$\Rightarrow x = 17 - 10 = 7\,cm$

Hence, the required other side

$=7 cm.$

Water is pouring into a cubiodal reservoir at the rate of $60\ litres\ per\ minute$ . if the volume of reservoir is $108\ m^{3}$ , find the number of hours it will take to fill the reservoir.

Given, Rate of flow of water

$= 60$ litres/minute.

Volume of the tank

$= 108 m^{3}.$

$1 m^{3} = 1000$ litres

$108 m^{3} = 108\times 1000 = 108000$ litres.

Time taken to fill the complete tank.

$= 108000/60$

$= 1800$ minutes.

$= 1800/60$ hours.

$= 30$ hours.

$\therefore$ The tank will be filled completely in

$30$ hours.

Area of a trapezium shaped field is $84\ m^{2}$ & height is $7\ m$ if one of them the parallel side is $\dfrac{3}{5}$ of the other find their length.

$A = 84 m^{2}$

h = 7 m

$a = \frac{3}{5}b$

$h = \frac{2A}{a+b}$

$\Rightarrow 7 = \frac{2\times 84}{3/5b+b} \Rightarrow 7\times \frac{8}{5}b=2\times 84$

$\Rightarrow 56 b =840$

$\Rightarrow b = \frac{840}{56} = 15$

$\therefore a= 9 m$

$b = 15 m$

1194773_1295518_ans_7cad2b45f2e14bb1a224dbff73db8c29.jpg

The distance between parallel sides of the trapezium is 12 cm and the distance between the midpoint of other sides is 18 cm. Find the area of the trapezium.

REF.Image

The mid points distance is the average hence,

$\frac{1}{2}\times$ sum of parallel sides = 18 cm

$\therefore 18\times 12=$ Area

Area =

$216 cm^{2}$

1213369_1306472_ans_7d407ebaee8d4dc0aba07cb9965d4cda.JPG

The parallel sides of a trapezium are $25\ cm$ and $11\ cm$ , while its nonparallel sides are $15\ cm$ and $13\ cm$ . Find the area of the trapezium.

$ABCD$ is a Trapezium whose sides

$AB = 25\ cm , BC = 13\ cm , CD = 11\ cm$ and

$DA = 15\ cm$

Now do some construction, Draw a line

$DE$ which is parallel to

$BC$ . Hence forming a parallelogram

$DEBC$ .

Hence

$DE = BC$

Now for finding the area of trapezium we will add area

$\triangle DAE$ to the

area of

$\parallel$ gm.

Now for

$area$ of

$\triangle DAE$

We have to find

$S$ first, then use Heron's formula.

$S = \dfrac{AE + DE + DA}{2} = \dfrac{(25-11) + 13 +15}{2} = \dfrac{42}{2} = 21$

Now Area of triangle

$= \sqrt{s(s - a) (s - b) (s - c)}$

$= \sqrt{21 (21 - 15) (21 - 14) (21 - 13)}$

$= 84\ cm^{2}$

Now draw

$DF$

$\perp$ to

$E$ hence

$\rightarrow$
Area of

$\triangle ADE = \dfrac{1}{2} \times$ Base

$\times$ height

$84\ cm^{2} = \dfrac{1}{2} \times 14 \times h$

$\boxed{h = 12\ cm}$

Now area of parallelogram

$= 11 \times 12 = 132\ cm^{2}$

Area of Trapezium

$= 216\ cm^{2}$

$\left \{i.e, 132 + 84 \right \}$

2093494_1299271_ans_5581d2dfe9304744b0fd4b07b96f6690.png

A show-piece shown in figure is made of two solids - a cube and a hemisphere. The base of the block is a cube with edge $7\ cm$ and the hemisphere fixed on the top has diameter $5.2\ cm$ . Find the total surface area of the piece.

The Surface area of

$5$ sides of cube other than with hemisphere is

$5\times 7\times 7=245cm^2$

The surface area of hemisphere is

$2 \pi\times 2.7\times 2.7=45.78cm^2$

The base area of circle of the hemisphere is,

$\pi r^2=3.14\times 2.7\times 2.7\ cm^2\\\ \ \ \ \ =22.89\ cm^2$

The area of upper face of cube is

$=(7^2-22.89)\ cm^2\\=(49-22.89)\ cm^2\\=26.11cm^2$

So total surface area of object is

$(245+26.11+45.78(\ cm^2=316.89\ cm^2$

The ratio between the curved surface area and the total surface area of a a cylinder is $1:2$ . Find the ratio between the height and the radius of the cylinder

Note: Curved surface area of cylinder

$= 2\pi rh$

Total surface area

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

Given:

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

$\rightarrow \dfrac{h}{h + r} = \dfrac{1}{2}$

$\Rightarrow 2h = h + r$

$\Rightarrow h = r$

$\dfrac{h}{r} = 1$

$\therefore$ required ratio

$h : r = 1 : 1$

The diameter of a garden roller is $1.4m$ and it is $2m$ long. How much area will it cover in $5$ revolution?

Diameter = 1.4 m

$r = \frac{d}{2}$

$r = \frac{1.4}{2}$

$r = 0.7$ m

Curved surface area of cylinder

$= \pi$

$r^2h$

$= \frac{22}{7}\times0.7\times0.7\times2$

$=8.8m^2$

No of revolutions

$= 5$

ATQ,

$=8.8 \times 5$

$= 44 m^2$

The breadth of a cuboid is doubled, the length is increased four times and the height is reduced to one third, find the new volume of the cuboid.

Let, the original length, breadth and height of the cuboid be

$l, b,$ and

$h$ respectively.

Therefore, new length =

$4l$

New breadth =

$2b$

New height =

$\dfrac{1}{3}h$

Initial Volume =

$lbh$

New Volume =

$\dfrac{1}{3} \times 2 \times 4 (lbh)$ =

$\dfrac{8}{3}lbh$

Find the area of the trapezium $ABCD$ in which $AB\parallel DC, AB=18\ cm, \angle B=\angle C=90^{o}, CD=12\ cm$ and $AD=10\ cm$ .

$\begin{array}{l} MD=18-12=6\, \, cm \\ AD=10\, \, cm \\ A{ M^{ 2 } }=A{ D^{ 2 } }-M{ D^{ 2 } }={ \left( { 10 } \right) ^{ 2 } }-{ \left( 6 \right) ^{ 2 } } \\ AM=8\, cm \\ AM=BC=8\, \, cm \end{array}$

area of trapezium

$= \dfrac{1}{2} \times \left( {sum\,\,of\,\,parallel\,\,sides} \right) \times Altitude$

$\begin{array}{l} =\dfrac { 1 }{ 2 } \left( { AB+CD } \right) \times BC \\ =\dfrac { 1 }{ 2 } \left( { 18+12 } \right) \times 8 \\ =\dfrac { 1 }{ 2 } \times 30\times 8=15\times 8=120\, \, c{ m^{ 2 } } \end{array}$

1208989_1311358_ans_a9daad93aa164133a5ae153656af9b90.PNG

A cylinder of circumference 8 cm and length 21 cm rolls without sliding for $4\dfrac { 1 }{ 2 }$ seconds at the rate of 9 complete rounds per second.
The area covered by the cylinder in $4\dfrac { 1 }{ 2 }$ seconds
Area covered by the cylinder in 1 round = its curved surface area

Given: Distance travelled by the cylinder in

$1$ round

$=$ it's circumference

Distance travelled in

$4.5=4.5\ \times$ Distance covered in

$1\ sec$

$=4.5 \times 9 \times 8$

$=324\ cm$

Calculate the area of the trapezium shown in the figure:

$CO^2 = CB^2 - OB^2$

$\implies CO^2 = 17^2 - 8^2$

$\implies CO = 15\text{ cm}$

Area of trapezium

$=\dfrac{1}{2}\times$ altitude

$\times$ sum of parallel sides

$\therefore$ Area

$= \dfrac{CO}{2} (DC + AB)$

Area

$= \dfrac{15}{2} (6 + 6 + 8)$

Area

$= 150 \text{ cm}^2$

2057376_1311085_ans_5315c213a220467a96662226c01993c1.png

A closed wooden box $80\mathrm\ { cm }$ long, $65\mathrm\ { cm }$ wide and $45\mathrm\ { cm }$ high, is made of $2.5\ \mathrm { cm }$ thick wood. Find the capacity of the box and its weight if $100\mathrm\ { cm } ^ { 3 }$ of wood weighs $8\ \mathrm { g }.$

$l'=80-2(2.5)=75$ cm

$b'=60$ cm

$h'=40$ cm

[Dimensions of inside of box]

$v=l'b'h'=(75)(60)(40)$

$=180000cm^3$

$100cm^3\rightarrow 8$ g

$180000cm^3\rightarrow \dfrac{8}{100}\times 180000$

$=14400$ g

Weight

$=14.4$ kg.

1203689_1327052_ans_7b46b460eee94b8780decdeb9057c436.jpg

The Area of trapezium is $16\ { cm }^{ 2 }$ . Length of one parallel side is $5\ cm$ and distance between two parallel sides is $4\ cm$ . Find the length of the other parallel side.

Given area of trapezium is

$16\ { cm }^{ 2 }$ .

Also given one parallel side is

$5$ cm and the distance between the parallel sides is equal to

$4$ cm.

Let the other side be of

$a$ cm.

$\text{Area of trapezium}=\dfrac {1}{2} × \text {Sum of parallel sides} × \text{Distance between them}.$

Substituting the values, we get

$\dfrac{1}{2}\times(a+5)\times 4=16$

$\Rightarrow a+5=8$

$\Rightarrow a=3$ .

Length of the other parallel side is

$3$ cm.

Find the volume of cuboid if its surface area is $208\ sq\ cm$ and the ratio of length, breadth and height is $2:3:4$ .

Let length, breadth and height of cuboid be

$2x, 3x$ and

$4x$ respectively.

Now,

Surface area of cuboid

$= 2 \left( lb + bh + hl \right) = 208 \; {cm}^{2} \quad \left( \text{Given} \right)$

$\therefore 2 \left( 2x \times 3x + 3x \times 4x + 4x \times 2x \right) = 208$

$6 {x}^{2} + 12 {x}^{2} + 8 {x}^{2} = 104$

$\Rightarrow {x}^{2} = \cfrac{104}{26}$

$\Rightarrow x = \sqrt{4} = 2$

As we know that,

Volume of cuboid

$= l \times b \times h \\ = 2x \times 3x \times 4x \\ = 24 {x}^{3} \\ = 24 \times {2}^{3} \\ = 24 \times 8 = 192 \; {cm}^{3}$

Find the area of the base of a box of height $4\ cm$ and lateral surface area $120\ cm^{2}$ , if its length is twice its breadth.

Let

$l$ and

$b$ be the length and breadth of box.

Therefore,

$l = 2b$

Height of box

$\left( h \right) = 4 \; cm$

Lateral surface area of box

$= 120 \; {cm}^{2} \quad \left( \text{Given} \right)$

As we know that,

Lateral surface area of cuboid

$= 2h \left( l + b \right)$

$\therefore 2 \times 4 \left( 2b + b \right) = 120$

$\Rightarrow 3b = \cfrac{120}{8}$

$\Rightarrow b = \cfrac{15}{3} = 5 \; cm$

Therefore,

Length of box

$= l = 2b = 2 \times 5 = 10 \; cm$

Therefore,

Area of base of box

$= l \times b \\ = 10 \times 5 = 50 \; {cm}^{2}$

The lateral surface area of a cylinder of length $14\ m$ is $220\ m^{2}$ . Find the volume of the cylinder.

Lateral surface area of a cylinder of radius r and height h is

$=2\pi rh=220$

Given, length

$=$ height

$=14m$

$\Rightarrow 2\pi r(14)=220$

$\Rightarrow 2\times \dfrac{22}{7}\times r\times 14=220$

$\Rightarrow r=\dfrac{10}{4}=\dfrac{5}{2}m$ .

and its volume

$=\pi r^2h=\dfrac{22}{7}\times \dfrac{25}{4}\times 14=275m^3$ .

1189694_1344472_ans_0a4923aa6d44455292ac4fce6762f83a.jpg

A village, having a population of $4000$ , requires a $150$ litre water per head per day. It has a tank measuring $20m\times25m\times15m$ . For how many days will the water of this tank last?

Tank measurement

$20m×25m×15m.$

Volume of water in tank

$=20\times 25\times15$

$=7500m^{3}$

$=7500000L$ ..........

$[1m^{3}=1000L]$

Water required

$=4000\times150L$

$= 600000L$

No of days

$=\dfrac{7500000}{600000}=12.5 days.$

The total surface area of a cubical box is $486 \ cm^2$ . Find the volume of the box.

Given that, the total surface area of a cubical box is

$486\ cm^{2}$

To find out: Volume of the box.

Let

$a$ be the side of the cubical box.

We know that the total surface area of a cuboid is

$6a^2$

$\therefore \ 6a^{2} = 486$

$\Rightarrow a^{2} = \dfrac{486}{6}$

$\Rightarrow a^{2} = 81$

$\therefore \ a = 9\ cm$

We also know that, the volume of a cuboid is

$a^{3}$

$\therefore \$ Volume of the box

$= 9^{3}$

$\Rightarrow 729 \ cm^{3}$

Hence, the volume of the given cuboidal box is

$729\ cm^3$ .

Find the cuboidal box if $l=15 \ cm;b=8\ cm;h=6\ cm$ .

Length of box Is

$=15cm$

Breadth of box Is

$=8cm$

Height of box Is

$=6cm$

Volume of box

$lbh=15(8)(6)\\=120(6)=720cm^3$

The circumference of the base of a cylinder is $132\ cm$ and its height is $72\ cm$ . Find its lateral surface area.

Base

$= 5x=50$ m & height

$= 2x=20$ m

CSA

$=2\pi rh=(2\pi r)\times h=132\times 72$

$= 9504 cm^{2}$ .

1187818_1353346_ans_414d888af2134aa698228ec26810d067.jpg

Two cubes have their volumes in the ratio 8:The ratio of their curved surface area is:

Let the Sides of cube be

$a,b$

Volume of cubes are given by

$a^3,b^3$

Ratio of Volumes is given by

$a^3:b^3=8:27$

$\left(\dfrac{a}{b}\right)^3=\left(\dfrac{2}{3}\right)^3$

$\dfrac{a}{b}=\dfrac{2}{3}$

$a=\dfrac{2}{3}b$

The surface area of Cubes is given by

$6a^2,6b^2$

$=6a^2:6b^2$

$=a^2:b^2$

Put

$a=\dfrac{2}{3}b$

$=\left(\dfrac{2}{3}b\right)^2:b^2$

$=\dfrac{4}{9}:1$

$=4:9$

Find the total surface area of the following cylinders

$\textbf{Step -1: Find the total surface area of first cylinder.}$

$\text{Radius, }r=14\text{ cm}$

$\text{Height, }h=8\text{ cm}$

${\because\text{Total surface area of cylinder}=\pi r (2h+r)}$

$\therefore \text{Total surface area of first cylinder}=\pi (14)(2\times8+14)$

$=14\pi\times30$

$=420\pi$

$=420\times\dfrac{22}{7}$

$=1320\mathrm{ cm^2}$

$\textbf{Step -2: Find the total surface area of second cylinder.}$

$\text{Diameter, }d=2\text{ m}$

$\text{Radius, }r=1\text{ m}$

$\text{Height, }h=2\text{ m}$

$\therefore \text{Total surface area of first cylinder}=\pi (1)(2\times2+1)$

$=\pi\times5$

$=\dfrac{22}{7}\times5$

$=\dfrac{110}{7}\mathrm{m^2}$

$\textbf{Hence, the total surface area of first and second cylinders are }\mathbf{1320 cm^2\text{ and }\dfrac{110}{7}m^2\text{ respectively.}}$

The area of a trapezium is $540\ cm^{2}$ . If the ratio of parallel sides is $7:5$ and the distance between them is $18\ cm$ , find the lengths of parallel sides.

Area of trapezium = 540

$cm^{2}$

Let the paraliel sides be 7x & 5x

& height = 18 cm

540

$= \frac{1}{2}h (a + b)$ where a,b are sids of trapezium

$\Rightarrow 1080 = 18 (7x+5x)$

$\Rightarrow 60 = 12 x$

$\Rightarrow x = 5$

Sides = 25 , 35

1185810_1349019_ans_ff1208ac55f048b6bb371f4c617b0380.jpg

The value of a cuboid is $972\ m^{2}$ . If the length and breadth be $16\ m$ and $13.5\ m$ respectively, find the height.

volume of cuboid be l b h

v = (16) (13.5) h

$h = \frac{972}{16\times 13.5} = 4.5$ m

1189493_1354497_ans_53f72e71708f43a09f18d9043112ec20.jpg

The area of a trapezium in $390\ sq.\ cm$ . One of its parallel sides is $5\ cm$ longer than the other and the distance between them is $12\ cm$ . Find the length of the parallel sides of the trapezium.

Let height be

$h=12$ cm, Area

$=390$ sq.cm

Let smaller parallel side be

$x$ and the other side be

$x+5$

So, Area

$=\dfrac{1}{2}h\left(x+x+5\right)$

$\Rightarrow 390=\dfrac{12}{2}\left(x+x+5\right)$

$\Rightarrow 390=12x+30$

$\Rightarrow 12x=360$

$\therefore x=\dfrac{360}{12}\\\ \ \ \ \ \ \ =30$

$\therefore$ Smaller side

$=x=30$ cm

other side is

$=x+5\ cm\\=30+5\\=35\ cm$ cm

Hence length of parallel sides are

$30$ cm and

$35$ cm.

Find the volume of the following cuboids

Given length of cuboid

$=8$ cm

Breadth of cuboid

$=3$ cm

Height of cuboid

$=2$ cm

Volume of cuboid

$=lbh=8\times 3\times 2$

$[$ Volume

$=48cm^3]$ .

1183016_1358161_ans_9fc2611c190c4b36bac5e75f4fec00af.JPG

A cylindrical pillar is $50\ \text{cm}$ in diameter and $3.5\ \text{m}$ in height. Find the cost of painting the curved surface of the pillar at the rate of $\text{Rs.}\ 12.50$ per $\text{m}^{2}$ .

Diameter

$=D=50\ \text{cm}$

Radius of cylinder

$=r=\dfrac{D}{2}=\dfrac{50}{2 }\ \text{cm}=25\ \text{cm}=0.25\ \text{m}$

Height of cylinder,

$h= 3.5\ \text{m}$

Curved Surface Area of cylinder

$=2\pi rh$

$=2\pi (0.25\times 3.5)\ \text{m}^2=5.5\ \text{m}^2$

Cost of painting per meter square

$= \text{Rs.}\ 12.50$

Total cost of painting

$5.5\ \text{m}^2=\text{Rs.}\ 12.50\times 5.5=\text{Rs.}\ 68.75$

A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? $\left( 1{ m }^{ 3 }=1000 l \right)$

Length of cuboidal water tank

$l= 6 cm$
breadth

$b = 5 m$
height

$h = 4.5 m$

$\therefore$ volume of cuboidal water tank

$V = l \times b \times h$

$= 6 \times 5 \times 4.5$

$V = 1335 m^3$

$1 m^3 = 1000 l$

$135 m^3 = 135 \times 1000$

$= 135000 lit$

$\therefore$ Number of litres of water tank

$= 1350000 Litres$

Find the area of trapezium $ABCD$ .

1211732_1362963_ans_651670c624ca4c9c9aa522371f27d2dd.PNG

Find the volume of the rectangular solids whose dimensions are as follows:
(a) Length = 15 m, Breadth = 12 m and Thickness = 75 cm
(b) Length = $12\frac{1}{2}$ m, Breadth = 5.2 m and Thickness = 80 cm

Given are two sets of dimensions of a rectangular solid.

To find out: Volume of the solid.

$(a)$ Given dimensions are:

$l=15\ m,\ b=12\ m$ and thickness

$=75\ cm$ .

Hence, volume of the solid

$=l\times b\times \text{thickness}$

$=15\times 12\times 0.75\quad \quad [75\ cm=0.75\ m]$

$=180\times 0.75$

$=135 \ m^3$

$\therefore \$ Volume of the solid

$=135 \ m^{3}$

$(b)$ Given dimensions are:

$l=12\frac12=\frac{25}{2}\ m,\ b=5.2\ m$ and thickness

$=80\ cm=0.80\ m$ .

Hence, volume of solids

$=l\times b\times \text{thickness}$

$=\dfrac{25}{2}\times 5.2\times 0.80\\$

$=12.5\times 4.16\\$

$= 52\ m^3\\$

$\therefore \$ Volume of the solid

$=52 \ m^{3}$ .

The area of a trapezium is $432 m^{2}$ the perpendicular distance between the two parallel sides is $18m$ . If the difference of the parallel sides is $18 m$ . Find the length of the parallel sides.

Let the parallel sides be

$x$ and

$x+18$

Area

$=432{m}^{2}$

$\Rightarrow \dfrac{1}{2}\left(x+x+18\right)\times 18=432$

$\Rightarrow \dfrac{2x+18}{2}=\dfrac{432}{18}=24$

$\Rightarrow x+9=24$

$\Rightarrow x=24-9=15$ m

$\therefore$ one side is

$15$ m and the other side is

$15+18=33$ m

A closed box is made up of wood 2 cm thick and its length, width and height externally and 80 cm, 60 cm, 50 cm. Find the volume of wood.

volume of wood

$=80\times 60\times 50=240,000‬{cm}^{3}$

Dimensions of inner wood because to calculate the volume of wood required we have to subtract the inner dimensions from outer dimensions.

Thickness of wood

$=2$ cm

Inner dimension

$=\left(80-4\right)\times\left(60-4\right)\times\left(50-4\right)$

$=76\times 56\times 46=1,95,776 \ {cm}^{3}$

volume of wood

$=2,40,000‬-1,95,776=44,224\ {cm}^{3}$

The dimensions of a car petrol tank are $50$ cm $\times$ $32$ cm $\times$ $24$ cm, which is full of petrol. If car's average consumption is $15$ km per litre, find the maximum distance that can be covered by the car.

Given dimension of petrol tank =

$50 cm \times 32 cm \times 24 cm$

Volume of tank =

$50\times 32\times 24=38.400 cm^{3}$

We know, 1 liter =

$100 cm^{3}$

Now volume of tank = 38.4 L

Given average consumption of car = 15 km/L

Now

total distance covered is given volume

$=15\times 38.4$

Distance = 576 km.

1202249_1384535_ans_c0555746062142c18e8e9c6e0d7602fe.JPG

A goodown measure $40m\times25m\times15m$ . Find the maximum number of wooden crates each measuring $1.5m\times1.25m\times0.5m$ that can be stored in the godown.

No of wooden crates

$=\frac{volume of godown}{volume of crate}$

$=\frac{40\times25\times16 }{1.5\times1.25\times0.5}$ = 16000 crates.

1190702_1377207_ans_96d6cf0d5be748b891ef12a560e9a327.jpg

The internal measurements of a box are $20$ cm long, $16$ cm wide and $24$ cm high. How many $4$ cm cubes could be put into the box?

Given Dimensions of a box are $$20$cm long.

$16$ cm wide and

$24$ cm high.

Volume of box

$=$ lbh

$=16cm\times 24cm\times 20cm$

$=320\times 24$

Volume of box

$=7680cm^3$

Given edge of cuboidal box

$=4cm$

Volume of

$1$ box

$=(4)^3=64cm^3$

No. of boxes fit inside cuboidal box

No. of boxes

$=\dfrac{7680}{64}$

No. of boxes fit inside large box is

$120$ .

1204181_1393515_ans_106d148809e5413283fae74c77b18872.JPG

The curved surface area of a right circular cylinder is $660\ \text{cm}^{2}$ and base radius is $5$ cm. Find its height.

Curved surface area of cylinder with radius

$r$ and height

$h$ is

$=2\pi r h$

$\Rightarrow 2\pi rh=660$

$\Rightarrow h=\dfrac{660\times 7}{2\times 22\times 5}=21$ cm.

The outer dimensions of a closed wooden box are $10\ cm$ by $8\ cm$ by $7\ cm$ Thickness of the wood is $1\ cm$ . Find the total cost of wood to make box, if $1\ cm^{3}$ of wood cost $Rs.2.00$ .

The outer dimension of Closed window is

$10cm;8cm;7cm$

Width of wooden box is

$1cm$

The inner dimensions are given by

$8cm;6cm;5cm$

The volume occupied by Wooden box is given by

$\text{Volume of box with outer dimensions-Volume of box with inner dimensions}\\10\times 8\times 7-8\times 6\times 5\\560-240\\320cm^3$

Cost of

$1cm^3$ is

$2Rs.$

Cost of

$320cm^3$ is

$2\times 320=640Rs.$

The length of the fence of a trapezium- shaped field ABCD is 130 m and side AB is perpendicular to each of the parallel sides AD and BC. If BC=54 m, CD=19 m and AD=42 m, find the area of the field.

The perimeter of field is

$130$

$\implies AB+BC+CD+AD=130\\AB=130-54-19-42\\AB=15m$

Area is given as

$\dfrac 12 AB (AD+BC)\\\dfrac 12 \times 15(54+42)\\15\times 48=720m^2$

Find the area of trapezium whose parallel sides are $25$ cm, $13$ cm and other sides are $15$ cm and $15$ cm.

1196642_1397866_ans_ddf99ebd9d1c4668bc11ae0eee7c3ba0.jpg

The volume of a 343 $cm^{3}$ . Find the length of an edge of cube.

Given

Volume of a cube

$=343{ cm }^{ 3 }$

Let's consider

$a$ to be the edge of the cube.

Then,

$V={ a }^{ 3 }=343={ 7 }^{ 3 }\quad$

$\therefore a=7cm$

Calculate the area of a trapezium whose length of the parallel sides are $30\text{ cm}$ and $20\text{ cm}$ and distance between the parallel lines is $15\text{ cm}$ .

Area of a trapezium

$=\dfrac{1}{2}\times$ (distance between two parallel sides)

$\times$ (sum of two parallel slides)

$=\dfrac{1}{2}\times 15\times (30+20) \text{ cm}^2$

$=\dfrac{1}{2}\times 15\times 50$

$=\dfrac{750}{2}= 375 \text{ cm}^2$

$PQRS$ is an isosceles trapezium. $l\left(PQ\right) = 7$ cm. Segment $PM$ is perpendicular to segment $SR$ , $l\left(SM\right) = 3$ cm, Distance between two parallel sides is $4$ cm, Find the area of $PQRS.$

Length of

$PQ =7$ cm

Length of

$PM =$ distance between two parallel sides

$= 4$ cm

Length of

$SM = 3$ cm

Length of

$SR = SM + MN + NR$

because

$PQRS$ is isosceles trapezium,

so,

$SM = NR = 3$ cm and

$MN = PQ = 7$ cm

so, Length of

$SR = 3$ cm

$+ 7$ cm

$+ 3$ cm

$= 13$ cm

Now, area of trapezium

$PQRS$

$=\dfrac{1}{2}\left(PQ+SR\right)\times PM$ sq.cm

$=\dfrac{1}{2}\left(7+13\right)\times 4$ sq.cm

$=20\times 2$ sq.cm

$=40$ sq.cm

Find the lateral or curved surface area of a closed cylindrical petrol storage tank that is $4.2\ m$ in diameter and $4.5\ m$ high.

REF.Image

Height (h) of a cylinder tank

$= 4.5$ m

Radius (r) of the circular end of cylindrical tank =

$\dfrac{4.2}{2}m$

$=2.1 \ m$

Curved surface Area of tank =

$2\pi rh$

$=\left(2.\dfrac{22}{7}.\dfrac{21}{10}.\dfrac{45}{10}\right)m^{2}$

$=\left(\dfrac{297}{5}\right)m^{2}$

$= 59.4 m^{2}$

1201497_1411318_ans_bf38c7466bf04ef5aeb7afd1ea487ec0.JPG

In the figure, $AB\parallel DC$ is perpendicular to $AB.$ Further, $DC=7$ cm, $CB=10$ cm and $AB=13$ cm. find the area of the trapezium $ABCD$ .

1213593_1437338_ans_b3621ae7aaac4113b2d9c02583220fd6.jpg

What will happen to the volume of a cube if the length of its edge is reduced to half ? Is the volume get reduced? If yes, how much?

$\textbf{Step-1: Finding the new volume}$

$\text{Let the original side of the cube be}$

$a.$

$\text{Hence, the new side will be}$

$\dfrac{a}{2}.$

$\text{Volume of a cube}$

$V=(side)^3$

$V=\dfrac{a^3}{8}$

$\textbf{Step-2: Finding the decrease in volume}$

$\text{The original volume of the cube was}$

$a^3$

$\text{Change in volume}$

$=a^3-\dfrac{a^3}{8}$

$\text{Change}$

$=\dfrac{7a^3}{3}$

$\textbf{Hence, when the side of a cube is halved its volume reduces by}$

$\mathbf{\dfrac{1}{8}.}$

If the areas of three adjacent faces of a cuboid are 8 ${ cm }^{ 2 },$ $\quad 18\quad { cm }^{ 2 },$ and $\quad 25\quad { cm }^{2 },$ Find the volume of the cuboid .

Given:

${A}_{1}=lb=8$ sq.cm

${A}_{2}=bh=18$ sq.cm

${A}_{3}=lh=25$ sq.cm

$\Rightarrow lb\times bh\times lh=8\times 18\times 25$

$\Rightarrow {\left(lbh\right)}^{2}=2\times 2\times 2\times 3\times 3\times 2\times 5\times 5$

$\Rightarrow lbh=\sqrt{\left(2\times 2\times 2\times 3\times 3\times 2\times 5\times 5\right)}$

$\therefore lbh=2\times 2\times 3\times 5=60$ cubic cm

$\therefore$ Volume of the cuboid

$=lbh=60$ cubic cm

The diameter of a roller is $84$ cm and it's length is $120$ cm. It takes $500$ complete revolutions to roll once the play ground to level. Find the area of the play ground in $m^2$ .

Roller is of the form cylinder

Radius

$=\dfrac{Diameter}{2}=\dfrac{84}{2}=42$ cm

Area of roller

$=$ curved surface of cylinder

$=2\pi rh$

$=\left(2\cdot \dfrac{22}{7}\cdot 42\cdot 120\right)cm^2$

$=31680cm^2$

Area of

$1$ revolution

$=$ Area of roller

$=31680 cm^2$

Area of

$500$ revolution

$=500\times 31680cm^2$

Area of playground

$=15840000cm^2$

$=15840000\times \dfrac{1}{100}\times \dfrac{1}{100}m^2$

$=1584m^2$ .

The area of a trapezium with equal non-parallel sides is $168{ m }^{ 2 }$ . If the lengths of the parallel sides are 36m and 20 m, find the length of each non-parallel side.

$\dfrac{1}{2} \times$ (sum of 11 sides)

$\times$ height

$= 168$

$\therefore 336 = (36 + 20) \times height$

$\therefore height = \dfrac{336}{56} = 6m$

$\therefore$ length of non parallel side

$= \sqrt{6^2 + 8^2}$

$= \sqrt{100}$

$\therefore$ Answer

$= 10 cm$

1309506_1442223_ans_f6b80651e00e44529d888c182797be23.png

The cross- section ABCD of a swimming pool is a trapezim.It's width AB =14 m depth at the shallow end is 1.5 m and at the deep end is 8 m Find the area of the cross section.

From the figure,Area of cross section=area of

$rectangle \ ABED+$ area of

$\triangle DEC$

$=14\times 1.5+\cfrac { 1 }{ 2 } \times 6.5\times 14$

$=14\times \cfrac { 3 }{ 2 } +\cfrac { 1 }{ 2 } \times 6.5\times 14$

$=21+7\times 6.5$

$=21+45.5$

$\therefore$ Area

$=66.5 \ m^2$

1235417_1525590_ans_24dcdcc075cd46d4887f4f2d15b4edd3.JPG

An iron pipe $10\ cm$ long has a diameter equal to $7\ cm.$ Find the total surface area of the pipe.

Pipe is an open cylinder,

$\therefore$ Total surface area

$= 2\pi rh$

$= 2\times \dfrac{22}{7}\times \dfrac{7}{2}\times 10\,{cm}^{2}$

$= 220\,cm^{2}$

If the volume of a cylinder is $3080\text{ cm}^3$ and the base radius is $7$ cm, find the curved surface area of the cylinder.

Given,

Volume of the cylinder

$=3080\ cm^3$

Radius of base

$r=7\ cm$

Let the height be

$h.$

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

Volume of cylinder

$=3080\ cm^3$

$\pi r^2h=3080$

$\Rightarrow \dfrac{22}{7}\times 7\times 7\times h=3080$

$\Rightarrow 154\times h=3080$

$\Rightarrow h=\dfrac {3080}{154}$

$\Rightarrow h=20$

Height of the cylinder

$h=20\ cm$

Curved surface area of the cylinder

$=2\pi rh$

$=2\times \dfrac{22}{7}\times 7\times 20$

$=2\times 22\times 20$

$=880cm^2$

Hence, the curved surface area of the ccylinder is

$880\ cm^2$

Ratio of $CSA$ & $TSA$ of cylinder is $1:2$ , find volume If $T.S.A\ 616\ cm^2$ .

As we know that,

TSA of cylinder

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

CSA of cylinder

$= 2 \pi rh$

Volume of cylinder

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

Now,

$\cfrac{CSA}{TSA} = \cfrac{1}{2} \quad \left( \text{Given} \right)$

$\because TSA = 616 \; {cm}^{2} \quad \left( \text{Given} \right)$

$\therefore CSA = \cfrac{1}{2} \times 616$

$\Rightarrow CSA = 308 \; {cm}^{2}$

Now,

$TSA = CSA + 2 \pi {r}^{2}$

$616 - 308 = 2 \pi {r}^{2}$

$\Rightarrow {r}^{2} = \cfrac{308 \times 7}{22 \times 2} = 49 \; cm^2$

$\Rightarrow r = 7 \; cm$

Also,

$2 \pi rh = 2 \pi {r}^{2} = 308 \; cm^2$

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

$\Rightarrow r = h$

Therefore,

Volume of cylinder

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

$= \cfrac{22}{7} {r}^{3} \quad \left( \because r = h \right)$

$= \cfrac{22}{7} \times {\left( 7 \right)}^{3} = 1078 \; {cm}^{3}$

The shape of the cross-section of a canal is a trapezium if the canal is $10$ m wide at the top, $6$ m wide at the bottom and the area of its cross-section is $72\text{ m}^{2 },$ find the depth of the canal.

1670482_1463204_ans_fc481cc5faac4262abdac138c417d41e.jpg

The area of trapezium is $34{ cm }^{ 2 }$ and the length of one of the parallel sides is 10 cm and its height is is 4 cm. Find the length of the other parallel side.

1318649_1490383_ans_d9bf9a824d4240a2aa50f6513d7a6fb3.JPG

The parallel sides of a trapezium are 25 cm and 13 cm;its non-parallel sides are equal,each being 10 cm. Find the area of the trapezium.

1894534_1468297_ans_88630dde82784c379107ed85c5c8f769.jpg

The length , breadth and height of a rectangular side are in the ratio $5:4:2$ . If its total surface area is $1216 cm^2$ , find the volume of the solid.

Given that the ratio in length, breadth and height of a rectangular solid

$=5:4:2$

Total surface area

$1216{cm}^{2}$

Let's assume the length

$=5x$ , breath

$=4x$ , height

$=2x$

Total surface area

$=2\left[ 5x\times 4x+4x\times 2x+2x\times 5x \right] =\left[ 20{ x }^{ 2 }+8{ x }^{ 2 }+10{ x }^{ 2 } \right] =2\times 38{ x }^{ 2 }=76{ x }^{ 2 }$

So,

$76{ x }^{ 2 }=1216\Rightarrow { x }^{ 2 }=1216/76=16={ 4 }^{ 2 }\Rightarrow x=4\quad$

Hence the dimensions of the rectangular solid are

Length

$=5\times 4=20cm$

Breadth

$=4\times 4=16cm$

Height

$=2\times 4=8cm$

and volume

$=lbh=20\times 16\times 8=2560{ cm }^{ 3 }$

If two cubes of dimensions 2 cm by 2 cm by 2 cm are placed side by side, what would the dimensions of the resulting cuboid ?

The side of cube is

$2cm$

The resultant length of cuboid is

$2+2=4cm$

The breadth of cuboid is

$2cm$

The height of cuboid is

$2cm$

A sweet of cuboidal shape measures $4{ cm }\times 3{ cm }\times 2{ cm }.$ How many such sweets can be placed in a card board box measuring $27{ cm }\times 21{ cm }\times 8{ cm }$ ?

Volume of sweet

$=4\times 3\times 2=24{ cm }^{ 3 }$

Volume of cardboard box

$=27\times 21\times 8=4536{ cm }^{ 3 }$

$\therefore$ No. of sweets

$=\dfrac { 4536 }{ 24 } =189$

Find the area of quadrilateral $RENT$ .

Area of

$RENT$

$=$ Area of

$\triangle{REN}+$ Area of

$\triangle{RTN}$

$=\dfrac { 1 }{ 2 } \times 28\times 28+\dfrac { 1 }{ 2 } \times 28\times 8=224\ { cm }^{ 2 }$

The perimeter of a trapezium is 52 cm. Its non-parallel sides are 10 cm each and the distance between two parallel sides is 8 cm. Find the area of the trapezium.

The perimeter of trapezium is

$52$

The length of Non parallel sides is

$10$

$(a+b)=52-20=32$

The height is

$8cm$

The area of trapezium is

$\dfrac 12(a+b)h\\\dfrac 12(32)\times 10=160cm^2$

Find a side of a cube whose total surface area is $486cm^2$ .

Let the side of a cube be

$a\ cm$

As we know that

total surface area of a cube with side

$a$ is

$6 a^2$

Given total surface area of a cube is

$486\ cm^2$

$\implies 6 a^2=486$

$\implies a^2=81$

$\implies a=9$

So side of a cube is

$9\ cm$

The dimension of a match box is $3 cm\times 2 cm\times 1 cm$ . Find the volume 12 packets.

Volume of one cubic box

$= 3\ cm \times 2\ cm \times 1\ cm$

$=6\ cm^3$

Volume of

$12$ cubic box packets

$= 12\times 6$

$= 72\ cm^3$

1300305_1609031_ans_1777e8752d7d4d539e4932d2d442e983.jpg

Find the total surface area of cube having side 3.4 cm ?

Given edge of a cube is

$3.4\ cm$

as we know that

Total surface area of a cube with edge

$a$ is

$6 a^2$

$\implies$ Total surface area of cube is

$6(3.4)^2=69.36\ cm^2$

In figure $ABCD$ is a trapezium in which $AB||DC$ and $DC=40\ cm$ and $AB=60\ cm$ . If $X$ and $Y$ are, respectively, the mid-point of $AD$ and $BC$ , prove that:
$ar(trap. DCYX)=\dfrac {9}{11}ar (trap . XYBA)$ .

1661034_1670495_ans_8cfbdd5844b8420ebee2e359201934fa.jpeg

If two cubes each of side 6 cm are joined face to face, then find the volume of the resulting cuboid.

1661447_1670419_ans_96ed1b97717342048cf5265e669ad2fd.jpg

Which of the following figure lie on the same base and between the same parallels.
In such a case, write the common base and the parallels:

Two figures are on the common base and between the same parallel if the base is one of the parallel lines and the opposite vertex lie on the line which is parallel to the common base.

Since

$\Rightarrow$ Trapezium

$ABCD$ and

$\triangle PAB$ have same base

$AB$

$\Rightarrow$

$AB\parallel DC$

$\Rightarrow$

$P$ lies on

$DC$

Hence, trapezium

$ABCD$ and

$\triangle PAB$ lie on same base

$AB$ and between same parallel lines

$AB$ and

$DC$ .

Find the area in square (correct upto one decimal places) meters of the trapezium whose bases and altitude are as under:
bases $=12\ dm$ and $20\ dm$ , altitude $=10\ dm$

Let the lengths of bases be

$b_1$ and

$b_2$

$(b_1)=12\space\mathrm{dm}\\=\dfrac{12}{10}\space\mathrm{m}\\=1.2\space\mathrm{m}\quad$

$(b_2)=20\space\mathrm{dm}\\=\dfrac{20}{10}\space\mathrm{m}\\=2\space\mathrm{m}$

$(h)=10\space\mathrm{dm}\\=\dfrac{10}{10}\space\mathrm{m}\\=1\space\mathrm{m}$

Now, Area of trapezium

$=\dfrac{1}{2}\times(b_1+b_2)\times h$

$=\dfrac{1}{2}\times(1.2+2)\times1$

$=1.6\space\mathrm{m^2}$

In figure $ABCD$ is a trapezium in which $AB||DC$ and $DC=40\ cm$ and $AB=60\ cm$ . If $X$ and $Y$ are, respectively, the mid-point of $AD$ and $BC$ , prove that:
$DCYX$ is a trapezium

1661031_1670493_ans_33ad5a4da6da4ab0aa8cbdd9460d3723.jpeg

In figure $ABCD$ is a trapezium in which $AB||DC$ and $DC=40\ cm$ and $AB=60\ cm$ . If $X$ and $Y$ are, respectively, the mid-point of $AD$ and $BC$ , prove that:
$XY=50\ cm$

1660928_1670492_ans_abd1fc80e672402bbcc905f4d3eb24c5.jpeg

Find the area in square meters of the trapezium whose bases and altitude are given as below:
bases $=8\ dm$ and $60\ dm$ , altitude $=40\ dm$

Given: Base of trapezium

$(b_1)=8\space\mathrm{dm}=\frac{8}{10}\space\mathrm{m}=0.8\space\mathrm{m}\quad[\because 1\space\mathrm{dm}=\frac{1}{10}\space\mathrm{m}]$

Base of trapezium

$(b_2)=60\space\mathrm{dm}=\frac{60}{10}\space\mathrm{m}=6\space\mathrm{m}$

Height of trapezium

$(h)=40\space\mathrm{dm}=\frac{40}{10}\space\mathrm{m}=4\space\mathrm{m}$

Now, Area of trapezium

$=\frac{1}{2}\times(b_1+b_2)\times h$

$=\frac{1}{2}\times(0.8+6)\times4$

$=\frac{1}{2}\times6.8\times4$

$=13.6\space\mathrm{m^2}$

Find the area in square meters of the trapezium whose bases and altitude are as under:
bases $=28\ dm$ and $3\ dm$ , altitude $=25\ dm$

Given: Base of trapezium

$(b_1)=28\space\mathrm{dm}$

$=\dfrac{28}{10}\space\mathrm{m}=2.8\space\mathrm{m}\quad[\because 1\space\mathrm{dm}=\dfrac{1}{10}\space\mathrm{m}]$

Base of trapezium

$(b_2)=3\space\mathrm{dm}$

$=\dfrac{3}{10}\space\mathrm{m}=0.3\space\mathrm{m}$

Height of trapezium

$(h)=25\space\mathrm{dm}$

$=\dfrac{25}{10}\space\mathrm{m}=2.5\space\mathrm{m}$

Now, Area of trapezium

$=\dfrac{1}{2}\times(b_1+b_2)\times h$

$=\dfrac{1}{2}\times(2.8+0.3)\times2.5$

$=\dfrac{1}{2}\times3.1\times2.5$

$=3.875\space\mathrm{m^2}$

The shape of the top surface of a table is trapezium. Its parallel sides are $1\ m$
and $1.4\ m$ and the perpendicular distance between them is $0.9\ cm$ . Find its area

We are given that the length of the parallel sides is

$1m$ and

$1.4m$ and height

$(h)=0.9m$

By using the formula Area of trapezium

$=\dfrac{1}{2} \times (sum\ of\ parallel\ side )\times height$

$=1/2 \times (1+1.4) \times 16$

$=1.08m^2$

Find the area of a trapezium whose parallel sides are of length $16\ cm$ and $22\ dm$ and whose height is $12\ dm$ .

Given: Base of trapezium

$(b_1)=16\space\mathrm{dm}$

Base of trapezium

$(b_2)=22\space\mathrm{dm}$

Height of trapezium

$(h)=12\space\mathrm{dm}$

Now, Area of trapezium

$=\frac{1}{2}\times(b_1+b_2)\times h$

$=\frac{1}{2}\times(16+22)\times12$

$=\frac{1}{2}\times38\times12$

$=228\space\mathrm{dm^2}$

$=\frac{228}{100}\space\mathrm{m^2}\quad[\because 1\space\mathrm{dm^2}=\frac{1}{100}\space\mathrm{m^2}]$

$=2.28\space\mathrm{m^2}$

Find the volume in cubic metre:
length = 4 m, breadth = 2.5 m, height = 50 cm.

Given that,

Length

$=4 \space\mathrm{m}$

Breadth

$=2.5\space \mathrm{m}$

Height

$=50\space \mathrm{cm}$

To find out,

The volume in

$m^3$

Height

$=\cfrac{50}{100} \space\mathrm{m}\quad(\because 1\space \mathrm{m}=100\space \mathrm{cm})$

$=0.5 \space\mathrm{m}$

We know that, volume of a cuboid

$= \text{length} \times \text{breadth} \times \text{height}$

$=4 \times 2.5 \times 0.5$

$=5\space \mathrm{m}^{3}$

Hence, the volume of the given cuboid is

$5\ m^3$ .

Find the height of a trapezium, the sum of the lengths of whose bases (parallel sides ) is $60\ cm$ and whose area is $600\ cm^2$

Let

$h$ be the height of trapezium.

Given: Sum of bases of trapezium,

$(b_1+b_2)=60\space\mathrm{cm}$

Area of trapezium

$=600\space\mathrm{cm^2}$

$\Rightarrow 600=\dfrac{1}{2}\times(b_1+b_2)\times h$

$\Rightarrow 600=\dfrac{1}{2}\times60\times h$

$\Rightarrow h=\dfrac{600}{60}\times2$

$\Rightarrow h=20$

$\therefore$ Height of trapezium

$=20\space\mathrm{cm}$

Enter 1 if it is true else enter 0.
The area of a trapezium is $384\ cm^2$ . Its parallel sides are in the ratio $3:5$ the perpendicular distance between them is $12\ cm$ . The difference between the length of each one of the parallel sides is $16$ cm.

Given:

Area of the trapezium

$=384 \space\mathrm{cm}^{2}$

The parallel sides are in the ratio

$3:5$ and the perpendicular height between them is

$12 \mathrm{cm}$ .

Suppose that the sides are in

$x$ multiples of each other.

Then, length of the shorter side

$=3x$

Length of the longer side

$=5x$

Area of a trapezium

$=\dfrac{1}{2}\times$ (Sum of parallel sides)

$\times$ ( Height)

$\Rightarrow 384=\dfrac{1}{2} \times(3 \mathrm{x}+5 \mathrm{x}) \times(12)$

$\Rightarrow 384=\dfrac{12}{2} \times(8 \mathrm{x})$

$\Rightarrow 384=6 \times(8 \mathrm{x})$

$\Rightarrow 8 x=\dfrac{384}{6}=64$

$\Rightarrow x=\dfrac{64}{8}=8\space \mathrm{cm}$

$\therefore$ Length of the shorter side

$=3 \times \mathrm{x}=3 \times 8=24 \space\mathrm{cm}$

And, length of the longer side

$=5 \times \mathrm{x}=5 \times 8=40\space \mathrm{cm}$

Differnece between the length is

$16$ cm

The diameter of the base of a right circular cylinder is $42\ cm$ and its height is $10\ cm$ . find the area of the curved surface and total surface area.

Diameter of the base of the cylinder

$= 42\ cm$

$\therefore$ r = radius of the base of the cylinder

$= 21\ cm$

$=$ height of the cylinder

$= 10\ cm$

$\therefore$ curved surface area of the cylinder =

$2 \pi rh$

$2 \times \dfrac {22}{7} \times 10\ cm^2$

$2 \times 22 \times 3 \times 10\ cm^2 =1320\ cm^2$

total surface area of the cylinder =

$2 \pi r ( h+ r)$

$2 \times \dfrac {22}{7} \times 21 \times (10 +21 )\ cm^2$

$2 \times 22 \times 3 \times 31 cm^2 = 4092\ cm^2$

Find the area of trapezium with base $15\ cm$ and height $8\ cm$ , if the side parallel to the given base is $9\ cm$ long.

Given: Base of trapezium

$(b_1)=15\space\mathrm{cm}$

Base of trapezium

$(b_2)=9\space\mathrm{cm}$

Height of trapezium

$(h)=8\space\mathrm{cm}$

Now, Area of trapezium

$=\dfrac{1}{2}\times(b_1+b_2)\times h$

$=\dfrac{1}{2}\times(15+9)\times8$

$=\dfrac{1}{2}\times24\times8$

$=96\space\mathrm{cm^2}$

Find the volume, lateral surface area and the total surface area of the cuboid whose dimension are:
Length $=48\ cm$ , breadth $=6\ dm$ and height $=1\ m$

Given Length

$=48\ cm=0.48\ m$ , breadth

$=6\ dm=0.6 \ m$ and height

$=1\ m$

We know that volume of cuboid =length

$\times$ breadth

$\times$ height

So,

$V=(0.48\times 0.6\times 1)$

$V=0.288\ m^{3}$

We know that total surface area of cuboid

$=2(lb+bh+hl)$

So,

Surface area

$=2(0.48\times 0.6+0.6\times 1+1\times 0.48)$
Surface area

$=2(.288+0.6+0.48)$

Total Surface area

$=2.736\ m^{2}$

We know that lateral area of cuboid

$=2[(l+b)\times h]$

So,

Lateral surface area

$=2[(0.48+0.6)\times 1]$
Lateral surface area

$=261\ m^{2}$

Find the area in square meters of the trapezium whose bases and altitude are given as below:
bases $=150\ dm$ and $30\ dm$ , altitude $=9\ dm$

Given: Base of trapezium

$(b_1)=150\space\mathrm{dm}=\frac{150}{10}\space\mathrm{m}=15\space\mathrm{m}\quad[\because 1\space\mathrm{dm}=\frac{1}{10}\space\mathrm{m}]$

Base of trapezium

$(b_2)=30\space\mathrm{dm}=\frac{30}{10}\space\mathrm{m}=3\space\mathrm{m}$

Height of trapezium

$(h)=9\space\mathrm{dm}=\frac{9}{10}\space\mathrm{m}=0.9\space\mathrm{m}$

Now, Area of trapezium

$=\frac{1}{2}\times(b_1+b_2)\times h$

$=\frac{1}{2}\times(15+3)\times0.9$

$=\frac{1}{2}\times18\times0.9$

$=8.1\space\mathrm{m^2}$

The area of a trapezium is $1586\ cm^{2}$ and the distance between the parallel sides is $26\ cm.$ If one of the parallel sides is $38\ cm,$ find the other.

Given:

Area of the trapezium

$=1586 \space\mathrm{cm}^{2}$

Distance between the parallel sides

$=26 \space\mathrm{cm}$

And, length of one parallel side

$=38 \space\mathrm{cm}$

Let us suppose the length of the other side to be

$x \space\mathrm{cm}.$

Now, area of the trapezium

$=\dfrac{1}{2} \times \space\text{(Sum of the parallel sides)} \times \space \text{(Distance between the parallel sides)}$

$\Rightarrow 1586=\dfrac{1}{2}\times (38+x)\times 26$

$\Rightarrow 1586=\dfrac{26}{2}\times (38+x)$

$\Rightarrow 13 \times(38+x)=1586$

$\Rightarrow 38+x=\dfrac{1586}{13}$

$\Rightarrow 38+x=122$

$\Rightarrow x=122-38=84 \space\mathrm{cm}$

Hence, the length of the other parallel side is

$84 \space\mathrm{cm}$ .

A closed cylindrical tank of radius $1.5\ m$ and height $3\ m$ is made from sheet metal. How much sheet of metal is required?

Clearly, total area of the sheet of metal required in making closed cylindrical tank is equal to the total surface area of the cylinder of radius

$r = 1.5\ m$ and height,

$h = 3\ m$

$\therefore$ total area of metal sheet

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

$= 2 \times \dfrac {22}{7} \times 1.5 \times ( 3 + 1.5 )\ m^2$

$= 2 \times \dfrac {22}{7} \times 1.5 \times 4.5 m^2 = 42.43\ m^2$

The volume of a cuboid is given by the product of its length, breadth and height. The length of a cuboid is $2x^2$ times its breadth and the height is $\dfrac{3}{2}xy$ times of length. Find the volume of the cuboid if its breadth is $6y^2$ .

Let, the length of the cuboid be

$l$ .

The breadth of the cuboid be

$b$ .

The height of the cuboid be

$h$ .

Given,

$b = 6y^{2}$

$l = 2x^{2} \times b$

$=2x^{2}.6y^{2}$

$=12x^{2}y^{2}$

$h= \dfrac{3}{2}xy \times l$

$=\dfrac{3}{2}xy \times 12x^{2}y^{2}$

$=18x^{3}y^{3}$

Volume of Cuboid

$= l\times b\times h$ .

$=6y^{2}\times12x^{2}y^{2}\times18x^{3}y^{3}$

$= 1296 x^{5}y^{7}$

In given figure, a parallelogram is drawn in a trapezium, the area of the parallelogram is $80\ cm^{2}$ , fin the area(in $cm^2$ ) of the trapezium.

1872115_1673761_ans_de65983b21a840f89960764ab8ecd315.jpeg

The curved surface area of a cylinder is $1320 cm^2$ and its base has diameter 21 cm . find the volume of the cylinder.

Given: Diameter of cylinder (d)

$=21\space\mathrm{cm}$

$\Rightarrow$ Radius of cylinder (r)

$=\dfrac{d}{2}=\dfrac{21}{2}=10.5\space\mathrm{cm}$

Curved Surface area of cylinder

$=1320\space\mathrm{cm^2}$

$\Rightarrow 2\pi rh=1320$

$\Rightarrow 2\times\dfrac{22}{7}\times10.5\times h=1320$

$\Rightarrow h=1320\times\dfrac{1}{2}\times\dfrac{7}{22}\times\dfrac{1}{10.5}$

$\Rightarrow h=20\space\mathrm{cm}$

Now, Volume of cylinder

$=\pi r^2h=\dfrac{22}{7}\times(10.5)^2\times20=6930\space\mathrm{cm^3}$

Find the area of trapezium whose parallel sides are $38.7\ cm$ and $22.3\ cm$ and the distance between them is $16\ cm$ .

we know that the length of the parallel sides is

$38.7cm$ and

$22.3cm$ and height

$(h)=16cm$

By using the formula

$Area\ of\ trapezium$

$=1/2 \times (sum\ of\ parallel\ side )\times height$ , we have

$=1/2 \times (38.7+22.3) \times 16$

$=488cm^2$

It is required to make a closed cylindrical tank of height $1\ m$ and base diameter $140\ cm$ from a metal sheet. How many square meters of the sheet is required for the same?

It is given that:

Diameter of the cylinder

$=140 cm$

Radius of the cylinder

$=\dfrac{140}{2}=70 cm$

Height of the cylinder

$=1m=100 cm$

We know that:

Area of sheet required = TSA of cylinder

$=2\pi r(r+h)=2\times \dfrac{22}{7}\times 70 (100+70)$

On further calculation,

Area of sheet required

$=2\times \dfrac{22}{7}\times 70\times 170$

So we get,

Area of sheet required

$=74800\ cm^2= 7.48\ m^2$

Therefore, the area of the sheet required is

$7.48\ m^2$ .

The area of a trapezium is $475 cm^2$ and its height is 19cm. Find the lengths of its two parallel sides if one side is 4cm greater than the other.

Consider x cm as the smaller side of the trapezium

So the larger parallel can be written as (x + 4) cm

We know that the area of trapezium =

$\dfrac{1}{2} \times$ sum of parallel sides

$\times$ height

By substituting the values

$475 = \dfrac{1}{2} \times (x + (x + 4)) \times 19$

On further calculation

$25 =\dfrac{1}{2} \times (2x + 4)$

So we get

50 = 2x + 4

By subtraction

2x = 50 - 42

x = 46

By division

x = 23cm

So we get Larger parallel side (x + 4) = 23 + 4 = 27cm

Therefore, the length of two parallel sides is 23cm and 27cm.

Find the area of a trapezium whose parallel sides are 11m and 25m long, and the nonparallel sides are 15m and 13m long.

From the point C construct CE || DA

We know that ADCE is a parallelogram having AE || DC and AD || EC with AD = 13m and D = 11m

It can be written as

AE = DC = 11m and EC = AD = 13m

So we get

BE = AB - AE

By substituting the values

BE = 25 –-11 = 14m

Consider

$\Delta$ BCE

We know that BC = 15m, CE = 13m and BE = 14m

Take a = 15m, b = 13m and c = 14m

So we get

$s=\dfrac{a+b+c}{2}$

By substituting the values

$s=\dfrac{15+13+14}{2}$

So we get

s=21m

We know that

Area=

$\sqrt{s(s-a)(s-b)(s-c)}$

By substituting the values

Area=

$\sqrt{21(21-15)(21-13)(21-14)}$

So we get

Area=

$\sqrt{21\times6\times8\times7}$

It can be written as

Area=

$\sqrt{7\times3\times2\times3\times4\times2\times7}$

We get

Area=

$7\times3\times2\times2$

So we get

Area=

$84cm^2$

We know that

Area of △ BCE =

$\dfrac{1}{2} \times BE \times CL$

By substituting the values

$84 =\dfrac{1}{2} \times 14 \times CL$

On further calculation

$84 = 7 \times CL$

By division

CL = 12m

We know that

Area of trapezium ABCD =

$\dfrac{1}{2}\times$ sum of parallel sides

$\times$ height

It can be written as

Area of trapezium ABCD =

$\dfrac{1}{2}\times (AB + CD) \times CL$

By substituting the values

Area of trapezium ABCD =

$\dfrac{1}{2} \times (11 + 25) \times 12$

On further calculation

Area of trapezium ABCD =

$36 \times 6$

By multiplication

Area of trapezium ABCD =

$216 m^2$

Therefore, the area of trapezium ABCD is

$216 m^2$ .

The difference between the lengths of the parallel sides of a trapezium is 8cm, the perpendicular distance between these sides is 24cm and the area of the trapezium is $312cm^2$ . Find the length of each of the parallel sides.

Consider x cm as the smaller parallel side
So the longer parallel side can be written as (x + 8) cm
It is given that height = 24cm
Area =

$312 cm^2$
We know that
Area of trapezium =

$\dfrac{1}{2} \times$ sum of parallel sides

$\times$ height
By substituting the values

$312 =\dfrac{1}{2} \times (x + x + 8) \times 24$
On further calculation

$312 = 12 \times (2x + 8)$
By division
2x + 8 = 26
So we get
2x = 26 – 8
By subtraction
2x = 18
By division
x = 9cm
So we get
x + 8 = 9 + 8 = 17cm
Therefore, the length of each of the parallel sides is 9cm and 17cm.

From the adjoining sketch, calculate
(i) the length $AD$
(ii) the area of trapezium $ABCD$
(iii) the area of triangle $BCD$

(i) In right angled

$\triangle ABD$

${ BD }^{ 2 }={ AD }^{ 2 }+{ AB }^{ 2 }\qquad$ [by Pythagoras Theorem ]

$\Rightarrow{ AD }^{ 2 }={ BD }^{ 2 }-{ AB }^{ 2 }$

$\Rightarrow{ AD }^{ 2 }={ (41) }^{ 2 }-{ (40) }^{ 2 }$

$\Rightarrow{ AD }^{ 2 }=1681-1600=81$

$\Rightarrow AD=\sqrt{81}=9cm$

(ii) Area of trapezium

$ABCD=\cfrac 12$ (sum of opposite parallelogram lines)

$\times$ height

$=\cfrac12 (AB+CD)\times AD$

$=(40+15)\times 9=247.5{ cm }^{ 2 }$

(iii) Area of traingle

$BCD$ = Area of trapezium

$ABCD$ - Area of

$\triangle ABD$

$=(247.5-\cfrac12 \times 40\times 9){ cm }^{ 2 }$

$=67.5{ cm }^{ 2 }$

The perimeter of a trapezium is $52cm$ . If its non-parallel sides are $10cm$ each and its altitude is $8cm$ , find the area of the trapezium.

Given,

Perimeter of a trapezium

$=52cm$

Length of each non-parallel side

$=10cm$

Altitude

$DL=8cm$

Let the sum of parallel sides be

$S$ .

Then,

Perimeter

$=$ Sum of parallel sides

$+$ Sum of non-parallel sides

$\Rightarrow 52=$ Sum of parallel sides

$+10+10$

$\Rightarrow$ Sum of parallel sides

$=52-20=32$

Thus, area of trapezium

$=\cfrac { 1 }{ 2 } \times$ (Sum of parallel sides )

$\times$ altitude

$=\cfrac { 1 }{ 2 } \times 32\times 8=128{ cm }^{ 2 }$

Find the height of a cuboid whose volume is $312{cm}^{3}$ and base area is $26{cm}^{2}$

Given

Volume of a cuboid

$=312{ cm }^{ 3 }$

Base area

$=26{ cm }^{ 2 }$

We know that,

Volume of a cuboid

$=$ Base area

$\times$ height

$\Rightarrow 312=26\times h\\ \Rightarrow h=\dfrac{312}{26}=12\ cm$

The cross-section $ABCD$ of a swimming pool is a trapezium. Its width $AB=14cm$ , depth at the shallow end is $1.5m$ and at the deep end is $8m$ . Find the area of the cross-section

Here,

$AD$ and

$BC$ are the two parallel sides of the trapezium.

And the distance between them is

$14\ m$

$\therefore$ Area of trapezium

$=\cfrac { 1 }{ 2 } \times (1.5+8)\times 14$

$=\cfrac { 1 }{ 2 } \times 9.5\times 14$

$=66\ {m}^{2}$

From the given diagram, calculate
(i) the area of trapezium $ACDE$
(ii) the area of parallelogram $ABDE$
(iii) the area of triangle $BCD$

(i) Area of trapezium

$ACDE=\cfrac { 1 }{ 2 } \times (AC+DE)\times h=\cfrac { 1 }{ 2 } \times (13+7)\times 6.5=\cfrac { 1 }{ 2 } \times 20\times 6.5=65{ m }^{ 2 }$

(ii) Area of parallelogram

$ABDE=\cfrac { 1 }{ 2 } \times base\times height=\cfrac { 1 }{ 2 } \times 6\times 6.5=15.5{ m }^{ 2 }$

(iii) From the figure,

$BC=AC-AB=13-7=6m$

$\therefore$ Area of

$\triangle BCD=\cfrac { 1 }{ 2 } \times base\times height=\cfrac { 1 }{ 2 } \times 6\times 6.5=19.5{ m }^{ 2 }$

The total surface area of a cube is $384cm^2$ calculate the side of the cube.

$T.S.A$ of cube

$=384cm^2$

$6l^2=384$

$l^2=\dfrac{384}{6}$

$l^2=64$

$l=\sqrt{64}$

$l=8cm$

The L.S.A a cube is $64m^2$ , calculate the side of the cube.

L.S.A of cube

$=4l^2$

$64=4l^2$

$\dfrac{64}{4}=l^2$

$\therefore l=\sqrt{16}=4m$
The side of the cube is

$4m$ .

Fill in the following blanks:
Volume of cuboid Length Breadth Height
(i) $90\ { cm }^{ 3 }$ --- $5\ cm$ $3\ cm$
(ii) --- $15\ cm$ $8\ cm$ $7\ cm$
(iii) $62.5\ {m}^{3}$ $10\ cm$ $5\ cm$ ---

We know that,

Volume of cuboid

$=length\times breadth\times height$

$(i)$

$90\ { cm }^{ 3 }=length\times 5\ cm\times 3\ cm$

$\Rightarrow length=\dfrac{90}{5\times 3}$

$=\dfrac{90}{15}$

$=6\ cm$

Hence, length is

$6\ cm$

$(ii)$

$Volume=15\ cm\times 8\ cm\times 7\ cm$

$=840\ { cm }^{ 3 }$

Hence, volume is

$840\ cm^3$

$(iii)$

$62.5\ { m }^{ 3 }=10\ m\times 5\ m\times height$

$\Rightarrow height =\dfrac{62.5}{10\times 5}$

$=1.25\ m$

Hence, height is

$1.25\ m$ .

The volume of a cuboid is $3456 ^{3}cm$ . If its length =24 cm and breadth =18 cm; find its height.

Volume of the given cuboid $=3456^{3}cm$ .

Length of the given cuboid $=24 cm$

Breadth of the given cuboid $=18 cm$

Here

$Length × Breadth × Height$ = Volume of a cuboid

Substituting the values

$24×18×$ Height =3456

By further calculation

Height = $\dfrac{3456}{24\times 18}$

So we get

Height = $\dfrac{3456}{432}$

Height $=8cm$

The volume of a cuboid is $7.68 ^{3}$ . If its length = 3.2m and height =1.0m; find its breadth.

Volume of a cuboid = $7.68 m^{3}$

Length of a cuboid $=3.2 m$

Height of a cuboid $=1.0 m$

Here

$Length \times Breadth \times Height = Volume \,\,of \,\,a \,\,cuboid$

Substituting the values

$3.2 \times Breadth \times 1.0=7.68$

By further calculation

$\rightarrow Breadth=\dfrac{7.68}{3.2\times 1.0}$

so we get

$\rightarrow Breadth=\dfrac{7.68}{3.2}$

$\rightarrow Breadth= 2.4 cm$

The length, breadth and height of a cuboid are in the ratio 6:5:3. If its total surface area is $504 c m^{2}$ ; find its dimensions. Also, find the volume of the cuboid.

Consider length of the cuboid $=6x$

Breadth of the cuboid $=5x$

Height of the cuboid $=3x$

We know that

Total surface area of the given cuboid = $2(1\times b+b\times h+h\times l)$

Substituting the values

= $2(6x\times 5x+5x\times 3x+3x\times 6x)=x(30\times 2+15\times 2+18\times 2)$

We get

$=(60+30+36)x=126x^{2}$

It is given that

Total surface area of the given cuboid $504cm^3$

Substituting the values

$126x^{2}=504cm^{2}$

By further calculation

$\Rightarrow x^{2}=\dfrac{504}{126}$

So we get

$\Rightarrow x^{2}=4\Rightarrow x=\sqrt{4}$

$x = 2 cm$

Here

Length of the cuboid $=6x=6\times 2=12cm$

Breadth of the cuboid $=5x=5\times 2=10cm$

Height of the cuboid = $=3x=3\times 2=6cm$

We get

Volume of the cuboid $=l \times b \times h=12 \times 10 \times 6=720cm^{3}$

Find the volume and total surface area of a cube whose edge is:
8 cm

Edge of the given cube =8cm

We know that

Volume of the given cube $=( Edge )^{3}=(8)^{3}=8\times 8\times 8=512cm^{3}$

Total surface area of a cube = $6( Edge )^{2}=6 \times (8)^{2}=384cm^{2}$

Find the volume and the total surface area of a cuboid, whose:
l = 3.5m, b = 2.6m and h = 90cm,

Length

$= 3.5m$
breadth

$= 2.6m,$

height

$= 90 cm =\dfrac{90}{100}m = 0.9 m$

We know that

Volume of a cuboid

$= l \times b \times h = 3.5 \times 2.6 \times 0.9 = 8.19^{3}cm$
Here
Total surface area of a cuboid =

$2 ( l \times b + b \times h + h \times l ) = 2 ( 3.5 \times 2.6+ 2.6\times 0.9+0.9 \times 3.5 )$
By further calculation

$= 2 ( 9.10 + 2.34 + 3.15 ) = 2 ( 14.59) = 29.18m^{2}$

For given trapezium find its area.

Area

$= \dfrac12$ x Sum of parallel sides x Distance between the parallel sides

Area of ABCD =

$\dfrac{1}{2} \times (8 + 14) \times \left ( \sqrt{10^{2} - 6^{2}}\right )$

$\dfrac{1}{2} \times 22 \times 8$

$88 sq.cm$

1782028_1845265_ans_43cfc1d4bde342e2b45ee6a424cd1f74.png

The cross-section of a tunnel perpendicular to its length is a trapezium $ABCD$ as shown in the following figure; also given that :
$AM=BN; AB=7\ m; CD=5\ m$ . The height of the tunnel is $2.4m$ .The tunnel is $40\ m$ long
Calculate:
The cost of paving the floor at the rate of Rs. $18$ per $m^2$ .

Given,

$AB=7m,\ CD=5\ m$ and

$AM=BN$
The height is

$2.4\ m$ and its length is

$40m$
Area of floor of tunnel

$l \times b=40 \times 7=280m^2$
Rate of cost of paving

$=Rs 18\ per\ m^2$
Total cost

$=280 \times 18=Rs 5040$

Find the area of given trapezium.

Draw DE and CF perpendiculars to AB.

Area of the parallelogram is

$30 = AE + 18 + FB$

$\Rightarrow 30 = AE + 18 + AE$

$\Rightarrow 2AE + 18 = 30$

$\Rightarrow 2AE = 30 - 18$

$\Rightarrow 2AE = 12$

$\Rightarrow AE = 6 cm$

Now, consider the right angled triangle ADE.

$AD^{2} = AE^{2} + DE^{2}$

$\Rightarrow 12 = 6^{2} + DE^{2}$

$\Rightarrow 144 = 36 + DE^{2}$

$\Rightarrow DE^{2} = 144 - 36$

$\Rightarrow DE^{2} = 108$

$\Rightarrow DE^{2} = \sqrt{36 * 3}$

$\Rightarrow DE = 6\sqrt{3}$

$Area (\square ABCD) = Area(\triangle ADE) + Area(\square DEFC) + Area(\triangle CFB)$

$\Rightarrow Area (\square ABCD) = \dfrac{1}{2} * 6 * 6\sqrt{3} + 18 * 6\sqrt{3} + \dfrac{1}{2} * 6 * 6\sqrt{3}$

$\Rightarrow Area (\square ABCD) = 6 * 6\sqrt{3} + 18 * 6\sqrt{3}$

$\Rightarrow Area (\square ABCD) = 144\sqrt{3} = 249.41 cm^{2}$

1782510_1845303_ans_89f55209fd1940a18fab10c8c6b94d45.png

Find the area of given trapezium.

For the triangle

$EBC,$

$S = 19$ cm

Area of

$\Delta EBC = \sqrt{19 \times (19 - 16) \times (19 - 12) \times (19 - 10)}$

$\sqrt{19 \times 3 \times 7 \times 9}$

$59.9 sq.cm$

Let

$h$ be the height.

Area of

$\Delta EBC = \dfrac{1}{2} \times 12 \times h$

$\Rightarrow 59.9 = 6h$

$\Rightarrow h = \dfrac{59.9}{6} = 9.98 cm$

Area of trapezium

$= \dfrac12$ x Sum of parallel sides x Distance between the parallel sides

Area of ABCD =

$\dfrac{1}{2} \times (20 + 32) \times 9.98$

$\dfrac{1}{2} \times 52 \times 9.98$

$259.48\ cm^{2}$

1782423_1845287_ans_88e3319134754e19b76626ef827d2819.png

Take a small wooden rectangular block and find its volume (volume = length * breadth * height).
Take a big measuring jar and fill three-fourth of it with water and mark the water level. Then dip the block in water in the jar. (To prevent floating, nails can be inserted in the block). Mark the difference in the water level.
a. Is there any relation between the difference in the water level and the volume of the block?
b. Which property of matter is revealed by this experiment?

a. Yes, the volume of water rises is the volume of wooden block.
b. The substance that occupies a space.

Using the information in the given figure, find its area.

In right-angled triangle

$CMB$ ,

$\mathrm{BM}^{2}=\mathrm{BC}^{2}-\mathrm{CM}^{2}=(15)^{2}-(9)^{2}$

$=225-81=144$

$\Rightarrow \mathrm{BM}=12 \mathrm{m}$

Now,

$A B=A M+B M=23+12$

$=35 \mathrm{m}$

Area of trapezium

$A B C D=\dfrac{1}{2} \times($ sum of parallel sides

$) \times$ Height between parallel sides

$\begin{array}{l}=\dfrac{1}{2} \times(A B+C D) \times A D \\=\dfrac{1}{2} \times(23+35) \times 9 \\=\dfrac{1}{2} \times 58 \times 9 \\=261 \mathrm{m}^{2}\end{array}$

1782818_1844473_ans_dd92aa4128f749aeb32d5ff53a0a2497.png

A filed is in the shape of a trapezium whose parallel sides are $25m$ and $10 m$ . The non-parallel sides are $14m$ and $13 m$ . Find the area of the field.

$ABCD$ is trapezium

$AB || DC$
Draw

$AE || BC$ and

$AF \bot DC$

$\therefore ABCE$ is parallelogram.

$DE =DC-EC = 25-10$

$DE = 15cm$
Sides of

$\Delta ADE$

$a= 14 cm , b= 13 cm , c = 15 cm$

$\therefore$ perimeter =

$a+b+c$

$= 14 + 13 + 15 = 42 cm$

$\therefore s =\frac {a+b+c}{2} =\frac {42}{2} = 21 cm$

$\therefore$ Area of

$\Delta ADE , A$

$A = \sqrt {s(s-a)(s-b)(s-c) }$

$= \sqrt { 21 (21-14)(21-13)(21-15) }$

$= \sqrt { 21 \times 7 \times 8 \times 6 }$

$= \sqrt { 7056}$

$\therefore$ Area of

$\Delta ADE , A = 84 sq cm$
Area of

$\triangle ADE ,A = \frac {1}{2} \times b \times h$

$= \frac {1}{2} \times DE \times AF = 84$

$\frac {1}{2} \times 15 \times h = 84$

$h = 84 \times \frac {2}{15}$

$h = 11.2 cm$

$\therefore AF = 11.2 cm$
(ii) Area of quadrillateral ABCE = base

$\times$ height

$=EC \times AF$

$= 10 \times 11.2$

$= 112.0$

$= 112 sq.cm$

$\therefore$ Complete area of trapezium

$ABCD, A$
A = Area of

$\Delta ADE$ + Area of quadrilateral

$ABCE$

$= 84 +112$

$= 196 sq.cm.$

A solid cuboid of metal has dimension 24 cm cm , 18 cm and 4 cm .Find its volume

It is given that
Length of the cuboid

$= 24 cm$
Breadth of the cuboid

$= 18 cm$
Height of the cuboid

$= 4 cm$
We know that
Volume of the cuboid

$= l \times b \times h = 24 \times 18 \times 4 = 1728 cm ^{3}$

Find the volume and total surface area of a cube whose edge is:
2 m 40 cm .

Edge of the given cube $=2 m 40 cm=2.40 m$

We know that

Volume of a cube $=( Edge )^{3}$

Substituting the values

Volume of the given cube $=(2.40)3=2.40 \times 2.40 \times 2.40=13.824m^{3}$

Total surface area of the given cube $=6\times 2.4\times 2.4=34.56m^{2}$

The base radius and height of a cube are $7 \mathrm{cm}$ and $24 \mathrm{cm}$ respectively. Find it slant height, lateral surface area, total surface area and volume.

Given, The base radius of cone

$=7 \mathrm{cm}$ Height

$(\mathrm{h})=24 \mathrm{cm}$

$\therefore \quad$ Slant height

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

$\begin{array}{l}=\sqrt{(7)^{2}+(24)^{2}} \\=\sqrt{49+576} \\=\sqrt{625} \\=25 \mathrm{cm}\end{array}$
Curved surface area of cone

$=\pi r l$

$\begin{array}{l}=\dfrac{22}{7} \times 7 \times 25 \\=22 \times 25 \\=550 \mathrm{cm}^{2}\end{array}$
Total surface area of cone

$=\pi r(l+r)$

$\begin{array}{l}=\dfrac{22}{7} \times 7(25+7) \\=22 \times 32 \\=704 \mathrm{cm}^{2}\end{array}$
Volume of the cone

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

$\begin{array}{l}=\dfrac{1}{3} \times \dfrac{22}{7} \times 7 \times 7 \times 24 \\=\dfrac{3696}{3}=1232 \mathrm{cm}^{3}\end{array}$
Hence slant height of cone 25 cm, curved surface area=

$550 cm^2$ , total surface area=

$704 cm^2$ and volume=

$1232 cm^2$

The dimension of rectangular parallelopiped are in the ratio $6: 5: 4$ and its total surface area is $33300$ square meters. Find its volume.

Let length, breadth, and height of a cuboid are

$6x$ ,

$5x$ , and

$4x$ respectively.

Its total surface area;

$= 2(lb + bh + hl)$

$= 2(6x \times 5x + 5x \times 4x + 4x \times 6x)$

$= 2(30 \times x^2 + 20 \times x^ 2 + 24 \times x^2)$

$= 2 \times 74 \times x^2$

$= 148 \times x^2$

But it is given that total surface area

$= 33300\,sq.m$

i.e.

$148 \times x^2 = 33300$

$\Rightarrow x^2 =\dfrac{33300}{148}$

$\Rightarrow x^2 = 225\,m^2$

$\Rightarrow x = \sqrt{225}$

$\Rightarrow x = 15\,m$

Dimensions of cuboid are;

$6x = 6 \times 15 = 90\,m$

$5x = 5 \times 15 = 75\,m$

and

$4x = 4 \times 15 = 60\,m$

Volume of cuboid

$= 90 \times 75 \times 60$ cubic metres

$=405000$ cubic metres.

Find the area of a trapezium-shaped field, the lengths of whose parallel sides are $42$ metre and $30$ metre and other sides are $18$ metre and $18$ metre.

In the figure, ABCD is a trapezium-shaped field in which parallel sides

$AB = 42\,m$ and

$CD = 30\,m$ and non-parallel sides AD and BC equal to

$18\,m$ .
Draw EC || AD and CF ⊥ AB

Therefore

$EB=AB-AE=42\,m-30\,m$

$=12\,m$ and

$EC=18\,m$

$\therefore$ Area of

$\Delta BCE=\dfrac{b}{4}\sqrt{4a^2-b^2},$

where

$a=18\,m$ and

$b=12\,m$

$=\dfrac{12}{4}\sqrt{4(18)^2-(12)^2}$

$=3\sqrt{4 \times 324-144}$

$=3\sqrt{1296-144}=3\sqrt{1152}$

$=3 \times 33.94=101.82\,m^2$

Now, area of

$\Delta BCE=\dfrac{1}{2} \times BE \times CF$

$\Rightarrow 101.82=\dfrac{1}{2} \times 12 \times CF$

$\therefore CF=\dfrac{101.82}{6}=16.9\,m$

Area of parallelogram

$AECD=AE \times CF$

$=30 \times 16.9=507\,m^2$

$\therefore$ Area of trapezium ABCD = area of parallelogram AECD + area of ABCE

$= 507 + 101.82 = 610.92\,m^2$ .
1882326_1880538_ans_fba4fde6effb45558c21b7161173f8c3.jpg

A table is in the shape of a trapezium. Its parallel sides are $8\,m$ , and $16\,m$ respectively. Area of table is $108\,sq.\,m$ . Find the width of the table i.e. Distance between the parallel sides.

Let distance between the parallel sides be h.

Area of trapezium

$=\dfrac{1}{2} \times (sum\,of \,the\,parallel\,sides) \times$ distance between the parallel sides

$\Rightarrow 108=\dfrac{1}{2} \times (8+16) \times h$

$\therefore x=\dfrac{108 \times 2}{24} \Rightarrow x=9\,m$

Hence, distance between the parallel sides be

$9 \,m$ .

A closed box is made of wood $2.5\,cm$ thick. Its external dimensions are $1.25\,m$ long, $0.80\,m$ wide, and $0.55\,m$ high. If $1$ cubic meter of water weighs $250\,kg$ , find the weight of the box.

Outer length of the box (L)

$= 1.25\,m = 125\,cm$

Outer breadth of box (B)

$= 0.80\,m = 80\,cm$

Outer height of box (H)

$= 0.55\,m = 55\,cm$

Thickness of the wood

$=2.5\,cm$

Inner length of the box

$= 125 - 2 \times 2.5 = 120\,cm$

Inner breadth of the box

$= 80 - 2 \times 2.5 = 75\,cm$

Inner height of the box

$= 55 - 2 \times 2.5 = 50\,cm$

Outer volume

$= l \times b \times h = 125 \times 80 \times 55 = 550000\,cm^3$

Inner volume

$= l \times b \times h = 120 \times 75 \times 50 = 450000\,cm^3$

Volume of wood = outer volume – inner volume

$= 550000 - 450000 = 100000$ cubic cm

$= \dfrac{1}{10}$ cubic m

Weight of

$1$ cubic metre of wood is

$250\,kg$

Weight of cubic metre of wood is

$\dfrac{1}{10} \times 250 = 25\,kg$

The diagonal of a cube is $6\sqrt{3}$ cm. Find its surface area.

Let edge of the cube be a.

$\sqrt{3}a = 6\sqrt{3} \Rightarrow a = 6$

Surface area

$= 6a^2 = 6 \times (6)^2 = 6 \times 36 = 216\,cm^2$

If area of trapezium is $350\,sq.\,cm$ and its parallel sides are $26$ cm and $44$ cm then find the distance between the parallel sides.

Let distance between the parallel sides be x.

Area of trapezium

$=\dfrac{1}{2} \times (sum\,of \,the\,parallel\,sides) \times$ distance between the parallel sides

$\Rightarrow 350=\dfrac{1}{2} \times (26+44) \times x$

$\therefore x=\dfrac{350 \times 2}{70} \Rightarrow x=10\,cm$

Hence, distance between the parallel sides be

$10 \,cm$ .

Length of the fence of a trapezium shaped field $ABCD$ is $120\ m$ . If $BC= 48\ m$ , $CD = 17\ m$ and $AD = 40\ m$ , find the area of this field. Side $AB$ is perpendicular to the parallel sides $AD$ and $BC$ .

Length of fence of trapezium

$ABCD = AB +17 +48 + 40$

$\Rightarrow 120 = 105 + AB$

$\Rightarrow AB = 15$ m
Area of trapezium

$= \dfrac12\times\text{Sum of Parallel sides}\times\text{Distance between Parallel sides}$

$= \dfrac12 [ (40+48)\times15] = 660$ sq. metres

Find the area of trapezium $ABCD$ , where $AB$ is parallel to $DC$ . $AB = 77 cm, BC = 28 cm, CD = 60 cm, DA = 26 cm$

Find the area of trapezium ABCD where AB is parallel

to DC.

$AB = 77 cm \ BC = 28 cm \ CD = 60 cm \ DA = 26 cm$

Draw a line BC parallel to AD and draw a

perpendicular BE on CD. It can be observed

that

$ABCD$ is a parallelogram

$BC = AD = 28 cm$

$CD = AB = 60 cm$

$CF = 77 cm - CD$

$= 77 - 60 = 17 cm.$

For

$\triangle BCF$

Perimeter of a triangle

$25-(28 + 26 + 17)$

$S = \dfrac{71}{2} = 35.5$

By Heron's formula

$\sqrt{S }(S-a)(S-b)(S-c)$

$\triangle BCF = \sqrt{35.5(35.5-28)(35.5-26)(35.5-17)}$

$= 216.3 cm^{2}$

area of

$\triangle BCF = \dfrac{1}{2}\times BE \times CF$

$216.3 cm^{2} = \dfrac{1}{2}\times BE\times 17$

$BE = 432.6/17 = 25.4 cm$

Area of

$ABCD = BE\times CD = 25.4\times 60$

$= 1524 cm^{2}$

Area of field

$= 216.3 + 1524$

$= 1740.3\,cm^{2}$

Find the difference in volume of two tanks of the following dimensions, in $cm^{3}$ and in litres. Tank1; $l=35\ cm, b=20\ cm, h=20\ cm$ ; Tank 2: $l=40\ cm, b=24\ cm, h=24\ cm$ .

$\begin{array}{l} { V_{ 1 } }=35\times 20\times 20=400\times 35=14000=14\, \, litre \\ { V_{ 2 } }=40\times 24\times 24=23040=23.04\, \, \, litre \\ { V_{ 2 } }-{ V_{ 1 } }=23.04-14\, \, \\ =9.04\, \, litre \end{array}$

A metal pipe is 77 cm long. The inner diameter of a cross-section is 4 cm, the outer diameter being 4.4 cm, then find its outer curved surface area.

Dimensions be :

$\Rightarrow R=$ outer radius

$=\dfrac{4.4}{2}=2.2cm$

$r=$ Inner radius

$=\dfrac{4}{2}=2cm$

$\Rightarrow$ Height

$h=77cm$

$\therefore$ Outer curved surface area

$=2\pi Rh$

$=2\times \dfrac{22}{7}\times 2.2\times 77$

$=1064.8cm^2$

$\therefore$ Outer curved surface area

$=1064.8cm^2.$

Hence, the answer is

$1064.8cm^2.$

A box with a lid is made of wood which is $3\ \mathrm { cm }$ thick. Its external length, breadth and height are $5\ \mathrm { cm } , 39\ \mathrm { cm }$ and $30\ \mathrm { cm }$ respectively. Find the capacity of the box. Also find the volume of wood used to make the box.

Volume

$=\left(56-6\right) \times \left( 39-6\right)\times \left(30-6\right)$

$=39600cm^3$

Volume of wood

$=3cm\times SA\;of\;external\;box$

$\Rightarrow 3\times 2\left( 56\times 39+56\times 30\times 39\times 30\right)$

$\Rightarrow 30204cm^3.$

Hence, the answer is

$30204cm^3.$

PQRS is a trapezium. OR = 9 cm, $\angle RQS = 90^o$ , $PQ = 5 cm$ and $\displaystyle area (\triangle QRS) = \frac{81}{2\sqrt{3}} cm^2$ . Find the area of the trapezium

$QR=9cm,PQ=5cm$

$\Rightarrow ar\left(\triangle QRS\right)=\dfrac{81}{2\sqrt{3}}cm^2$

$\Rightarrow \dfrac{1}{2}\times QR\times QS=\dfrac{81}{2\sqrt{3}}$

$\Rightarrow \dfrac{1}{2}\times 9\times x=\dfrac{81}{2\sqrt{3}}$

$\Rightarrow x=\dfrac{9}{\sqrt{3}}=3\sqrt{3}cm$

$\Rightarrow SR=\sqrt{QR^2+QR^2}=\sqrt{27+81}=\sqrt{108}=6\sqrt{3}cm$

Also

$ar\left(\triangle QRS\right)=\dfrac{1}{2}\times QT\times SR=\dfrac{81}{2\sqrt{3}}$

$\Rightarrow \dfrac{1}{2}\times y\times 6\sqrt{3}=\dfrac{81}{2\sqrt{3}}$

$\Rightarrow y=\dfrac{81}{\sqrt{3}\times 6\sqrt{3}}cm$

$y=\dfrac{81}{6\times 3}cm$

$y=\dfrac{9}{2}cm=QT$

$\Rightarrow area(trap)=\dfrac{1}{2}\times \left(PQ+SR\right)\times QT$

$=\dfrac{1}{2}\times 4.5\left(5+6\sqrt{3}\right)$

$=\dfrac{4.5}{2}\times 15.38$

$=4.5\times 7.19$

$=32.355cm^2$

Hence, the answer is

$32.355cm^2.$

1166607_878882_ans_575d9fa88e8f427b94fb054c71cd0108.png

Volume of a cube is 3375 $\mathrm { cm } ^ { 3 }$ . Complete the following activity to findthe total surface area of the cube.

2040858_1433170_ans_7d8bb3151d98407fb08ac93d3feef122.jpg

the perimeter of a trapezium is 104m,its non-parallel sides are 18m and 22m, and its altitude is 16m.find the area of trapezium

2041260_1458018_ans_589b19df1d46423ca1e152e3a7acbbf0.jpg

A matchbox measures $4 \mathrm{cm} \times 2.5 \mathrm{cm} \times 1.5 \mathrm{cm} .$ What will be the volume of a packet containing $12$ such boxes?

VOLUME:

The space occupied by an object is called the volume of the particular object.

Volume is always measured in cubic unit.

Given:

Dimensions of the matchbox

$= 4 \text{ cm} \times 2.5 \text{ cm} \times 1.5 \text{ cm}$

So,

$l = 4 \text{ cm}$ ,

$b = 2.5 \text{ cm}$ and

$h = 1.5 \text{ cm}$

Volume of cuboid

$= (l × b × h)$

Volume of one matchbox = Volume of cuboid

Volume of one matchbox

$=4 × 2.5× 1.5$

$= 15 \text{ cm}^3$

Volume of a packet containing

$12$ such boxes

$= 12\; \times$ volume of one match box

$= (12 ×15) \text{ cm}^3 = 180\text{ cm}^3$

Hence, the volume of a packet containing

$12$ such boxes

$= 180$

$\text{cm}^3$

Diagram of the adjacent picture frame has outer dimensions $=24$ cm $\times 28$ cm and inner dimensions $16$ cm $\times 20$ cm. Find the area of each section of the frame, if the width of each section is same.

Area of trapezium

$=\dfrac{1}{2} \times$ (sum of parallel sides)

$\times$ height.

Height of section

$I \& II =\dfrac{28-20}{2}=4\ cm$
Height of section

$III \& IV =\dfrac{24-16}{2}=4\ cm$

Area of section

$I=$ Area of section

$II$

$=\dfrac{1}{2}\times 4 \left( 20+28\right)$

$=2\left( 48\right)$

$=96cm^2$

Area of section

$III=$ Area of section

$IV$

$=\dfrac{1}{2}\times 4\times \left( 16+24\right)$

$=40\times 2$

$=80cm^2$

2105477_1347679_ans_dd8e99a22c9b48b3b3ba9e9c64de0897.png

The dimensions of a rectangular box are in the ratio of $2:3:4$ and the difference between the cost of covering it with sheet of paper at the rates of Rs. $8$ and Rs. $9.50$ per ${m}^{2}$ is Rs. $1248$ . Find the dimensions of the box,

Let

$l=2x,b=3x$ and

$h=4x$ in metres.

Surface area

$=2\left(lb+bh+lh\right)=52\ x^{2}m^{2}$

Difference of costs at the two rates

$=Rs\left\{52\ x^{2}\times 9.5- 52\ x^{2}\times8\right\}=78 \ x^{2}$

$\therefore 78\ x^{2}=1248\Rightarrow x=4$

dimensions of the box are

$l=8,b=12,h=16$

Find the lateral surface area and total surface area of a cube of edge $10cm$

we have,

$a=10\ cm$

lateral surface area

$=4{a}^{2}=4\times {10}^{2}{cm}^{2}=400{cm}^{2}$

total surface area

$=6{a}^{2}=6\times {10}^{2}{cm}^{2}=600{cm}^{2}$

A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio $8:5$ , show that the radius of each is to the height of each as $3:4$ .

For cylinder, we have

base radius

$=$ r, height

$=$ h

$\therefore S_1=$ Curved surface

$=2\pi rh$

For cone, we have

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

$S_2=\pi rl=\pi r\sqrt{r^2+h^2}$

We have,

$\dfrac{S_1}{S_2}=\dfrac{8}{5}$

$\Rightarrow \dfrac{2\pi rh}{\pi r\sqrt{r^2+h^2}}=\dfrac{8}{5}$

$\Rightarrow \dfrac{2h}{\sqrt{r^2+h^2}}=\dfrac{8}{5}$

$\Rightarrow \dfrac{4h^2}{r^2+h^2}=\dfrac{64}{25}$

$\Rightarrow 25h^2=16r^2+16h^2\Rightarrow 9h^2=16r^2\Rightarrow 3h=4r$

$\Rightarrow \dfrac{r}{h}=\dfrac{3}{4}$ .

A trapezium with its parallel sides in the ratio $16: 5$ , is cut from a rectangle whose sides are $63$ metre and $5$ metre respectively. The area of the trapezium is of the rectangle. Find the length of the parallel sides of the trapezium.

1911046_1880607_ans_68b6a09abb8d4707bc0bd2cb92405ca1.jpg

	Radius(cm)	height(cm)	Curved surface	Area of the base	Total surface
i)	7	9
ii)	14	10
iii)	21	5

Mensuration - Class 8 Maths - Extra Questions

A cuboidal vessel is 12 cm12\ \text{cm} long and 8 cm8\ \text{cm} wide. How high should it be made so that it may hold 480480 cubic centimetre of a liquid?

Find the side of a cube, whose total surface area is 216 cm2cm^{2} ?

Find the volume of a cuboid of length 4 cm4\ cm,breadth 3 cm3\ cm and height 1 cm1\ cm

Find the capacity of a closed rectangular cistern whose length is 88m,breadth 66m and depth 2.52.5.

Find the area of trapezium whose parallel sides are 24\ cm24\ cm and 20\ cm20\ cm and the distance between them is 15\ cm15\ cm.

The parallel sides of a trapezium are 8.48.4 cm and 12.312.3 cm. If its height is 7.27.2 cm; then the value of its area is 74.52\,\,cm^274.52\,\,cm^2If true then enter 11 and if false then enter 00

If a solid cuboid of metal has dimensions 24 cm, 18 cm and 4 cm, then its volume 1728 cm^3^3.If true enter 1 else 0.

How much length of the iron sheet 1111 cm wide is required for making an open cylinder 1515 cm high and 77 cm as base radius \displaystyle \left ( \pi =22/7 \right )\displaystyle \left ( \pi =22/7 \right )?

A cuboid is 33 cm high, 44 cm wide and 55 cm long. Its volume is ____ (in \text{cm}^3\text{cm}^3).

The parallel sides of trapezium are 9cm9cm and 7cm7cm long and the distance between them is 6cm6cm. Find the area of the trapezium

Find the area of a trapezium whose parallel sides are 6cm6cm and 10cm10cm long and the distance between them is 4cm4cm.

Find the total surface area of a cube with side 55 cmcm.

The radius of the base of a right circular cylinder is 55 cmcm and height is 21 21 cmcm, find the curved surface area.

A rectangular sheet of paper 44 cm \times\times 18 cm is rolled along the length to form a cylinder. Assuming that the cylinder is solid (Completely filled), find its total surface area and the radius.

Find the total surface area of a cube with side 7 7 cmcm.

A water tank is 6 m in length, 3 m in breadth, and 4 m in height. How much liter water can be stored in it?

The length, breadth, and height of a tank are 3 \text{ m},\;2 \text{ m}3 \text{ m},\;2 \text{ m} and 6 \text{ m}6 \text{ m} respectively. What is the volume of the tank in \text{m}^3\text{m}^3?

Salma bought a tank measuring 3 m X 3 m X 2 m for storing water.How many litres of water can be stored in it ?(in L)

The radius and height of a right circular cylinder are 44 and 1414cm respectively. Find the curved surface area of a cylinder.

Find the area of a trapezium whose parallel side are 24cm24cm and 20cm20cm B the distance them is 15cm15cm

The perimeter of a trapezium is 52cm52cm and its non-parallel sides are each equal to 10cm 10cm and its altitude is 8cm 8cm its Area =?=?

The are of a trapezium id 1080cm^21080cm^2 If the length of its parallel sides are 55.6cm55.6cm and 34.4cm34.4cm find the distance between them.

Find the volume in cm^3cm^3

Two cylinders of same radii 14\ cm14\ cm, and heights 20\ cm 20\ cm, 25\ cm25\ cm are placed one upon another then find the T.S.AT.S.A (Total Surface Area) of the shape formed.

If the ratio between the curved surface area and the total surface area of a right circular cylinder is 3:43:4, find the ratio of height to the radius of the cylinder.

A 4 cm4 cm cube is cut into 1 cm1 cm cubes. Calculate the total surface area of all the small cubes.

Find the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2\ m4.2\ m in diameter and 4.5\ m4.5\ m high.

Find the sum of length of the parallel sides of a trapezium whose area is 4.2 \ m^24.2 \ m^2 and whose height is 280 \ cm280 \ cm.

The lengths of the parallel sides of a trapezium are 3030 centimetres, 1010 centimetres and the distance between them is 2020 centimetres. What is its area?

Find the volume of cuboid , if its length , breadth and height are 6 cm , 5 cm and 3 cm respectively .

A rectangular sheet of dimensions 25\ cm\times 7\ cm25\ cm\times 7\ cm is rotated about its longer side. Find the whole surface area of the solid thus generated.

If the areas of three adjacent faces of a cuboid are X, YX, Y and ZZ respectively, then find the volume of a cuboid.

Find the area of a trapezium whose parallel sides are 24cm24cm and 20cm20cm and the distance between them is 15cm15cm

The length of an edge of a cube is ll. Find the formula for the sum of lengths of all the edges of the cube.

Find the area of a trapezium whose parallel sides are 1111 m and 2525 m and the non parallel sides are 1515 m and 1313 m.

Given equilateral ABCDABCD with right angles in BB and DD. We also know that BC=1, CD=4BC=1, CD=4 and DA=3\ unitsDA=3\ units. What is the area of ABCDABCD

A cuboid has volume four times the volume of a cube whose each edge is 7\ \mathrm{m}7\ \mathrm{m}. Find the volume of the cuboid.

ABCDABCD is a trapezium in which AB || CDAB || CD and AB=3cm ,DC=5cmAB=3cm ,DC=5cm. Find the of area if height is 4cm4cm

Find the ratio between TSA of a cylinder to CSA given that its height and radius are 7.57.5 and 3.53.5 cm.

Find the total surface area of cubes having the following sides.5\ cm5\ cm

Find the total surface area of cubes having the following sides.7.2\ m7.2\ m

The volume of a solid cylinder is 448 \pi \ cm^3448 \pi \ cm^3 and height 7 \ cm7 \ cm. Find its lateral surface area and total surface area.

The diameter of a cylindrical tin is 8080 cm cm and height is 1.51.5 m m . How many square metre will be its curved surface area? (\pi = 3.14 )(\pi = 3.14 )

A cuboid of dimensions 11 cm \times 2\times 2 cm \times 5\times 5 cm, then how many small cubes of dimension 0.750.75 cm can be cut.

The area of a rectangular stage is 750750 sq.m if its height is 1.51.5 m find its volume.

A small indoor green house (her barium) is made entirely of glass panes (including house) held together with tape.It is 3030 cm long, 2525 cm wide and 2525 cm high. Find the volume of the house.

If the dimensions of a room are 44 m \times 3.5\times 3.5 m \times 3\times 3 m. What is the volume of air contained in the room?

Find the volume of the container of measurement 5.5 m , 4 m ,3 m.

Lateral surface area of a cube whose edge is 3.5cm3.5cm is

Find the surface area and lateral surface area of a cube whose edge is 99m.

Find the volume of a cuboid which is 15\ \text{m}15\ \text{m} long, 12\ \text{m}12\ \text{m} wide and 4.5\ \text{m}4.5\ \text{m} high.

Find the volume of a cube if its total surface area is 294\ sq\ cm294\ sq\ cm.

Find the total surface area of cubes having the following sides.1) 3 cm2) 5 cm3) 5.5 cm

The edges of a cuboid are in the ration 3:2:1 and it's surface area is 88\,c{m^2}88\,c{m^2}. Find the volume of a cuboid.

Total surface area of cylinder is 60\pi\ cm^ {2}60\pi\ cm^ {2} and its height 7\ cm7\ cm. Find the curved surface area of cylinder.

The adjoining figure shows the cross-section of a concrete wall to be constructed. It is 1.9 m wide at the bottom, 90 cm wide at the top, and 5 m high. If its length is 20 m, find:(i) the cross-sectional area;.

Calculate the area of the isosceles trapezium drawn above .

ABCDABCD is a trapezium with AB \parallel CDAB \parallel CD. AD = BC= 5 cm, AB =12 cmAD = BC= 5 cm, AB =12 cm and CD =7 cmCD =7 cm, find the area of trapezium ABCDABCD.

If total surface area of a cube is 150\ cm^{2}150\ cm^{2}, find the edge.

Find the area of the following trapeziums

Find the area of the following trapeziums

Find the area of a rhombus whose diagonals are 12,1612,16 cmcm

The volume of cuboid with length breadth and height as 3\ cm,4\ cm,5\ cm 3\ cm,4\ cm,5\ cm respectively is

The volume of a tank, which is cuboid in shape, is 7.2 m^3^3. The dimensions of its base are 1.2 m\times\times 0.6 m. Find the depth of the tank.

If side of a cube is double then how many times its total surface area will increase?

Find the volume of the cuboid whose dimensions are given below. Length Breadth Height8.2 m5.3m2.6m

The diagonal of a cube is 16 \sqrt{3}16 \sqrt{3} cm. Find its surface area and volume.

Find the surface area of a cube whose edge is 14 cm.

Find the curved surface area and total surface area of a right circular cylinder of height 1515 cm and whose base radius is 77 cm.

Radius of the base of a cylinder is 20 cm and its height is 13 cm. Find its curved surface area of and total surface area.

Find the volume, curved surface area and total surface area of a cylinder whose dimensions are, radius of the base =7\ cm=7\ cm and height =50\ cm=50\ cm

Find the volume, curved surface area and total surface area of a cylinder whose dimensions are radius of the base =14\ dm=14\ dm and height =15\ m=15\ m

Find the area of the shaded portion in the following figures.

Fill in the blanks statements are true.The curved surface area of a cylinder is reduced by _________ peer cent if the height is half of the original height.

The area of a trapezium is 475 cm^2475 cm^2 and the height is 19 cm, then the lengths of smaller side of two parallel sides if one side is 4 cm greater than the other is

The following figure shows the cross-section ABCD of a swimming pool which is trapezium in shape. If the width DC of the swimming pool is 6464 m, depth (AD) at the shallow end is 8080 m and depth (BC) at the deepest end is 2424 m, find its area of the cross-section(m^2m^2).

The area of a trapezium is 279279 sq. cm and the distance between its two parallel sides is 1818 cm. If one of its parallel sides is longer than the other side by 55 cm, find the value of smallest lengths of its parallel sides.

The parallel sides of a trapezium are in the ratio 3 :If the distance between the parallel sides is 9 dm and its area is 126\,\,dm^2126\,\,dm^2, find the lengths of smallest its parallel sides.

The two parallel sides and the distance between them are in the ratio 3 : 4 :If the area of the trapezium is 175 \,\,cm^2175 \,\,cm^2; find its height.

A cuboidal vessel is $12\ \text{cm}$ long and $8\ \text{cm}$ wide. How high should it be made so that it may hold $480$ cubic centimetre of a liquid?

Find the side of a cube, whose total surface area is 216 $cm^{2}$ ?

Find the volume of a cuboid of length $4\ cm$ ,breadth $3\ cm$ and height $1\ cm$

Find the capacity of a closed rectangular cistern whose length is $8$ m,breadth $6$ m and depth $2.5$ .

Find the area of trapezium whose parallel sides are $24\ cm$ and $20\ cm$ and the distance between them is $15\ cm$ .

The parallel sides of a trapezium are $8.4$ cm and $12.3$ cm. If its height is $7.2$ cm; then the value of its area is $74.52\,\,cm^2$
If true then enter $1$ and if false then enter $0$

If a solid cuboid of metal has dimensions 24 cm, 18 cm and 4 cm, then its volume 1728 cm $^3$ .
If true enter 1 else 0.

How much length of the iron sheet $11$ cm wide is required for making an open cylinder $15$ cm high and $7$ cm as base radius $\displaystyle \left ( \pi =22/7 \right )$ ?

A cuboid is $3$ cm high, $4$ cm wide and $5$ cm long. Its volume is ____ (in $\text{cm}^3$ ).

The parallel sides of trapezium are $9cm$ and $7cm$ long and the distance between them is $6cm$ . Find the area of the trapezium

Find the area of a trapezium whose parallel sides are $6cm$ and $10cm$ long and the distance between them is $4cm$ .

Find the total surface area of a cube with side $5$ $cm$ .

The radius of the base of a right circular cylinder is $5$ $cm$ and height is $21$ $cm$ , find the curved surface area.

A rectangular sheet of paper 44 cm $\times$ 18 cm is rolled along the length to form a cylinder. Assuming that the cylinder is solid (Completely filled), find its total surface area and the radius.

Find the total surface area of a cube with side $7$ $cm$ .

The length, breadth, and height of a tank are $3 \text{ m},\;2 \text{ m}$ and $6 \text{ m}$ respectively. What is the volume of the tank in $\text{m}^3$ ?

The radius and height of a right circular cylinder are $4$ and $14$ cm respectively. Find the curved surface area of a cylinder.

Find the area of a trapezium whose parallel side are $24cm$ and $20cm$ B the distance them is $15cm$

The perimeter of a trapezium is $52cm$ and its non-parallel sides are each equal to $10cm$ and its altitude is $8cm$ its Area $=?$

The are of a trapezium id $1080cm^2$ If the length of its parallel sides are $55.6cm$ and $34.4cm$ find the distance between them.

Find the volume in $cm^3$

Two cylinders of same radii $14\ cm$ , and heights $20\ cm$ , $25\ cm$ are placed one upon another then find the $T.S.A$ (Total Surface Area) of the shape formed.

If the ratio between the curved surface area and the total surface area of a right circular cylinder is $3:4$ , find the ratio of height to the radius of the cylinder.

A $4 cm$ cube is cut into $1 cm$ cubes. Calculate the total surface area of all the small cubes.

Find the lateral or curved surface area of a closed cylindrical petrol storage tank that is $4.2\ m$ in diameter and $4.5\ m$ high.

Find the sum of length of the parallel sides of a trapezium whose area is $4.2 \ m^2$ and whose height is $280 \ cm$ .

The lengths of the parallel sides of a trapezium are $30$ centimetres, $10$ centimetres and the distance between them is $20$ centimetres. What is its area?

A rectangular sheet of dimensions $25\ cm\times 7\ cm$ is rotated about its longer side. Find the whole surface area of the solid thus generated.

If the areas of three adjacent faces of a cuboid are $X, Y$ and $Z$ respectively, then find the volume of a cuboid.

Find the area of a trapezium whose parallel sides are $24cm$ and $20cm$ and the distance between them is $15cm$

The length of an edge of a cube is $l$ . Find the formula for the sum of lengths of all the edges of the cube.

Find the area of a trapezium whose parallel sides are $11$ m and $25$ m and the non parallel sides are $15$ m and $13$ m.

Given equilateral $ABCD$ with right angles in $B$ and $D$ . We also know that $BC=1, CD=4$ and $DA=3\ units$ . What is the area of $ABCD$

A cuboid has volume four times the volume of a cube whose each edge is $7\ \mathrm{m}$ . Find the volume of the cuboid.

$ABCD$ is a trapezium in which $AB || CD$ and $AB=3cm ,DC=5cm$ . Find the of area if height is $4cm$

Find the ratio between TSA of a cylinder to CSA given that its height and radius are $7.5$ and $3.5$ cm.

Find the total surface area of cubes having the following sides.
$5\ cm$

Find the total surface area of cubes having the following sides.
$7.2\ m$

The volume of a solid cylinder is $448 \pi \ cm^3$ and height $7 \ cm$ . Find its lateral surface area and total surface area.

The diameter of a cylindrical tin is $80$ $cm$ and height is $1.5$ $m$ . How many square metre will be its curved surface area? $(\pi = 3.14 )$

A cuboid of dimensions $1$ cm $\times 2$ cm $\times 5$ cm, then how many small cubes of dimension $0.75$ cm can be cut.

The area of a rectangular stage is $750$ sq.m if its height is $1.5$ m find its volume.

A small indoor green house (her barium) is made entirely of glass panes (including house) held together with tape.It is $30$ cm long, $25$ cm wide and $25$ cm high. Find the volume of the house.

If the dimensions of a room are $4$ m $\times 3.5$ m $\times 3$ m. What is the volume of air contained in the room?

Lateral surface area of a cube whose edge is $3.5cm$ is

Find the surface area and lateral surface area of a cube whose edge is $9$ m.

Find the volume of a cuboid which is $15\ \text{m}$ long, $12\ \text{m}$ wide and $4.5\ \text{m}$ high.

Find the volume of a cube if its total surface area is $294\ sq\ cm$ .

Find the total surface area of cubes having the following sides.
1) 3 cm
2) 5 cm
3) 5.5 cm

The edges of a cuboid are in the ration 3:2:1 and it's surface area is $88\,c{m^2}$ . Find the volume of a cuboid.

Total surface area of cylinder is $60\pi\ cm^ {2}$ and its height $7\ cm$ . Find the curved surface area of cylinder.

The adjoining figure shows the cross-section of a concrete wall to be constructed. It is 1.9 m wide at the bottom, 90 cm wide at the top, and 5 m high. If its length is 20 m, find:
(i) the cross-sectional area;
.

$ABCD$ is a trapezium with $AB \parallel CD$ . $AD = BC= 5 cm, AB =12 cm$ and $CD =7 cm$ , find the area of trapezium $ABCD$ .

If total surface area of a cube is $150\ cm^{2}$ , find the edge.

Find the area of a rhombus whose diagonals are $12,16$ $cm$

The volume of cuboid with length breadth and height as $3\ cm,4\ cm,5\ cm$ respectively is

The volume of a tank, which is cuboid in shape, is 7.2 m $^3$ . The dimensions of its base are 1.2 m $\times$ 0.6 m. Find the depth of the tank.

Find the volume of the cuboid whose dimensions are given below.
Length Breadth Height

8.2 m 5.3m 2.6m

The diagonal of a cube is $16 \sqrt{3}$ cm. Find its surface area and volume.

Find the curved surface area and total surface area of a right circular cylinder of height $15$ cm and whose base radius is $7$ cm.

Find the volume, curved surface area and total surface area of a cylinder whose dimensions are, radius of the base $=7\ cm$ and height $=50\ cm$

Find the volume, curved surface area and total surface area of a cylinder whose dimensions are radius of the base $=14\ dm$ and height $=15\ m$

Fill in the blanks statements are true.
The curved surface area of a cylinder is reduced by _________ peer cent if the height is half of the original height.

The area of a trapezium is $475 cm^2$ and the height is 19 cm, then the lengths of smaller side of two parallel sides if one side is 4 cm greater than the other is

The following figure shows the cross-section ABCD of a swimming pool which is trapezium in shape. If the width DC of the swimming pool is $64$ m, depth (AD) at the shallow end is $80$ m and depth (BC) at the deepest end is $24$ m, find its area of the cross-section( $m^2$ ).

The area of a trapezium is $279$ sq. cm and the distance between its two parallel sides is $18$ cm. If one of its parallel sides is longer than the other side by $5$ cm, find the value of smallest lengths of its parallel sides.

The parallel sides of a trapezium are in the ratio 3 :If the distance between the parallel sides is 9 dm and its area is $126\,\,dm^2$ , find the lengths of smallest its parallel sides.

The two parallel sides and the distance between them are in the ratio 3 : 4 :If the area of the trapezium is $175 \,\,cm^2$ ; find its height.

If $A, B, C$ denote the area of the three continuous faces of a rectangular solid cuboid and $V$ be its volume, then prove that $\displaystyle V=\left ( ABC \right )^{1/2}$ .

The total surface of a rectangular block is $846$ $\displaystyle \text{cm}^{2}$ . Find the volume, if the dimensions are in proportion $5 : 4 : 3$ .

The three coterminous edges of a rectangular solid are $36\text{ cm}, \ 75 \text{ cm}$ and $80 \text{ cm}$ respectively, find the edge of a cube which will be of the same capacity.

Find the area of a trapezium whose parallel sides are of length 10 cm and 6 cm respectively and the perpendicular distance between the two parallel sides is 4.5 cm. ( in $cm^2$ )

The shape of the top surface of a table is trapezium. Find its area if it's parallel sides are $1\ m$ and $1.2\ m$ and perpendicular distance between them is $0.8\ m$ .

Sohan wants to buy a trapezium shaped field. Its sides along the river is parallel to and twice the sides along the road. If the area of this field is $10500$ sq metre and the perpendicular distance between the two parallel sides is $100$ m , find the length of the side along the river.

The area of the trapezium is $34$ sq cm and the length of one of the parallel side is $10$ cm and its height is $4$ cm. Find the length of the parallel side.

Calculate the surface area(in $cm^2$ ) of a solid cylinder with diameter $14$ cm and height $10$ cm. $\left ( \pi = \dfrac{22}{7} \right)$

In a trapezoid of area $20$ , the two bases measure $4$ and $6$ . What is the height of the trapezoid?

Circular discs of $5 mm$ thickness, are placed one above the other to form a cylinder of curved surface area $462 cm^2$ . Find the number of discs, if the radius is $3.5 cm$ .

The length of a cold storage is double its breadth. Its height is 3 metres. The area of its four walls ( including doors ) is 108 m $^2$ . Find its volume.

A tent is in the form of a cylinder surmounted by a cone having its diameter of the base equal to 24 m. The height of cylinder is 11 m and the vertex of the cone is 5 m above the cylinder. Find the cost of making the tent, if the rate of canvas is Rs. 10 per m $^2$ .