A piece of wire is $\dfrac { 3 } { 4 } m$ long. What will be the total length of 10 such pieces of wire?

Length of a piece

$= \dfrac{3}{4}\,m$

$\Rightarrow$ length of 10 such pieces

$= 10+\dfrac{3}{4}\,m$

$= \dfrac{30}{4} = 7.5\,m$

1061365_1176925_ans_5ec1b62ce3cd45899eb8a49989ef5263.jpg

Find the circumference of the circle whose radius is $28$ mm. Take $\pi=\dfrac{22}{7}$ .

Let

$r$ be the radius of the circle.

Then according to the problem we've

$r=28$ .

Then perimeter will be

$(P)=2\pi r$

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

$P=176$ mm.

So the perimeter of the circle is

$176$ mm.

If the length of longest chord of a circle is $8\ cm$ then find its radius.

The longest chord is the diameter of the circle

$\therefore$ Diameter

$=8$ cm

Radius

$=\dfrac{diameter}{2}=\dfrac{8}{2}=4$ cm

$\therefore$ Radius of the circle

$=4$ cm

Find the area of the circle having radius $7$ cm.

Given the radius of the circle

$r=7$ cm.

Now the area of the circle is

$\pi\times r^2=\dfrac{22}{7}\times 7^2=154$ cm

$^2$ .

Find area of circle with circumference $88cm$

Circumference of circle is

$2\pi r=2\times \dfrac{22}{7}r=88\\r=14cm$

Area of square is

$\pi r^2=\dfrac{22}{7}\times 14\times 14\\22\times 2\times 14=616cm^2$

Find the circumference of the circle having radius $7$ cm.

Given the radius of the circle is

$7$ cm.

Then its circumference

$=2 \pi r$

Circumference

$=2\times \pi\times 7$

Circumference

$=2\times 22=44$ cm.

The school ground is Circular in shape if its radius is 21 m Find its area

The radius of ground is

$21cm$

The area is given as

$\pi r^2\\\dfrac{22}{7}\times 21\times 21\\22\times 3\times 21=10626m^2$

Find the diameter of $a$ wheel whose is $176\ cm$ .

Let $r$ be the radius of the wheel.

Then according to the problem,

$2\pi r=176$

or, $2r=\dfrac{176}{\pi}$

or, $2r=56.022$ .

So diamerer of the wheel is $56.022$ cm.

In a triangle, the measures of the angles are x, x + 20 and 2x what is the value of x?

The angles of triangle is

$x,x+20,2x$

By angle sum property

Sum of all angles is equal to

$180$

$x+x+20+2x=180\\4x=180-20\\4x=160\\x=40$

Find the area of parallelogram whose base is 48 cm and height is 25 cm

Base of parallelogram

$b= 48\ cm$

Height of parallelogram

$h =25\ cm$

Area of parallelogram

$=b \times h \\= 48\times 25\\=1200\ cm^2$

The circle of radius 5cm has the area as..

The radius of circle is

$5cm$

The area of circle is

$\pi (5^2)=25\pi cm^2$

Find the area of a circle of radius $7$ cm.

Formula:
Area of circle with radius

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

Given

$r=7$ cm

Area of given circle

$=\dfrac{22}{7}\times {7}^{2}=22\times 7=154$ sq.cm

The minute hand of a circular clock is $15\ cm$ long. How far does the tip of the minute hand move in $1$ hour? (Take $\pi =3.14$ )

$1$ hour, minute hand completes one round means makes a circle.
Radius of the circle

$(r)=15$ cm
Circumference of circular clock

$=2\pi r$

$=2\times 3.14\times 15=94.2$ cm
Therefore, the tip of the minute hand moves

$94.2$ cm in

$1$ hour.

The circumference (correct upto three decimal places) of a circle is $3.14\ m$ , find its area.

Given: Circumference of circle

$=3.14\space\mathrm{m}$

$\Rightarrow 2\pi r=3.14$

$\Rightarrow 2\times\dfrac{22}{7}\times r=3.14$

$\Rightarrow r=3.14\times\dfrac{1}{2}\times\dfrac{7}{22}$

$\Rightarrow r=0.5 \space\mathrm{m}$

Now, Area of circle

$=\pi r^2=\dfrac{22}{7}\times(0.5)^2=0.785\space\mathrm{m^2}$

Find the circumference of a circle whose diametre is $10.5\ cm$

Given
Diameter

$=10.5\ cm$

Circumference of the circle,

$C=2\pi r$

$=\pi \times (2r)$

$=\pi \times$ diameter of the circle

$=22/7\times 10.5$

$=33\ cm$

Hence, the circumference of the circle is

$33\ cm$

Ratio of the circumference of a circle to its diameter is denoted by symbol _____ .

$\Rightarrow$ Circumference

$=2 \pi r$

$\Rightarrow C =\pi d$

$\Rightarrow \dfrac {C}{d}=\pi$
Therefore,

$\pi$ is the answer.

Area of a parallelogram with base $24\ cm$ and height $4\ cm$ is:

We know that area of parallelogram

$=base\times height$

So,

$A=24\times 4\ cm^2$

$A=96\ cm^2$

If area of a triangular piece of cardboard is $90\, cm^2$ , then the length of altitude corresponding to $20\, cm$ long base is _____ cm.

Area of triangle

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

$\Rightarrow 90=\dfrac{1}{2} \times 20\times h$

$\Rightarrow h = 9\, cm$
Therefore, height is

$9\, cm$ .

1 hectare $=$ _____ $m^2$

$\Rightarrow 1 \,\,hectare =10000\,m^2$

Circumference 'C' of a circle ean be found by multiplying diameter 'd' with _____ .

$\Rightarrow$ Circumference

$=2\pi r$ ,

$\Rightarrow diameter = 2r$

$\Rightarrow C=\pi \times d$
Therefor,

$\pi$ is the answer.

$1\, km^2 =$ _____ $m^2$ .

We know that

$1\ km = 1000\, m$

$\Rightarrow 1\, km^2 = (1000)^2 \, m^2$

$\Rightarrow 1\, km^2 = 1000000\, m^2$

Area of a triangle $=\dfrac {1}{2} \times base \times$ ______ .

$A=\dfrac {1}{2} \times base \times height$ .

Find the circumference of a circle whose diameter is $4.9 \,m$

Here, diameter of the circle is given

$d = 4.9 \,m = \dfrac{49}{10}$
Circumference of a circle when diameter is given

$= \pi d \,units$
Where,

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

$= \dfrac{22}{7} \times \dfrac{49}{10}$

$= 11 \times \dfrac{7}{5}$

$= 15.4 \,m$

Find the area of a circle whose diameter is $28 \text{ cm}$

Area of a circle of radius

$r$ is

$\pi r^2$

$\implies$ Area

$=\dfrac{22}{7}\times 14\times 14=616\text{ cm}^2$

The carpet for a room $6.6\ m$ by $5.6\ m$ costs $Rs\ 3960$ and it was made from a roll $70\ cm$ wide. Find the cost of the carpet per metre.

1683373_1676806_ans_cb32834be17f40eeb61d7184bd14c6d2.PNG

Find the circumference of a circle whose radius is $3.5\ m$

Given radius

$=3.5\ cm$
We know that, Circumference of the circle is

$C=2\pi r$

$=(2\times 22/7\times 3.5)$

$=(21\times 22/7)$

$=22\ cm$

Thus, the circumference of the circle is

$22\ cm$

The area of a circle is $(14 + 6\sqrt{5})\pi$ units, Find its radius.

Area of circle

$=(14+6\sqrt{5})\pi$ sq. units

$\pi r^2=(14+6\sqrt{5})\pi$

$r^2=(14+6\sqrt{5})$

$r=\pm\sqrt{14+6\sqrt{5}}$

r cann't be negative.

$\therefore$ Radius of circle

$=\sqrt{14+6\sqrt{5}}$ units.

The radius of a circle is $28cm$ find the circumference and area.

circumference

$=2\pi r=2\times \dfrac{22}{7}\times 28 =22\times 8=$

$=176\ cm$

Area

$=\pi r^{2}=\dfrac{22}{7}\times 28\times 28 =88\times 28$

$=2464\ cm^{2}$

The area of the circular ring included between two concentric circles of radii $14\ \text{cm}$ and $10.5\ \text{cm}$ to the closest integer is ...........

Given, radius of outer circle,

$R = 14\ \text{cm}$ and

radius of inner circle,

$r = 10.5\ \text{cm}$ .

Area of the circular ring

$=\pi \left( { R }^{ 2 }-{ r }^{ 2 } \right) =\frac { 22 }{ 7 } \left( { 14 }^{ 2 }-{ 10.5 }^{ 2 } \right)\ \text{cm}^2 = 269.5 ~\approx 269\ \text{ cm }^{ 2 }$

Find the circumference (in cm) of a circle whose diameter is 5.6 m.

Diameter

$= 5.6 m =>$ radius

$= 2.8 m$

Circumference of a circle of radius r

$= 2\pi r$

Circumference of circle of radius

$2.8 m = 2\pi \times (2.8 m)$

$\Rightarrow 2 \times \dfrac {22}{7} \times 2.8 = 17.6 \ m$

$\Rightarrow 1760cm$

A horse is tethered by a rope $10$ m long at a point. Find the area of the region where it can graze. $(\pi = 3.14)$

The area of the region, the horse can graze is circular with radius equal to length of the rope.

$\text{Area of a circle}= \pi { r }^{ 2 } \\= 3.14 \times{ 10 }^{ 2 } \\= 3.14 \times 100\\ = 314 \ m^2$

Hence, the area of region where horse can graze is

$314 {m}^{2}$

diameter is $\displaystyle 17\dfrac{1}{2} \,\,cm$ is $240.625\,\, cm^2$

If true then enter $1$ else 0

Diameter $= 17\frac {1}{2} cm = 17.5 cm =>$ radius $= 8.75 cm$

Area of a circle of radius r $= \pi {r}^{2}$

Area of circle of radius

$8.75 cm = \pi { r }^{ 2 } = \frac { 22 }{ 7 } \times 8.75\times 8.75 =240.625 { cm }^{ 2 }$

radius = 17.6 cm is $972.6464\,\,cm^2$

If true then enter $1$ and if false then enter $0$

Area of a circle of radius r

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

Area of circle of radius

$17.6 cm = \pi { r }^{ 2 }$

$= 3.14 \times 17.6\times 17.6$

$=972.6464 { cm }^{ 2 }$

diameter $21.6\ \text{m}$ is $366.2496 \,\text{m}^2$ .

If true then enter $1$ and if false then enter $0$

Diameter $= 21.6 \ \text{m}$
Radius $= 10.8\ \text{ m}$

Area of a circle of radius

$r = \pi {r}^{2}$
Area of circle of radius

$10.8\ \text{ m} = \pi { r }^{ 2 }$

$= 3.14 \times 10.8\times 10.8\ \text{m}^2$

$= 366.2496 \text{ m }^{ 2 }$

Hence, the given statement is true.

The radius of the circle is $7$ cm , Find the area of the circle in $\text{cm}$ $^2$ .

Given, radius of circle

$=7$ cm

Area of a circle

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

$= \dfrac {22}{7} \times { 7 }^{ 2 } \\= 154 \ \text{cm}^{2}$

$1$ sq m $\displaystyle =t$ sq cm.Find $t$

1 mt = 100cm

1 mt sq = 100 x 100 sq cm

$1$ sq cm $\displaystyle = k$ sq mm.Find $k$

1 cm=10 mm

To convert from square centimeters to square millimeters, we have to multiply our figure by 100 (or divide by 0.01) .

Thus, 1sq cm

$\displaystyle =$ 100 sq mm

So the correct answer to this question is,

$k=100$

Area of square is $10,000$ sq $m.$ Then its side measures(in m)

Area of square = 10,000 sq m

area of square = s x s

s x s = 10000

s=100

The area of a parallelogram is $9.6 {cm}^{2}$ , and the base is $3.2 cm$ . Find the height.

We use the formula area

$= b \times h$ . Here area

$= 9.6 {cm}^{2}$ ,

$b = 3.2 cm$ . Hence

$h = \dfrac{area}{b} = \dfrac{9.6}{3.2} = 3 cm$ .

Find the area of parallelogram in which base is $2.5 cm$ , and altitude is $1.4 cm$ .

We know area

$=$ base

$\times$ height.

Hence area

$= 2.5 \times 1.4 = 3.50 {cm}^{2}$ .

Find the area of the circle from given information:
No. Radius $(r)$ Diameter $(d)$ Area of circle
$(1)$ $42\ cm$
$(2)$ $14\ metre$
$(3)$ $1.4\ cm$
$(4)$ $28\ metre$
$(5)$ $35\ cm$
$(6)$ _ $5.6\ metre$ __

As we know

$d=2r\ and\ area=\pi r^2$

We get the answers as below:

$(1) 84\ cm,5544\ sq\ cm$

$(2) 7\ m,154\ sq\ m$

$(3) 2.8\ cm,6.16\ sq\ cm$

$(4) 56\ m,2464\ sq\ m$

$(5) 70\ cm,3850\ sq\ cm$

$(6) 2.8\ m,24.64\ sq\ m$ .

How many times will a wheel of radius $35\ cm$ be rotated to travel $660\ m$ ?
(Take $\pi = \dfrac {22}{7})$ .

Radius of wheel

$(r)=35\space\mathrm{cm}$

Distance covered in one rotation

$=2\pi r=2\times\dfrac{22}{7}\times35=220\space\mathrm{cm}$

So, Number of rotations covered in

$660\space\mathrm{cm}=\dfrac{\text{Total distance travelled}}{\text{Distance travelled in one rotation}}=\dfrac{660}{220}=3$

The two adjacent sides of a parallelogram are $4.8 cm$ and $3.6 cm$ . The distance between the longer sides is $2.4 cm$ . Find the distance between the other pair.

We use the formula

${b}_{1}{h}_{1} = {b}_{2}{h}_{2}$ . Here

${b}_{1} = 4.8 cm$ ,

${b}_{2} = 3.6 cm$ ,

${h}_{1} = 2.4 cm$ . Hence

${h}_{2} = \dfrac{{b}_{1}{h}_{1}}{{b}_{2}} = \dfrac{4.8 \times 2.4}{3.6} = 7.2 cm$ .

In the following figure, the radius of the circle is $7$ cm and $m(arc RYS) = 30^{\circ}$ , find find:
Area of the circle

In the given figure,
Radius of the circle with centre

$P$ is

$r = 7$ cm

$M(arc RYS) = 30^{\circ}$
Area of circle

$= \pi r^{2}$

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

$= 154\ cm^{2}$

Sarlaben stiches golden plate on edges of circular bed of $7$ metre diameter. Labour to stich golden plate is $Rs. 3$ per $1$ metre. Find the labour charge of Sarlaben in rupees.

Given: Diameter of circular bed

$(d)=7\space\mathrm{m}$

$\Rightarrow$ Radius of circular bed

$(r)=\dfrac{d}{2}=\dfrac{7}{2}\space\mathrm{m}$

Labour to stich golden plate on edges of circular bed

$=\mathrm{Rs.}\space3$ per

$\mathrm{m}$ .

Therefore, Total Labour charges

$=\mathrm{Rs.}\space3\times 2\pi r$

$=\mathrm{Rs.}\space(3\times 2\times\dfrac{22}{7}\times\dfrac{7}{2})$

$=\mathrm{Rs.}\space66$

Calculate circumference of a circle from given information:
No. Radius $(r)$ Diameter $(d)$ Circumference
$(1)$ $70\ cm$ ____
$(2)$ _ $14\ metre$
$(3)$ $49\ metre$ _
$(4)$ $3.5\ cm$ _
$(5)$ _ $42\ metre$
$(6)$ _ $5.6\ cm$ ____

As we know

$d=2r$ and

$circumference=2\pi r$

Therefore we get the answers as below

$(1) 140\ cm, 440\ cm$

$(2) 7\ metre, 44\ metre$

$(3) 98\ m, 308\ metre$

$(4) 7\ cm, 22\ cm$

$(5) 21\ metre, 132\ metre$

$(6) 2.8\ cm, 17.6\ cm$ .

The radius of a circle is $7\ cm$ . Find the circumference of the circle in cm.

Given,

$r = 7\text{ cm}$
Circumference of circle

$= 2 \pi r$

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

$= 44\text{ cm}$

On the wall of school's prayer hall, in a circle of $70\ cm$ radius, a picture of nature is to be drawn, if the cost of drawing is $Rs. 5$ per $100\ sq\ cm$ , then find the total cost in rupees.

Given: Radius of circular portion on the wall

$(r)=70\space\mathrm{cm}$

Cost of drawing

$=\mathrm{Rs.}\space5$ per

$100\space\mathrm{sq. cm}$ .

Therefore, Total cost of drawing

$=\mathrm{Rs.}\space\dfrac{5}{100}\times \pi r^2$

$=\mathrm{Rs.}\space[\dfrac{5}{100}\times\dfrac{22}{7}\times(70)^2]$

$=\mathrm{Rs.}\space770$

A heritage stepwell is secured by a circular wire fence. Radius of this circle is $17.5\ metre$ . Find the cost of fencing(in Rs) at the rate of $Rs. 50$ per metre.

Given: Radius of circular wire fence

$(r)=17.5\space\mathrm{m}$

Cost of fencing

$=\mathrm{Rs.}\space50$ per

$\mathrm{m}$ .

Therefore, Total cost of fencing

$=\mathrm{Rs.}\space50\times2\pi r$

$=\mathrm{Rs.}\space[50\times2\times\dfrac{22}{7}\times17.5]$

$=\mathrm{Rs.}\space5500$

Radius of the circular swimming pool is $70\ cm$ . Then find the area of the bottom of the pool in $sq\ cm$

Radius of circular swimming pool

$=70\space\mathrm{cm}$

Then,

Area of the bottom of the pool

$=\pi r^2=\dfrac{22}{7}\times70^2=15400\space\mathrm{cm^2}$

Diameter of a bangle is $7\ cm$ . Thin stripe of gold is to be fixed on this bangle, find the cost(in Rs) for it at the rate of $Rs. 100$ per $1\ cm$ .

Given: Diameter of bangle

$(d)=7\space\mathrm{cm}$

$\Rightarrow$ Radius of bangle

$(r)=\dfrac{d}{2}=\dfrac{7}{2}\space\mathrm{cm}$

Labour charges to fix thin stripe of gold on bangle

$=\mathrm{Rs.}\space100$ per

$\mathrm{cm}$ .

Therefore, Total Labour charges

$=\mathrm{Rs.}\space100\times2\pi r$

$=\mathrm{Rs.}\space[100\times2\times\dfrac{22}{7}\times\dfrac{7}{2}]$

$=\mathrm{Rs.}\space2200$

No. Radius $(r)$ Diameter $(d)$ Circumference Area of circle
$(1)$ $0.7\ m$ ____ _
$(2)$ ____ $84\ cm$ _
$(3)$ $140\ metre$ __ _
$(4)$ $4.9\ cm$ __ _
$(5)$ _ $70\ cm$ _

As we know

$d=2r , circumference=2\pi r\ and\ area =2\pi r^2$

We get the answers as below:

(1)

$1.4\ m, 4.4\ m, 1.54\ sq\ m$ .
(2)

$42\ cm, 264\ cm, 5544\ sq\ cm$
(3)

$280\ metre, 880\ metre, 61600\ sq\ m$
(4)

$9.8\ cm, 30.8\ cm, 75.64\ sq\ cm$
(5)

$35\ cm, 220\ cm. 3850\ sq\ cm$ .

Diameter of the wheel of horse-cart is $1.40$ metre. How much distance(in metres) is covered (correct upto one decimal place) by this wheel in one rotation?

Diameter of wheel of horse cart

$(d)=1.40\space\mathrm{m}$

$\Rightarrow$ Radius of wheel of horse cart

$(r)=\dfrac{d}{2}=\dfrac{1.40}{2}=0.7\space\mathrm{m}$

$\therefore$ Distance covered in one rotation

$=2\pi r=2\times\dfrac{22}{7}\times0.7=4.4\space\mathrm{m}$

Radius of the circular ground is $10.5\ m$ . What is the cost(in Rs) to fill up this ground with soil at the rate of $Rs. 150$ per sq m?

Given: Radius of circular ground

$(r)=10.5\space\mathrm{m}$

Cost of filling up soil

$=\mathrm{Rs.}\space150$ per

$\mathrm{sq. m}$ .

Therefore, Total cost of filling

$=\mathrm{Rs.}\space150\times \pi r^2$

$=\mathrm{Rs.}\space[150\times\dfrac{22}{7}\times(10.5)^2]$

$=\mathrm{Rs.}\space51975$

The radius of a circle is increasing at the rate of $0.7 \ cm/sec$ . What is the rate of increase of its circumference ?

Circumference,

$C=2\pi r$

$\dfrac{dC}{dt}=\dfrac{dC}{dr} \times \dfrac{dr}{dt}$

$=\dfrac{d}{dr}(2\pi r)\dfrac{dr}{dt}$

$=2\pi \dfrac{dr}{dt}=2\pi (0.7)=1.4 \pi \ cm/sec$

What is the area of the circle that can be inscribed in a square of side 6cm?

From figure,

$AB=BC=CD=DA=6\ cm$

$PR$ is the diameter of the circle, inscribed in the square

$ABCD$

$\therefore PR=side \ of \ the \ square=6\ cm$

$\therefore Radius=\dfrac{6}{2}=3$

$Area=\pi r^2=\dfrac{22}{7}\times 3^2$

$=\dfrac{22}{7}\times 9$

$\therefore Area \ of \ circle=28.28\ cm^2$

961089_1029496_ans_a4ea67325fc94eccb7c86b03be5c0006.JPG

On a square handkerchief, nine circular designs each of radius $7\ cm$ are made (see the given figure). Find the area of remaining portion of the handkerchief. $\left [Use\ \pi = \dfrac {22}{7}\right ]$ .

Diameter of circle

$=14\ \ cm$

Length of handkerchief

$=3\times14=42\ \ cm$

Area of handkerchief

$=42\times42=1764$

Area of circles

$=9\pi { r }^{ 2 }=9\times \dfrac { 22 }{ 7 } \times 7\times 7=1386$

Area of remaining portion

$=1764-1386=378$

The vertices A , B , C of the triangles are (2 , 1) , (5 , 2) and (3 , 4) respectively . Find the coordinates of the circumcentre and also the radius of the circumcircle

Let P ( x , y ) be the co -ordinates of the centre of the circumcircle

$\therefore \, \, PA^2 \, = \, PB^2 and PB^2 \, = PC^2$

$\therefore \, \, 3x \, + \, y \, = \, 12 \ and x \, + \, 3y \, = \, 10$
solving these two we get the centre as (13 / 4 , 9 / 5 ) Having found P we can say that

$PA^2 \, = \, r^2 \, = \, 50/16 \, \, \therefore \, \, r \, = \, 5 \sqrt{2}/4$

Find the area of a circle inscribed in a square of side $6$ cm.

From figure, Diametre =side of a square=

$6cm$

Hence radius

$=3cm$ ,

So, the Area of circle

$=\pi.(3)^2=9\pi cm^2$

$\implies 28.26cm^2$

1058310_1150518_ans_4afe2be022dd4790a61d533c2a63aab9.png

Find the area of the following circle, given that: $Radius=5cm$

Area

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

$=\cfrac{22}{7}\times 25$

$=\cfrac{550}{9}{cm}^{2}=78.57{cm}^{2}$

The diameters of the two circles are 32 cm and 24 cm. Find the radius of the circle having its area equal to the sum of the areas of the two given circle.

Let,

$d$ be the diameter and

$r$ be the radius of circle

For circle 1,

${d_1} = 32cm$

$\Rightarrow {r _1} = 16cm$

For circle 2,

${d_2} = 24cm$

$\Rightarrow {r_2} = 12cm$

Area of circle 1

$= \pi r_1^2 = \pi {(16)^2} = 256\pi$

Area of circle 2

$= \pi r_2^2 = \pi {(12)^2} = 144\pi$

It is given that,

Area of circle 3 = Area of circle 1 + Area of circle 2

Area of circle 3

$= 256\pi + 144\pi$

$\pi r_3^2 = (256 + 144)\pi$ where

$r_3$ is the radius of third circle

${r_3} = \sqrt {256 + 144}$

${r_3} = \sqrt {400} = 20$

${r_3} = 20 \, cm.$

Find the cicumference of the circle whose diameter is $4$ units.

Circumference of the circle

$=2\pi r$

Given,

diameter

$=4$ units

$\implies$ radius

$= 2$ units

circumference

$=2\pi \times 2$

$=4 \pi$ units

If the radius of a circle is $3\ cm$ , find its circumference.

$\Rightarrow$ Given the radius of the circle

$r=3\ cm$

$\Rightarrow$ circumference of the circle

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

$=2\times \pi\times 3=6{\pi}\ cm=18.84cm$

Find the area of a circular park whose radius is $4.5$ m.

Given that

Radius

$r=4.5m$

The area of circular park=

$\pi r^2$

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

$=\dfrac{891}{14}=63.642$

The diameter of the whell of a can is $77cm$ .How many revolutions will it make to travel $242$ km

$Given,$ Total distance travel

$=$

$242km$

$=$

$242\times$

$10^5cm$

Diameter of wheel (d) is

$77cm .$

The circumference of wheel

$=$

$\pi d$

$=77\pi$ cm

No. of revolutions

$=$

$\dfrac { Total\quad distance\quad travel\quad }{ circumference }$

$=$

$\dfrac { 242{ \times 10 }^{ 5 }cm }{ 77\pi cm }$

$=$

$10^5$

$So ,$ The no. of revolutions will be

$10^5$

Two adjacent sides of parallelogram are 24 cm and 18 cm. If the distance between the longer sides is 12 cm, find the distance between the shorter sides.

The length of side

$24cm$

Its corresponding height is

$12cm$

The area of parallelogram is

$24\times 12=288cm^2$

The length of another side is

$18cm$

The area of parallelogram is

$288cm^2$

Its corresponding height is

$\dfrac {288}{18}=16cm$

1318149_1066148_ans_a6d0445375f14b6fa3d0950ff09c8e4c.PNG

Two sides of the parallelogram PQRS are $6$ cm and $4$ cm. The height corresponding to the base RS is $3$ cm. Find area of parallelogram.

Area of parallelogram =(base

$\times$ corresponding height )

$=6\ cm \times 3\ cm$

$=18\ cm^2$

Write the formula to find the Area of a sector of angle $\theta$ of a circle with radius $r$ .

Area of the sector

$=\dfrac{\theta}{360} \times \pi r^2$

There are four circular windows in Nisha's house, each of radius $28 cm$ are to be fitted. Find the total area required for these windows.

Given that,

Radius of window $=28\,cm=0.28\,m$

Let $A$ be the area of each circular window

$A=\pi {{r}^{2}}\\=\dfrac{22}{7}\times {{\left( \dfrac{28}{100} \right)}^{2}}\\=\dfrac{22\times 28\times 28}{7\times 100\times 100}\\=0.2464{{m}^{2}}$

Let $A'$ be the area of 4 circular windows

$A'=4\pi {{r}^{2}}\\=4\times 0.2464\\=0.9856\,{{m}^{2}}$

Two sides of the parallelogram PQRS are $6$ cm and $4$ cm. The height corresponding to the base RS is $3$ cm. Find The height corresponding to the base PS.

1326026_1064581_ans_21b6d46f207c4334a035fab8889a404a.jpg

Find the circumference of Circle if Diameter = 4 cm.

Diameter

$=4$ cm

Radius

$=2$ cm

Circumference

$=2 \pi r=2 \pi (2)=4 \pi$ cm

Find the area of the following circle, given that:
$Radius=14mm ( \pi=\dfrac{22}{7})$

Area

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

$=\cfrac{22}{7}\times (14)(14){mm}^{2}$

$=44\times 14{mm}^{2}=616{mm}^{2}$

Find the area of a circular park of janakpuri whose circumference is $77m$ .

Circumference

$=2\pi r=77$

$\Rightarrow\,2\times\dfrac{22}{7} \times r=77$

$\Rightarrow\,r=\dfrac{7\times 77}{2\times 22}=\dfrac{7\times 7}{2\times 2}=\dfrac{49}{4}$

Area of the circle

$=\pi{r}^{2}=\dfrac{22}{7}\times \dfrac{49}{4}\times \dfrac{49}{4}=471.625\,sq.\,m$

Hence,area of a circular park of janakpuri is

$471.625\,sq.\,m$

Find the area of the following circle, given that:
$Diameter=49m$

Area

$=\cfrac{\pi {d}^{2}}{4}$

$=\cfrac{22}{7}\times \cfrac{49\times 49}{4}$

$=\cfrac{154\times 49}{4}=\cfrac{77\times 49}{2}=1886.5{m}^{2}$

A car has wheels each of which is 84 cm in diameter. Find how many complete revolutions must each wheel take to cover a distance of 792 m.

Let the radius of each wheel be r cm. Then,

Diameter = 84 cm

$2r=84$

$r=42 \ cm$

Thus, circumference of the wheel =

$2\pi r$

$2\pi r =2\times \dfrac{22}{7} \times 42 = 264 \ cm = 2.64 \ m$

Hence, each wheel covers 2.64 m in one complete revolution.

Therefore, total number of revolutions to cover 792 m =

$\dfrac{792}{2.64}=300$ .

Find the ratio of the circumferences of two concentric circles of radii $2 m$ and $3 m$ .

Circumference of 1st circle=

$2* \pi*r_1$

circumference of 2nd circle=

$2*pi*r_2$

Ratio of their circumference of circles=r1:r2

$\therefore$ Ratio of circumference of given circle =

$\dfrac{2}{3}$

Use $\pi =\frac{22}{7}$ , unless stated otherwise.
The circumference of a circle is 39.6 cm. Find its area.

1521180_1198659_ans_a926b34d1cfa415d992940d74ecf09d9.jpg

The circumference of a circle is $57.2cm$ . Find its radius.

$2\pi r=57.2cm$

$r=\dfrac{57.2}{2\pi}$

$r=9.1cm$

Hence the radius is

$9.1cm$

A circular park is surrounded by a rod $21 m$ wide. If the radius of the park is $105 m$ , find the area of the road.

Area of Road

$=Perimeter\times Width\ of \ road =2\pi r\times21=42*\pi*105=42*105*\dfrac{22}{7}=6*105*22=13860\ m^2$

Find the area of the following circles , given that:
(a) radius = 14 mm (take = $\frac{22}{7})$ .
(c) radius =5 cm.

Area of a circle

$=\pi r^2$

(a) Radius

$=14$ cm

Area

$=\pi(14)^2=\dfrac{22}{7}(14)(14)$

Area

$=616mm^2$

(b) Radius

$=5$ cm

Area

$=\pi(5)^2=\dfrac{22}{7}(25)=78.5cm^2$ .

1201347_1196299_ans_39008f93be4d4e6bbf1ed90f5542eb2c.jpg

Find the circumference of a circle whose area is $12474 cm^2$

The area of the circle $=12474\,c{{m}^{2}}$

We know that the area of the circle $=\pi {{r}^{2}}$

$\dfrac{22}{7} \times {{r}^{2}}=12474$

$r^2 =12474\times \dfrac{7}{22}$

${{r}^{2}}=3969$

$r=63\,cm$

We know that the circumference of the circle

$=2\pi r$

$=2\times \dfrac{22}{7} \times 63$

$=396\,cm$

How to find area of circle?

Let there be a circle of radius

$'a'$ .

Take a ring of radius

$r$ and thickness

$dr$

Area of ring = Perimeter

$\times$ thickness

$\Rightarrow dA=2\pi r dr$

Total area of circle

$\Rightarrow \displaystyle \int{dA}=\displaystyle \int_{0}^{a}{2\pi r dr}$

$\Rightarrow A=2\pi \left[\dfrac{r^2}{2}\right]_{0}^{a}$

$\Rightarrow A=\pi r^2$

1355561_1198660_ans_46cce3c9c8764ed3bda85a31b7c62274.png

In the figure given below, $AD = 8 cm$ and the area of the parallelogram $ABCD = 44$ $cm^2$ . Find the measure of the corresponding altitude.

Area of parallelogram

$=$ Base

$\times$ Altitude

Base

$=8\ cm$

Area

$=44\ cm^2$

Let altitude

$=x$

then,

$\Rightarrow44=8\times x$

$\Rightarrow x=\dfrac{44}{8}$

$\Rightarrow x=5.5\ cm$

Hence Altituude is

$5.5 cm.$

Area of a parallelogram is $400\ cm^{2}$ and its height is $16\ cm$ . Find its base.

Given

$\Rightarrow$

Area of parallelogram

$=400$

$cm^2$

Height

$=16 \ cm$

$base \times height=Area$

$base \times height=400 \ cm^2$

$base \times 16=400 \ cm^2$

$base=\dfrac{400cm^2}{16cm}$

$Base=25 \ cm$ .

Find the circumference of a circle of radius $175\ cm.\ \left(Take\ \pi=\dfrac{22}{7}\right)$

circumference of circle

$= 2\pi r$

$= 2 \times \dfrac{22}{7}\times 175$

$= 1100 cm$

The circumference of a circle is 97.2 cm. Find its diameter and radius.

1338085_1289605_ans_d0c8966c1e4042b1bb757b3ea035d40a.PNG

The area of a circle is $616\ cm^2$ . Find its circumference.

$A=\pi r^{2}= 616$

$r^{2}=\dfrac{616}{22}\times 7$

$r^{2}= 296$

$r=\sqrt{296}= 14\ cm$

circumference

$,2\pi r=2\times \dfrac{22}{7}\times 14$

$circumference= 88\ cm$

Find the area enclosed by the circle of radius $(a)$ .

Radius of circle

$= a$

Area of circle

$= \pi a^{2}$

$= \pi (a)^{2}$

Area =

$\pi a^{2}$

If the area of circle is $176cm^2$ .find the radius of the circle.

The area of circle is given by

$\pi r^2=176cm^2\\\dfrac{22}{7}\times r^2=176\\r^2=176\times \dfrac{7}{22}\\r^2=8\times 7\\r=\sqrt{56}=2\sqrt{14}$

Find the height of a parallelogram whose area is $54\ cm^{2}$ and the base is $15\ cm$ .

Area of a parallelogram

$A=$ base

$\times$ height

$\Rightarrow A=b\times h$

$\Rightarrow h=\dfrac{A}{b}=\dfrac{54}{15}=3.6‬$ cms

$\therefore$ height of the parallelogram

$=3.6$ cms.

Find the area of the circular park,whose circumference is 22 m.

Circumference of the circular park

$=22\ m$

We know that the circumference of the circle

$=2\pi r$

Therefore,

$2\pi r=22$

$r=\dfrac{7}{2}\ m$

Area of the circle

$=\pi r^2$

$=\pi\times (\dfrac{7}{2})^2$

$=\dfrac{7\times 7\times 22}{4\times 7}$

$=\dfrac{7\times 11}{2}$

$=\dfrac{77}{2}\ m^2$

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 ?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

In $\triangle ABC$ , $seg MN\parallel seg BC$ , if $BM=6$ , $AN=5$ , $AM=2$ , then find $NC$ .

$\dfrac{AM}{AB}=\dfrac{AN}{AC}$

$\dfrac{2}{2+6}=\dfrac{5}{5+x}$

$5+x=20$

$x=15$

$\therefore NC=15$

Find diameter & circumference with radius 7.7 cm

Radius

$=7.7\text{ cm}$

Diameter

$=2\; \times$ Radius

$=2\times 7.7\text{ cm}$

$=15.4\text{ cm}$

Diameter

$=15.4\text{ cm}$

Circumference

$=2\pi\;\times$ Radius

$=2\times \dfrac{22}{7}\times 7.7$

Circumference

$=48.4\text{ cm}$

The radius of circle is $7cm$ find its circumference and Area.

The radius of circle is

$7cm$

The circumference of circle is

$2\pi r\\2\dfrac {22}{7}\times 7\\44cm$

The area of circle is

$\pi r^2\\\dfrac {22}7\times 7\times 7=154cm^2$

Area of circle $=?\ r=0.129\ cm$ .

Radius of circle given = 0.129 cm

Area of circle = π x r x r

= (22/7) x 0.129 x 0.129

= 0.0523 cm2

If the diameter of a semi circular protractor is 14 cm, then find its perimeter .

Perimeter if semicircle

$=\pi r + 2r$

since

$d=14cm$

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

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

$= 22cm + 14cm$

$= 36cm.$

Find the perimeter of a circle with radius $6.3 \ \mathrm { m }.$ Given $\left( \pi = \frac { 22 } { 7 } \right)$

Given: Radius of circle

$=6.3\ m$

Circumference of a circle

$=2\pi r$

$=2\times \dfrac{22}{7}\times 6.3$

$=44\times 0.9$

$=39.6\ m$

A race track is in the form of a ring whose inner circumference is 352m. and the outer circumference is 396m. Find the width of the track.

Let the radius of inner track

$=r$

$2\pi r = 352$

$2 \times \dfrac{{22}}{7} \times r = 352$

$r = 56$

let the radius of outer track

$=R$

$2\pi R = 396$

$2 \times \dfrac{{22}}{7} \times R = 396$

$R=63$

width of track

$=$ radius of outer track

$-$ radius of inner track

$=63-56=7m.$

Find the area of circle of diameter $7\ cm$ .

Given that,

$d=7cm$

$\therefore r= \dfrac{7}{2} = 3.5cm$

$\therefore$ Area of circle

$=\pi {r^2}$

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

$= \dfrac{{77}}{2}c{m^2}$

$=38.5 cm^2$

Then,

We get Area pf circle is

$38.5cm^2$ .

The area of a circular field is $13.86\ m^{2}$ . Find the circumference of the field .

Given that,

Area of circular field is

$13.86 m^2$ .

But area of circle

$= \pi {r^2}$

$\therefore \pi {r^2}=13.86$

${r^2} = \dfrac{{\left( {13.86 \times 7} \right)}}{{22}} = 4.41m$

$\therefore r=2.1m$

Hence,

We get Circumference of the field

$2\pi r = 2 \times \dfrac{{22}}{7} \times 2.1 = 13.2m$

A circle is inscribed in a square whose each side is $36$ cm. find the area of the circular region

1190958_1311118_ans_fe33a3edbcc54a7a93c0d09fdc6c6dca.jpeg

The radius of a wheel is 0.25 m . Find the number of revolutions it will make to travel a distance of 11 kms

Radius

$=0.25m$

circumference

$=2\pi r = 2 \times \dfrac{{22}}{7} \times 0.25$

$No.\,of\,revolutions = \dfrac{{total\,dis\tan ce}}{{circumference}}$

$= \dfrac{{11000 \times 7}}{{2 \times 22 \times 0.25}} = 7000\,\,revolutions$

The area of parallelogram $PQRS$ is $88\ \text{cm}^2$ . A perpendicular from $S$ is drawn to intersect $PQ$ at $M$ . If $SM=8\ \text{cm}$ , then find the length of $PQ$ .

R.E.F image

Area (ABCD) = 88

$cm^{2}$

Extending it to

'T'

Hence, the area

of parallelogram

will be equal to area of rectangle

SRTM.

$\therefore$

$Area=88=SM \times MT=8\times MT$

MT= 11 cm

also, PQ = MT = 11 cm.

1212302_1294802_ans_41c4380c9c974665b988a7ec427294b2.JPG

if the perimeter of a protractor is 108 cm, then what is its radius
?

1316033_1488959_ans_e786aa3dae5e44fa869c5e00d2a6de40.JPG

The area enclosed between the concentric circles is 770 cm $^{2}$ . If the radius of the outer circle is 21 cm, find the radius of the inner circle.

Let the radius of the inner circle be r cm. Then,

$( \pi \times 21 \times 21)-\pi r^2=770$

$\pi(441-r^2)=770$

$\dfrac{22}{7} \times (441-r^2)=770$

$r^2=196$

$r=14$

$cm$

Find the area of the circle whose circumference is:
$62.8\ cm$

As we know that,

Circumference of circle

$= 2 \pi r = 62.8 \; cm \quad \left( \text{Given} \right)$

$\Rightarrow r = \cfrac{62.8}{2 \times 3.14} = \cfrac{62.8}{6.28} = 10 \; cm$

Therefore,

Area of circle

$= \pi {r}^{2} \\ = 3.14 \times 10 \times 10 = 314 \; {cm}^{2}$

Circumference of two circle =i) $374$ cm
ii) $396$ cm
Find the area of the shaded region.

1196943_1320545_ans_9acc26d502d44868ae5827198fc37d94.JPG

Find the perimeter of the following: (Given $\pi = \frac { 22 } { 7 } )$
A circle with diameter $8.4$ $\mathrm { m }$ .

Given,

$d=8.4m$

$r=\dfrac{8.4}{2}=4.2m$

Circumference

$=2\pi r$

$=2\times \dfrac{22}{7}\times 4.2$

$=44\times 0.6$

$=26.4m$

Find the circumference and area of a circle of diameter $56 \; m$ .

Diameter of circle

$= 56 \; m$

Radius of circle

$= \cfrac{56}{2}$

$= 28 \; m$

Therefore,

Circumference of circle

$= 2 \pi r \\ = 2 \times \cfrac{22}{7} \times 28 \\ = 176 \; {m}^{2}$

Also,

Area of circle

$= \pi {r}^{2} \\ = \cfrac{22}{7} \times {\left( 28 \right)}^{2} \\ = 2464 \; {m}^{2}$

Find the area of a circle of diameter $8.4\ m.$

As we know that,

Area of circle

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

Whereas

$r$ is the radius of circle.

Diameter of circle

$= 8.4 \; m \quad \left( \text{Given} \right)$

Radius of circle

$= \cfrac{8.4}{2} = 4.2 \; m$

Therefore,

Area of the circle

$= \cfrac{22}{7} \times 4.2 \times 4.2 \\ = 22 \times 4.2 \times 0.6 = 55.44 \; {m}^{2}$

Taking $\pi=3.14$ , find the circumference of a circle whose radius is
$8 cm$

As we know that,

Circumference of a circle

$= 2 \pi r$

Whereas,

$r$ is the radius of circle.

Given:-

Radius of circle

$= 8 \; cm$

Therefore,

Circumference of circle

$= 2 \times 3.14 \times 8 = 50.24 \; cm$

Find the circumference of a circle whose radius is equal to $15\text{ cm}$ . Use $\pi = 3.14$ .

As we know that, circumference of a circle is equal to

$2 \pi r$ whereas,

$r$ is the radius of circle.

So, by using this formula,

Circumference of circle

$= 2 \times 3.14 \times 15$

$= 94.2 \text{ cm}$

Taking $\pi=3.14$ , find the circumference of a circle whose radius is
$20\ cm$

As we know that,

Circumference of a circle

$= 2 \pi r$

Whereas,

$r$ is the radius of circle.

Given:-

Radius of circle

$= 20 \; cm$

Therefore,

Circumference of circle

$= 2 \times 3.14 \times 20 = 125.6 \; cm$

Prove that the area of an equilateral triangle is equal to $\dfrac{\sqrt{3}}{4}a^2$ , where a is the side of the triangle.

Given :

An equlateral triangle ABC such that AB=BC=CA=a.

To prove :

$ar(\triangle ABC) = \dfrac{\sqrt{3}}{4}a^2$

Construction :

Draw

$AD \perp BC$ .

Proof :

$\triangle s ABD$ and

$ACD$ , we have

$AB=AC$

$\angle ADB = \angle ADC = 90^{o}$

$AD=AD$

So, by RHS criterion of congruence,

$\triangle ABD \cong \triangle ACD$

$\implies BD=DC$

But,

$BD+DC=a$

$BD=DC=\dfrac{a}{2}$

Now, in right triangle ABD, we have

$AB^2=AD^2+BD^2$

$a^2=AD^2+(\dfrac{a}{2})^2$

$AD^2=a^2-\dfrac{a^2}{4}=\dfrac{3a^2}{4}$

$AD = \dfrac{\sqrt{3}a}{2}$

Therefore,

$ar(\triangle ABC) = \dfrac{1}{2}(BC \times AD) =\dfrac{1}{2}(a\times \dfrac{\sqrt{3}}{2}a)=\dfrac{\sqrt{3}}{4}a^2$

1301411_1350684_ans_1510038d8f6d47da8ecfdff6135235b0.png

The ratio of the radii of two circles is $5:6$ . What is the ratio of their circumferences?

Let

$r_{1}$ be the radius of one circle.

Let

$r_{2}$ be the radius of another circle.

$c_{1}$ be the circumference of one circle.

$c_{2}$ be the circumference of another circle.

$\frac{r_{1}}{r_{2}}=$

$\frac{5}{6}$

$\therefore$

$c_{1}=2\pi r_{1}$ and

$c_{2}=2\pi r_{2}$

$\therefore$

$\frac{c_{1}}{c_{2}}$ =

$\frac{2\pi r_{1}}{2\pi r_{2}}$

$\Rightarrow\frac{c_{1}}{c_{2}}$

$=\frac{r_{1}}{r_{2}}$

$\Rightarrow\frac{c_{1}}{c_{2}}$ =

$\frac{5}{6}$

If the Circumference of circle is $176cm$ Find its area

The circumference of circle is

$2\pi r=176\\2\times \dfrac{22}{7}\times r=176\\r=176\times \dfrac{7}{22}\dfrac 12\\r=7\times 4=28cm$

Area of circle is

$\pi r^2=\dfrac {22}{7}\times 28\times 28=88\times 28=2464cm^2$

Find the area of the circle whose circumference is:
$88\ cm$

As we know that,

Circumference of circle

$= 2 \pi r = 88 \; cm \quad \left( \text{Given} \right)$

$\Rightarrow r = \cfrac{88 \times 7}{22 \times 2} = 2 \times 7 = 14 \; cm$

Thererfore,

Area of circle

$= \pi {r}^{2} \\ = \cfrac{22}{7} \times 14 \times 14 = 616 \; {cm}^{2}$

The radii of two circles are 77 cm and 91 cm respectively. How much more is the circumference of the second circle than the first?

1309603_1532937_ans_d0c5a93572674c63982e07652ca61ca3.jpeg

The cost of fencing a rectangular field at $Rs 18$ per metre is $Rs 1980$ . If the width of the field is $23 m$ find its length.

The perimeter of the rectangle field

$=\dfrac{1980}{18}=110\ m$

The width of the field

$=23\ m$

We know that

$2(length +width)=110$

Therefore,

$2(length +23)=110$

$length +23=55$

$length =32\ m$

Hence, this is the answer.

Find area of circle whose radius is $7\ m$ .
$\pi = 3.14$

We have,

Radius of the circle

$=7\ m$

Since,

$Area =\pi r^2$

Therefore,

$Area =3.14\times 7^2$

$Area =3.14\times 49$

$Area =153.86\ m^2$

Hence, this is the answer.

A path of width $3.5 m$ runs all around a circular pool having an area of $962.5m^{2}$ . Find the area of the path.

$r_1 + 3.5 = r_2$

$\pi r_{1}^{2}= 962.5 m^{2}$

$\therefore r_{1}= 17.5 m$

$\Rightarrow r_{2}= 21 m$

Area of path =

$\pi r_{2}^{2}-\pi r_{1}^{2}$

Area of path =

$\pi (21)^{2}- 962.5$

Area of path =

$422.94 m^{2}$

1172956_1338133_ans_e07cb1b923a742309736f5c2f72d632f.jpg

The moon is about $384000\ km$ from the earth and is nearly circular. Find the circumference of the path described by the moon in one complete revolution about the earth. (use $\pi=3.14$ )

$2\pi r=2\times \dfrac{22}{7}\times 384000=2413714.28$ km.
1189540_1354926_ans_cf79176e9825407a900eed0bdd4772ef.jpg

$AB\parallel CD, AD\parallel BC, DF\bot AB. \, l(AB)=9.9\ cm.,\ l(DF)=4.2\ cm$ . Find the area of $\Box ABCD$ .

As we know that,

Area of the parallelogram is given as-

$A = b \times h$

Here,

$b = AB = 9.9 \; cm$

$h = DF = 4.2 \; cm$

Therefore,

$A = 9.9 \times 4.2 = 41.58 \; {cm}^{2}$

Find the area of the circle whose circumference is:
$396\ cm$

As we know that,

Circumference of circle

$= 2 \pi r = 396 \; cm \quad \left( \text{Given} \right)$

$\Rightarrow r = \cfrac{396 \times 7}{22 \times 2} = 9 \times 7 = 63 \; cm$

Thererfore,

Area of circle

$= \pi {r}^{2} \\ = \cfrac{22}{7} \times 63 \times 63 = 12474 \; {cm}^{2}$

find the area of a quadrant of a circle whose circumference is 88 cm.

Circumference = 88 cm

$2 \pi r = 88$

$\pi r = 44$

$r=14$ cm

Area of quadrant =

$\dfrac{1}{4} \pi r^{2}$ =

$\dfrac{1}{4} \times \dfrac{22}{7} \times 14 \times 14$ =

$154$ sq. cm .

The area of a circle inscribed in an equilateral triangle is 154 cm $^2$ . Find the perimeter of the triangle.

Let r be the radius of the inscribed circle.Then,

Area = 154 cm

$^2$

$\pi r^2=154$

$r=7 \ cm$

Let h be the height of the triangle. Then,

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

$h=3r=21 \ cm$

If a is the side of the triangle. Then,

$h=\dfrac{\sqrt{3}}{2}a$

$a=\dfrac{2h}{\sqrt{3}}=\dfrac{42}{\sqrt{3}}=14\sqrt{3} \ cm$

Hence, perimeter =

$3a=3\times 14 \sqrt{3}=72.7 \ cm$

Find the circumference of a circle of radius $5\ cm$ .

Given radius of circle =5 cm

circumference of a circle =

$2\pi r$

$2\pi \times 5$

$10\pi=10\times 3.14$

circumference=31.4 cm

1215400_1368942_ans_c80420b5968f41d5b1379eb423b85263.JPG

A bicycle wheel makes 5000 revolutions in moving 11 km. Find the diameter of the wheel.

Let the radius of the wheel be r cm.

Distance covered by the wheel in one revolution =

$\dfrac{11}{5000}$

$km$ =

$220$

$cm$

$2\pi r = 220$

$2\times \dfrac{22}{7} \times r=220$

$r=35$

$cm$

Diameter =

$2r=70$

$cm$

Two adjacent sides of a parallelogram have lengths 12 cm and 8 cm. The distance between the two longer sides is 6 cm. Find the area of the parallelogram and the distance between the two shorter sides.

Area of Parallelogram = Base

$\times$ Height =

$8\times 6=48\ \mathrm{cm}^{2}$

So Distance between two shorter sides =

$\dfrac{48}{12}=4\ \mathrm{cm}$

The length, breadth and height of a room are $850$ cm, $625$ cm and $475$ cm. respectively. Find the length of the longest rod that cam measure the dimensions of the room exactly.

Length of the longest rod can be placed along the diagonal of the given room.

Length of longest rod

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

$=\sqrt{850^2+625^2+475^2}$

$cm$

$=\sqrt{722500+390625+225625}$

$cm$

$=\sqrt{1338750}\\=1157.04\ cm$

If the circumference of a circle is $33{ cm } ,$ find its diameter.

Let the diameter be d.

We know that circumference

$=\pi d = 33cm$

$\therefore d = \dfrac{33}{\pi} = 10.5 cm$

Area of circle is given below find their diameters.
$176\ sq\ cm$

Area of a circle

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

$\Rightarrow \pi{r}^{2}=176$ sq.cm

$\Rightarrow \dfrac{22}{7}\times {r}^{2}=176$ sq.cm

$\Rightarrow {r}^{2}=176\times \dfrac{7}{22}=56$ cm

$\Rightarrow r=\sqrt{56}=\sqrt{4\times 14}=2\sqrt{14}$ cm.

$\therefore$ Diameter

$=2r=2\times 2\sqrt{14}=4\sqrt{14}$ cm.

Area of circle is given below find their diameters.
$394.24\ sq\ cm$

Area of a circle

$\pi{r}^{2}$

$\Rightarrow \pi{r}^{2}=394.24$ sq.cm

$\Rightarrow \dfrac{22}{7}\times {r}^{2}=394.24$ sq.cm

$\Rightarrow {r}^{2}=394.24\times \dfrac{7}{22}=56$ cm

$\Rightarrow r=\sqrt{125.44‬}=11.2‬$ cm.

$\therefore$ Diameter

$=2r=2\times 11.2‬=22.4‬$ cm.

The circumference of a circular sheet is $154\ cm$ then find its radius and its area too.

Let

$r$ be the radius of the circular sheet.

Then the circumference of the sheet is

$2\pi r$ .

Then according to the problem we've,

$2\pi r=154$

or,

$2\times \dfrac{22}{7}\times r=154$

or,

$r=\dfrac{49}{2}\ cm$ .

Then the area will be

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

$=\dfrac{22}{7}\times \dfrac{49}{2}\times \dfrac{49}{2}$

$=1886.5\ cm^2$ .

The radius of a circular field is $24.5\ m$ . Find the distance run by a boy in making $4$ complete rounds.

Radius of a circular field

$=24.5$ m

Distance to cover the circular field for once

$=$ Circumference of the circle

$=2\pi\times r=2\times\dfrac{22}{7}\times 24.5=154$ m

Distance run by the boy in making

$4$ rounds

$=4\times 154$ m

$=616$ m

Given that two circles having equal area. Are their radii equal?

Yes, their radii are equal.

Suppose their radii be

$r_1$ and

$r_2$ .

As their areas are same.

Then, we've,

$\pi r_1^2=\pi r_2^2$

or,

$r_1^2=r_2^2$

or,

$r_1=r_2$ . [ Since

$r_1, r_2>0$ ].

Hence their radii are same.

The ratio of radius is $5:6$ Find ratio of areas

The ratio is

$5:6$

Let the common factor be

$x$

The area is

$\pi r^2$

The ratio of areas is

$\dfrac{\pi r_1^2}{\pi r_2^2}=\left(\dfrac{r_1}{r_2}\right)^2=\left(\dfrac{5}{6}\right)^2=\dfrac{25}{36}$

The radius of a wheel is $35$ cm. How far will the wheel travel in one complete revolution?

In one revolution, wheel travels a distance equal to its circumference

$=2\pi r=2\times \dfrac{22}{7}\times 35=220cm$ .

The area of circle if the the circumference is $30\pi$

The circumference of circle is

$2\pi r=30\pi\\r=15cm$

The area of circle is

$\pi r^2\\\pi \times 15 \times 15\\225\pi cm^2$

The diameter of a flowerbed in a garden is $2.8$ m. Find the cost of fencing it at the rate of Rs. $55\ \text{per} \ \text{m}$ .

Fencing would be at circumference of garden

$=2\pi r$

$=2\times \dfrac{22}{7}\times \dfrac{154}{10}=8.8m$

$\Rightarrow$ Cost

$=$ Rs.

$55\times 8.8=$ Rs.

$484$ .

1216126_1397180_ans_a263441a12954b299230911dde602245.jpg

Find the area of a quadrant of a circle whose radius is $14$ cm.

Given the radius of the circle is

$14$ cm.

Then the area of the circle

$=\pi r^2=\dfrac{22}{7}\times (14)^2=\dfrac{22}{7}\times 4\times 7^2=4\times 154$ cm

$^2$ .

Then the area of a quadrant of the circle is

$\dfrac{4\times 154}{4}=154$ cm

$^2$ .

Chord of a circle is $8\ cm$ in length & its distance from the centre is $3\ cm$ . Find the radius of the circle.

Length of chord

$8$

Distance from center

$3$

Length of chord is given by-

$2\sqrt{r^2-d^2}$

$2\sqrt{r^2-3^2}=8$

$\sqrt{r^2-3^2}=4$

$r^2-9=16\\r^2=16+9\\r^2=25\\r=\sqrt{25}\\r=5$

The area of circle made by rotating a rope of length $176cm$ is

The length of rope acts as circumference of circle.

The circumference of circle is

$2\pi r=176\\\dfrac{22}{7} r=88\\r=28cm$

The area of circle is

$\pi r^2\\\dfrac{22}{7}\times 28\times 28=2,464cm^2$

The radius of a circular garden is 56m. What would it cost to put a 4 -round fence around this garden at a rate of 40 rupees per metre ?

Given radius of circular garden

$=56$ m

Perimeter of garden

$=2\pi r=2\pi\times 56$

Perimeter

$=112\pi m$

Now for four rounds

$=112\pi \times 4$

$4$ perimeter

$=448\pi m$ .

Cost of fencing

$=40$ Rs./m

Total cost of fencing

$=40\times 448\pi$

$=17920\pi$ Rs.

Total cost

$=56,267.8$ Rs.

Total cost of fencing is

$56,268.8$ Rs.

1204131_1399889_ans_22b549a20e8345aea65b7f37cf0010a1.JPG

Find area of parallelogram $ABCD$ if $AB=8\ cm, BC=4\ cm$ and distance between the two parallel lines $AB$ and $CD$ is $3\ cm$ . Find the distance between $AD$ and $BC$ .

Given,

$AB=8\,cm;\,BC=4\,cm$

Perpendicular distance between

$AB$ and

$BC$ is

$DL=3\,cm$

The area of the parallelogram

$=base \times height$

$=AB\times DL$

$=8\times3=24\,cm^2$

Area of the parallelogram

$=base \times height$

$=AD\times BM$

$24=4\times height$

$Height=6\,cm$

So the perpendicular distance between

$AD$ and

$BC$ is

$6\,cm$

1263701_1400607_ans_ebb2c5bbdd354917b757aeb124c4daec.png

Find the circumference of the circle with the following radius. (Take $\pi=22/7$ )
$10.5\ cm$

Given the radius of the circle is

$10.5$ cm.

Then, circumference of the circle is

$2\times \dfrac{22}{7}\times 10.5=66$ cm.

The circumference of a circle is 10 $\pi$ cm , find the radius and the area of the circle . (Take $\pi$ = 3.14)

Circumference of circle =

$2\pi r$

$2\pi r=10\pi\\2r=10\\r=5$

Area of circle is

$\pi r^2$

$3.14\times 5\times5$

$3.14\times 25$

$78.5cm^2$

Find area of circle whose circumference is $\pi$

Circumference of circle is

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

Area of circle is

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

The area of circle with radius $21cm$ is

The radius of circle is

$21cm$

The area of circle is

$\pi r^2\\\dfrac {22}7\times 7\times 7=154cm^2$

Draw $2$ concentric circles.

1310117_1405904_ans_c1e533609ca04f3abdaf5a4551ef26ae.jpg

Work out the area the circumference of each circle. Give your answers correct to one decimal place (1 d.p)
Radius = 15 cm

$C=2\pi r=2\times 3.14\times 15$

$=94.2$ cm

$A=\pi r^2=(3.14)\times (15)^2$

$=706.5cm^2$ .

1199917_1445251_ans_fef43f613ad344ed8f31d1dd7717e7ff.jpg

Work out the area the circumference of each circle. Give your answers correct to one decimal place (1 d.p)
diameter = 12 cm

Diameter

$=12$ cm

Radius

$=6$ cm

$c=2\pi r=2\times 3.14\times 6$

$=37.68$ cm

$A=\pi r^2=(3.14)(6)^2$

$=113.04cm^2$ .

1199919_1445254_ans_681e546c7c434c9c95deb11d389203b8.jpg

Work out the area the circumference of each circle. Give your answers correct to one decimal place (1 d.p)
Radius = 3.5 cm

$C=2\pi r=2\times 3.14\times 3.5$

$=21.98$ cm

$A=\pi r^2=\pi(3.5)^2$

$=38.46cm^2$ .

1199918_1445252_ans_9d3a2585275747028655e1bc3248c719.jpg

Find the area of circle whose radius is $49cm$ .

We know that,

Area of circle with radius

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

Given:-

$r = 49 \; cm$

Therefore,

Area of circle

$= \cfrac{22}{7} \times {\left( 49 \right)}^{2} = 22 \times 7 \times 49 = 7546 \; {cm}^{2}$

Find the radius of given diameter :
(i) d = 15 (ii) d = 26 (iii) d = 40

As we know that

Radius

$=\dfrac{\mathrm{diameter}}{2}$

(i)Given

$d = 15$

$\implies$ Radius

$=\dfrac{d}{2}=\dfrac{15}{2}=7.5$

(ii)Given

$d = 26$

$\implies$ Radius

$=\dfrac{d}{2}=\dfrac{26}{2}=13$

(iii)Given

$d = 40$

$\implies$ Radius

$=\dfrac{d}{2}=\dfrac{40}{2}=20$

Find the area of the circles, given that
diameter = 49 m

$A=\pi \left(\dfrac{d}{2}\right)^2$

$a=\dfrac{22}{7}\times \dfrac{49\times 49}{2\times 2}$

$=1886.5m^2$ .

1199965_1445571_ans_67f72d053dd044b09ec2057fb8479d9f.jpg

Find the area of the circles, given that
radius = 14 mm (Take $\pi =\dfrac { 22 }{ 7 }$

$A=\pi r^2$

$=\dfrac{22}{7}\times 14\times 14$

$=616mm^2$ .

1199964_1445566_ans_3cc710dc90d14a20ba7275f6ac3d75da.jpg

Find the diameter of a circle whose circumference is $1256\text{ cm}$ . Use $\pi =3.14$

Circumference of the circle is equal to

$2\pi r$ where

$r$ is the radius of the circle or

$\pi d$ where

$d$ is the diameter of the circle.

$\begin{aligned}{}\pi d &= 1256\\3.14d& = 1256\\d &= 400\text{ cm}\end{aligned}$

Hence, diameter of the circle is equal to

$400 \text{ cm}$ .

If the area of a circle is 78.5 ${ cm }^{ 2 }$ , Finds its circumference . $(Take\pi =3.14)$

$A=\pi r^2$

$\Rightarrow 78.5=3.14r^2$

$\Rightarrow 25=r^2$

$r=5$

$c=2\pi r$

$=2\times 3.14\times 5$

$=31.4$ cm.

1199967_1445576_ans_015c9a809b2b424d94896e3134b6da7b.jpg

Find the diameter of a circle whose circumference is 66 m.

Let the diameter of a circle be

$d\ m$

Given circumference of the circle is

$66\ m$

$\implies \pi d=66$

$\implies \dfrac{22}{7} \times d=66$

$\implies d=\dfrac{66\times 7}{22}=21\ m$

So the diameter is

$21\ cm$

In the given figure,points $A,B,C$ and $D$ are centre of four circles,each having a radius of length $1$ cm. If a point is chosen at random from the interior of the square $ABCD$ , what is the probability that the point will be choosen the shaded region?

Find the area of a circle of diameter $8.4m.$

1325987_1565702_ans_204ae96f85c84f5fb0f06bbe64f9106e.jpeg

If the radius of a circle is 5 cm,What will its diameter be ?

1317389_1502908_ans_2c0d1d9f07624363b6613d4cea8e71bc.jpeg

The radius of the circular pipe is 10 cm. What length of a tape is required to wrap once around the pipe? ( $\text{Take }\pi =3.14$ )

Given that,

Radius of the circular pipe,

$r$ is

$10\ cm$ .

To find out,

Length of tape that can be wrapped around the pipe once

The pipe is circular in shape.

Hence, the length of tape required to wrap around the pipe will be equal to the circumference of the pipe.

We know that,

$\text{Circumference of a circle}=2\pi r\text{ units}$

Here,

$r=10\ cm\text{ and }\pi=3.14$

$\therefore \ C=2\pi r=2\times 3.14\times 10=62.8 \ cm$

Hence, the length of the tape required to wrap once around the given pipe is

$62.8\ cm$ .

The circumference of a circle is 264 cm.Find its area.

1325754_1565543_ans_72c5430f91f94266b4c3d55374efa5e1.jpeg

Find the circumference of a circle whose diameter is $10.5\ m$

Given diameter of a circle is

$d=10.5\ m$

As we know that

Circumference of a circle with diameter

$d$ is

$\pi d$

$\implies$ Circumference of a circle is

$\dfrac{22}{7}\times 10.5=33\ m$

The minute hand of a circular clock is 15 cm. How far does the tip of the minute hand move in 1 hour?
(take $\pi =3.14$ )

The length of minute hand is

$15cm$

The minute hand make one complete revolution in

$1$ hr

So the circumference of circle is

$2\pi r\\2\times 3.14\times 15\\30\times 3.14=94.2cm$

Find the area of circle having : (Take $\pi =3.14)$
(i) radius = 10 cm

The radius of circle is

$10cm$

Area of circle is

$=\pi r^2\\=3.14\times 100=314cm^2$

Find the circumference of the circle with radius $=14\ cm\ \bigg($ Take $\pi =\dfrac{22}{7}\bigg)$

Circumference of the circle

$=2\pi r$

$=2\times \dfrac{22}{7}\times 14$

$=88cm.$

If the circumference of a circle is $264$ cm, find its radius. Take $\pi =\dfrac{22}{7}$ .

Let the radius be

$r\ cm$

Given Circumference of a circle is

$264\ cm$

$\implies 2\pi r=264$

$\implies 2\times \dfrac{22}{7}\times r=264$

$\implies r=\dfrac{264\times 7}{44}=42$

So the radius is

$42\ cm$

Find the area of the above triangle.

We know that, the area of triangle

$=\dfrac{1}{2}\times base\times height$
Here,

$base=3$ cm and

$height=4$ cm

$\therefore$ Area of the triangle

$=\dfrac{1}{2}\times 3\times 4\ cm^2=6cm^2$ .

Hence, the area of the triangle is

$6cm^2$

DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD. If the area of the parallelogram is $1470$ $cm^2$ , AB $=35 \ cm$ and AD $=49\ cm,$ find the length of BM and DL.

Given: Area of parallelogram

$=1470$

$cm^2$
Base(AB)

$=35$ cm and base(AD)

$=49$ cm

Since, Area of parallelogram

$=$ base

$\times$ height

$\Rightarrow 1470=35\times$ DL

$\Rightarrow$ DL

$=\dfrac{1470}{35}$

$\Rightarrow$ DL

$=42$ cm

Again, Area of parallelogram

$=$ base

$\times$ height

$\Rightarrow 1470=49\times$ BM

$\Rightarrow$ BM

$=\dfrac{1470}{49}$

$\Rightarrow$ BM

$=30$ cm

Thus, the lengths of DL and BM are

$42$ cm and

$30$ cm respectively.

Find the circumference of the circle with radius $\bigg($ Take $\pi =\dfrac{22}{7}\bigg)$

Circumference of the circle

$=2\pi r=2\times \dfrac{22}{7}\times 21=132$ cm.

PQRS is a parallelogram. QM is the perpendicular height from Q to SR and QN is the height from Q to PS. If SR $=12\ cm$ and QM $=7.6\ cm$ then find the area of the parallelogram PQRS.

Given:

$SR$ = 12

$cm, QM$ = 7.6

$cm.$

We know that,

Area of parallelogram

$=$

$base\times height = SR\times QM$

$=12\times 7.6=91.2\ cm^2$ .

Find the diameter of the circle whose circumference is 44 cm.

Let the diameter be

$d\ cm$

As we know that

Circumference of circle with diameter

$d$ is

$\pi d$

Given Circumference of a circle is

$44\ cm$

$\implies \pi d=44$

$\implies \dfrac{22}{7}\times d=44$

$\implies d=\dfrac{44\times 7}{22}=14$

So the diameter is

$14\ cm$

Taking $\pi =3.14$ , find the circumference of a circle whose radius is $8 \ cm$

Given : The radius of circle

$8 \ cm$

The circumference of circle

$=2\pi r$

The circumference of circle

$=2\times 3.14\times 8$

The circumference of circle

$=16\times 3.14$

The circumference of circle

$=50.24 \ cm$

Find the circumference of the circle with radius $\bigg($ Take $\pi =\dfrac{22}{7}\bigg)$

Circumference of the circle

=2πr=2×(22/7)×28=176=2πr=2×227×28=176mm.

Find the circumference of a circle whose radius is
$14\ cm$

1683404_1676816_ans_8016d65274fb4b7fa14bf078e5109e1a.PNG

A circular flower bed is surrounded by a path $4\ m$ width. The diameter of the flower bed is $66\ m.$ What is the area of this path? (Take $\pi =3.14$ )

Diameter of the circular flower bed

$=66$ m

$\therefore$ Radius of circular flower bed

$(r)=\dfrac{66}{2}=33$ m

$\therefore$ Radius of circular flower bed with

$4$ m wide path

$(R)=33+4=37$ m

According to the question,
Area of path

$=$ Area of bigger circle

$-$ Area of smaller circle

$=\pi R^2-\pi r^2=\pi(R^2-r^2)$

$=\pi[(37)^2-(33)^2]$

$=3.14[(37+33)(37-33)]$

$[\because a^2-b^2=(a+b)(a-b)]$

$=3.14\times 70\times 4$

$=879.20m^2$ .

Therefore, the area of the path is

$879.20m^2$ .

1346200_1646536_ans_d3a3cf80718641b8af369cdeec8f2021.png

If the circumference of a circular sheet is $154\ m,$ find its radius. Also find the area of the sheet.
$\bigg($ Take $\pi=\dfrac{22}{7}\bigg)$

Circumference of the circular sheet

$=154$ m

$\Rightarrow 2\pi r=154$ m

$\Rightarrow r=\dfrac{154}{2\pi}$

$\Rightarrow r=\dfrac{154\times 7}{2\times 22}=24.5$ m
Now, Area of circular sheet

$=\pi r^2=\dfrac{22}{7}\times 24.5\times 24.5$

$=22\times 3.5\times 24.5=1886.5m^2$ .
Thus, the radius and area of circular sheet are

$24.5$ m and

$1886.5m^2$ respectively.

Find the area of a circle with d $\bigg($ Take $\pi =\dfrac{22}{7}\bigg)$

Diameter

$=49$ m

$\therefore$ radius

$=\dfrac{49}{2}=24.5$ m

$\therefore$ Area of circle

$=\pi r^2=\dfrac{22}{7}\times 24.5\times 24.5=22\times 3.5\times 24.5=1886.5m^2$ .

Find the area of the circle whose radius $\bigg($ Take $\pi =\dfrac{22}{7}\bigg)$

Area of circle

$=\pi r^2=\dfrac{22}{7}\times 14\times 14=22\times 2\times 14=616mm^2$ .

Find the circumference of the inner and the outer circles, shown in the adjoining figure. (Take $\pi =3.14$ )

Radius of outer circle

$(r)=19$ m

$\therefore$ Circumference of outer circle

$=2\pi r=2\times 3.14\times 19$

$=119.32$ m
Now radius of inner circle

$(r')=19-10=9$ m

$\therefore$ Circumference of inner circle

$=2\pi r'=2\times 3.14\times 9$

$=56.52$ m
Therefore the circumferences of inner and outer circles are

$56.52$ m and

$119.32$ m respectively.

A circle of radius $2\ cm$ is cut out from a square piece of an aluminum sheet of side $6\ cm.$ What is the area of the left over aluminum sheet? (Take $\pi =3.14$ )

Radius of circle

$=2$ cm and side of aluminium square sheet

$=6$ cm

According to question,
Area of aluminium sheet left

$=$ Total area of aluminium sheet

$-$ Area of circle

$=$ side

$\times$ side

$-\pi r^2$

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

$=36-12.56$

$=23.44 cm^2$ .

Therefore, the area of aluminium sheet left is

$23.44 cm^2$ .

Find the area of the circle, if r $\bigg($ Take $\pi =\dfrac{22}{7}\bigg)$

Area of circle

$=\pi r^2=\dfrac{22}{7}\times 5\times 5=\dfrac{550}{7}cm^2$ .

$1\ sq\ cm =$ _______ $cm\times 1\ cm$ .

$1\ sq\ cm = 1\ cm\times 1\ cm$ .

Find the area (correct upto two decimal places) of a circle whose diameter is
$7\ km$

Given: Diameter of circle

$=7\space\mathrm{km}$

Then, radius of cirlce

$=r=\dfrac{\text{Diameter}}{2}=\dfrac{7}{2}=3.5\space\mathrm{km}$

$\therefore$ Area of a circle

$=\pi r^2=\dfrac{22}{7}\times(3.5)^2=38.50\space\mathrm{km^2}$

Find the area of a circle whose radius is
$7\ km$

Given:

Radius of circle

$r=7\space\mathrm{km}$

$\therefore$ Area of a circle

$=\pi r^2$

$=\dfrac{22}{7}\times(7)^2=154\space\mathrm{km^2}$

Find the area (correct upto two decimal places) of a circle whose diameter is
5.6 m

Given:

Diameter of circle

$=5.6\space\mathrm{m}$

Then, radius of cirlce

$=r=\dfrac{\text{Diameter}}{2}=\dfrac{5.6}{2}=2.8\space\mathrm{m}$

$\therefore$ Area of a circle

$=\pi r^2=\dfrac{22}{7}\times(2.8)^2=24.64\space\mathrm{m^2}$

Find the circumference of a circle whose diametre is $14\ cm$

Given

Diameter

$=14\ cm=2\times radius=2r$

Circumference of the circle,

$C=2\pi r$

$=\pi \times (2r)$

$=\pi \times$ diameter of the circle

$=44\ cm$

Hence, the circumference of the circle is

$44\ cm$

$1\ sq\ m =$ ____ $sq\ cm$ .

$1\ sq\ m = 1\ m\times 1\ m=100\ cm\times 100\ cm=10000\ sq\ cm$ .

Find the circumference of a circle whose radius is
$10\ cm$

Given radius of a circle is

$(r) =10cm.$

Now, we know that circumference of circle is =

$\pi d$ , where

$d$ is diameter of circle.

So, here diameter of circle id

$d=2\times r$

$\Rightarrow d=2\times10=20cm$

circumference is =

$\pi\times20$ cm

$\dfrac{22}{7}\times20$ cm

Circumference =

$62.8571\,cm$

A rectangular grassy lawn measuring $40\ m$ by $25\ m$ is to be surrounded externally by a path which is $2\ m$ wide. Calculate the cost of levelling the path at the rate of $Rs\ 8.25\ per$ square metre.

1683416_1676820_ans_b73731af47f248ef9e36fb0ee5ff0015.PNG

Find the circumference of a circle whose radius is $10.5\ cm$

Given radius

$=10.5\ cm$

We know that, Circumference of the circle is

$C=2\pi r$

$=(2\times 22/7\times 10.5)$

$=(21\times 22/7)$

$=66\ cm$

Thus, the circumference of the circle is

$66\ cm$

A boy is cycling such that the wheels of the cycle are making 140 revolutions per hour. If the diameter of the wheel is 60 cm. calculate the speed in km/h with which the boy is cycling.

Circumference of a circle =

$2\pi r$
=

$2\times 22/7\times 30$
=

$\frac{44\times 30}{7}$ = 188.57 cm
Distance cover in 140 revolutions =

$140 \times 188.57$ = 26400 cm
Therefore speed = distance/time
=26400/100000 km/h = 0.264 km/h.

A bicycle wheel makes 500 revolutions in moving 1 km. Find the diameter of the wheel.

A bicycle wheel makes 500 revolutions in moving 1 km.
In 1 revolution, the bicycle wheel covers
= 1/500 km = 1000/500 m = 2m
1 revolution distance = circumference of the wheel

$\Rightarrow$

$2\pi r$ = 2

$\Rightarrow$

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

$\Rightarrow$

$r= \frac{2\times 7}{2\times 22}$
=

$\frac{7} {22}$
Therefore, diameter (d) = 2r =

$2\times 7/22$ = 7/11 = 0.636 m.

The areas of two circles are in the ratio 49:Find the ratio of their circumferences.

Ratio of the area of two circles given = 49:64
Area of circle =

$\pi r^{2}$
And the area of the first and second circle =

$\pi r^{2}_{1} and \pi r^{2}_{2}$
Now,

$\frac{49}{64}=\frac{\pi r^{2}_{1}}{\pi r^{2}_{2} }\Rightarrow \frac{49}{64}=\frac{r^{2}_{1}}{r^{2}_{2}}$

$\Rightarrow \dfrac{(7)^{2}}{(8)^{2}}=\dfrac{r^{2}_{1}}{r^{2}_{2}}\Rightarrow \left ( \dfrac{7}{8} \right )^{2}=\left ( \dfrac{r_{1}}{r_{2}} \right )^{2}$

$r_{1}=7,r_{2}=8$
The ratio of circumferences of these two circles-
=

$\dfrac{2\pi r_{1}}{2\pi r_{2}}=\dfrac{r_{1}}{r_{2}}=\dfrac{7}{8}= 7:8$

Circumference 'C' of a circle is equal to $2\pi \times$ _____.

We know that

$C=2\pi \times r$

where

$r=radius$

If the cost of fencing a square plot at the rate of 2.50 per metre is 200, then find the length of each side of the field.

Total cost of fencing a square plot

$= Rs. 200$
Rate of fencing

$= Rs 2.50$

$\therefore$ Perimeter of square =

$\dfrac{Total \ cost}{Rate} = \dfrac{200}{2.5} = 80 m$
Since, we know that
Perimeter of square

$= 4a$

$\Rightarrow 80 m = 4 a$

$\Rightarrow a = 20 m$

$\therefore$ Length of square plot

$= 20 m$

What is the locus of a runner, running around a circular track and always keeping a distance of $1.5m$ from the inner edge.

The locus of the runner, running around a circular track and always keeping a distance of

1.5m1.5m from the inner edge will be the circumference of a circle where the radius is equal to the radius of the inner circular track and exceeding

1.5m1.5m

1796992_1832822_ans_d4fc6bbe7af3456a9c94535ad7d9473b.png

What is the locus of the moving end of the minute hand of a clock.

The locus of the moving end of the minute hand of the clock will be a circle whose radius will be equal to the length of the minute hand.
1796987_1832815_ans_bdf47ed197c84aec99969f1d4ead3545.png

1796987_1832815_ans_bdf47ed197c84aec99969f1d4ead3545.png

The radius of a circle is $21$ cm. Find the circumference (Take $\pi = 22/7$ ).

It is given that,
Radius of a circle =

$21$ cm.
We know that

$\pi = 22/7$ ,
So the circumference of a circle =

$2 \pi r$
Substituting the values,

$= 2 * 22/7 * 21$
So we get,

$= 2 * 22 * 3$

$= 132$ cm.

What is the locus of the door-handle, as the door opens.

The locus of the door handle will be the circumference of a circle with centre as the axis of rotation of the door and the radius equal to distance between the door handle and the axis of rotation of the door.
1797012_1832823_ans_d261955d863f4160a9e7f526c537d034.png

1797012_1832823_ans_d261955d863f4160a9e7f526c537d034.png

The diameter of a circular field is $56$ m. Find its circumference and cost of fencing it at the rate of $80$ per m. (Take $\pi = 22/7$ ).

It is given that,
Diameter of a circular field =

$56$ m.
So the radius =

$56/2$ =

$28$ m.
We know that,
Circumference of the circle =

$2 \pi r$
Substituting the values,

$= 2 * 22/7 * 28$
So we get,

$= 2 * 22 * 4$

$= 176$ m.
Here the cost of fencing

$176$ m =

$176 * 80$ = ₹

$14,080$ .

The circumference of a circle is $440$ cm. Find its radius and diameter. (Take $\pi = 22/7$ ).

It is given that,
Circumference of a circle =

$440$ cm.
So the radius =

$C/ 2\pi$
Substituting the values,

$= (440 * 7)/ (2 * 22)$
So we get,

$= 3080/44$

$= 70$ cm.
Diameter of the circle =

$2 * radius$
So we get,

$= 2 * 70$

$= 140$ cm.

One side of a parallelogram is $20cm$ and its distance from the opposite side is $16 cm$ . Find the area of the parallelogram.

We know that,
Area of parallelogram =

$base \times height$
Here,
Area of parallelogram =

$AB \times DE$
Substituting the values,

$= 20 \times 16$

$= 320 cm^{2}$ .

1748781_1831346_ans_597d7f1422d1483bb15f4107ceb67411.png

There is a path of uniform width 7 m round and outside a circular garden of diameter 210 m. Find the area of the path.

Radius of the circular garden,

$r = \frac {210}{2} = 105 m$

Radius of the region covering the garden and the path,

$R = 105 m + 7 m = 112 m$

Area of the region between two concentric circles
with radius of outer circle R, and inner circle r

$= \pi \left( { R }^{ 2 }-{ r }^{ 2 } \right)$

Hence, the area of the path $= \pi \left( { 112 }^{ 2 }-{ 105 }^{ 2 } \right) = \frac { 22 }{ 7 } \left( 12544-11025 \right) = \frac {33418}{7} = 4774 {m}^{2}$

Figure depicts an archery target marked with its five scoring areas from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide, then match the two columns I and II:

We know that area of circle of radius

$r$

$unit$ is

$πr ^2$

$unit^2$

A) Area of Gold score =

$\pi (10.5)^2 = \frac{22}{7} \times (10.5)^2 = 346.5 cm^2$
We will use the same formula that area of region between two circles = area of bigger circle - area of smaller circle.Thus,
B) Area of Red score =

$\pi 21^2 - \pi (10.5)^2= 1039.5 cm^2$
C) Area of Blue score =

$\pi (31.5)^2 - \pi 21^2 = 1732.5 cm^2$
D) Area of Black score =

$\pi (42)^2 - \pi (31.5)^2 = 2425.5$

A circular field of radius 105 m has a circular path of uniform width 5 m along its boundary. then the area of the path is $\displaystyle 3221 \dfrac{3}{7}\,\, m^2$

If true then enter $1$ and if false then enter $0$

Radius of the inner circular region

$= 105 - 5 = 100 m$

Area of the region between two concentric circles with radius of outer circle R, and inner circle

$r= \pi\left( { R }^{ 2 }-{ r }^{ 2 } \right)$

So $R = 105$ $m$ & $r = 100$ $m$ .

Hence, the area of the path $= \pi \left( { 105 }^{ 2 }-{ 100}^{ 2 } \right) = \cfrac { 22 }{ 7 } \left( 11025-10000 \right) = \cfrac {22550}{7} = 3221 \cfrac{3}{7} {m}^{2}$

The circumference of a circular table is 88 m. Find the value of its area.

As we know that circumference=2

$\pi$

$\times$ r

$\Rightarrow 88$ = 2

$\times$

$\pi$

$\times$ r

$\Rightarrow r=14$

Now, area=

$\pi$

$\times$ r

$^2$

$\Rightarrow$ =

$\pi$

$\times$ 14

$^2$

$\Rightarrow$ = 616 m

$^2$

The radii of the inner and outer circumferences of a circular running track are $63$ $m$ and $70$ $m$ respectively. Find the area of track

Area of the region between two concentric circles with radius of outer circle R, and inner circle r $= \pi \left( { R }^{ 2 }-{ r }^{ 2 } \right)$

Hence, the area of the circular track $= \pi \left( { 70 }^{ 2 }-{ 63 }^{ 2 } \right) = \frac { 22 }{ 7 } \left( 4900-3969 \right) = 2926 {m}^{2}$

Radii of two circles are 3 m and 4 m respectively. Find the radius of a circle having area equal to the sum of the areas of these two circles

We know the area of a circle

$=\pi r^2$

Hence, area of a circle

$(C_1)$ of radius

$3\ m = \dfrac {22}{7} \times { (3) }^{ 2 }$

Area of a circle

$(C_2)$ of radius

$4\ m = \dfrac {22}{7} \times { (4 )}^{ 2 }$

Given, area of big circle

$=$ sum of the areas of circles

$C_1$ and

$C_2$

$= \dfrac {22}{7} \times { 3 }^{ 2 } + \dfrac {22}{7} \times { 4 }^{ 2 }$

$= \dfrac {22}{7} \times ( 9 + 16)$

$= \dfrac {22}{7} \times ( 25)$

Let the radius of the big circle be

$R.$

Area of big circle

$= \pi { R }^{ 2 } = \dfrac {22}{7} \times ( 25)$

$\Rightarrow { R }^{ 2 } = 25$

$\Rightarrow R = 5\ m$

Can a paralleogram be contructed with

$\displaystyle AB= 4.5$ cm, $\displaystyle\angle B= 120^{\circ}$ and the distance between
AB and $\displaystyle DC= 3.1$ cm.

Given:

$AB = 4.5cm, \ \angle B = 120^o$ and distance between

$AB$ and

$DC = 3.1cm$

Now, to construct a parallelogram with the above given values we will follow the below mentioned steps:

1) Draw

$AB = 4.5cm$

2) from point A, draw a line AM at angle

$90^o$

3) from point B, draw a line BN at angle

$90^o$

4) from point A and point B take an arc of length 3.1cm. Join point of intersecting of arc with line AM and BN, which gives line XY.

5) Where XY line is parallel to AB.

6) from point B draw a line at an angle of

$120^o$ with the line AB and intersect with the line XY at point. Which give point C.

7) Since ABCD is parallelogram, sum of two adjacent angle is

$180^o$ .

8) So,

$\angle A = 180^o - 120^o$

$\Rightarrow \angle A = 60^o$

9) from point A draw a line at an angle of

$60^o$ with the line AB and intersecting with line XY at a point. Which give point D.

10) Now, ABCD is required parallelogram.

Hence, ABCD is a parallelogram.

2104249_180536_ans_19b479ba60f24bd6b635d3b28003db40.png

Find the area of a flat circular ring formed by two concentric circles(circles with same centre) whose radii are 9 cm and 5 cm.

Area of the region between two
concentric circles with radius of outer circle R, and inner circle r $= \pi \left( { R }^{ 2 }-{ r }^{ 2 } \right)$

Hence, the area of a flat circular ring formed by two concentric circles whose radii are 9 cm and 5 cm

$= \pi \left( { 9 }^{ 2 }-{ 5 }^{ 2 } \right) = \frac { 22 }{ 7 } \left( 81-25 \right) = 176 {cm}^{2}$

Find the area of the shaded portion in the above figure:

Let us assume $R = 4.5$ $m$ & $r = 2.5$ $m$ .

Area of the region between two concentric circles

with radius of outer circle R, and inner circle r $= \pi \left( { R }^{ 2 }-{ r }^{ 2 } \right)$

Hence, the area of the region formed by two
concentric circles whose radii are 4.5 m and 2.5 m $= \pi \left( { 4.5 }^{ 2 }-{ 2.5 }^{ 2 } \right) = \frac { 22 }{ 7 } \left(20.25-6.25\right) = 44 {m}^{2}$

Find the area of the flower-beds.

As three sides of lawn are having flower-beds of 2m,

Length of the lawn

$=30-2-2$ m

$=26$ m
Breath of the lawn

$=12-2$ m

$=10$ m
Area of the lawn

$=26\times 10$ sq. m

$=260$ sq. m
Area of the lawn + area of the flower bed

$=12\times 30$ sq. m

$=360$ sq. m

$\therefore$ Area of the flower bed

$=(360-260)$ sq. m

$=100$ sq. m

A roller of diameter $0.9\ m$ . and length $1.8\ m$ is used to press the ground. Find the area of the ground pressed by it, in $500$ revolutions.(Take $\pi =3.14$ )

Given,

Diameter of roller

$d= 0.9\ m$

Length

$\ell = 1.8\ m$

$r =\dfrac{d}{2}=0.45$
To find area it rolled in

$1$ revolution we have to find curved surface area.
Curved surface area:

$=2\pi r\ell\\ =2\times 3.14\times 0.45\times 1.8\\ =5.0868\ { m }^{ 2 }$

Area covered in

$500$ revolutions

$=500\times 5.0868\\ =2543.4\ { m }^{ 2 }$

Find the area of the triangle(in $cm^2$ ) with base $6\ cm$ and height $4\ cm$

Here, In

$\triangle ABC,$

$BC$ and

$AD$ are base and height of triangle.

$\therefore$

$BC=6\,cm$ and

$AD=4\,cm$

$\Rightarrow$ Area of

$\triangle ABC=\dfrac{1}{2}\times BC\times AD$

$=\dfrac{1}{2}\times 6\times 4$

$=3\times 4$

$=12\,cm^2$

1356435_187611_ans_38d8c3a8a0194537b959bb2491044009.jpeg

Find the area of a circle whose radius is equal to the side of an equilateral triangle of area $\displaystyle 9\sqrt{3}cm^{2}$

Area of equilateral triangle

$\displaystyle =\sqrt{3/4}$

$\displaystyle a^{2}$

$\displaystyle =9\sqrt{3}$

$\displaystyle \Rightarrow a=6cm$

$\displaystyle \therefore$ Area of circle

$\displaystyle =\pi \times 6\times 6=36\pi cm^{2}$

A flooring tile has the shape of a parallelogram whose base is $24$ cm and the corresponding height is $10$ cm. How many such tiles are required to cover a floor of area $1080$ sq. metre.

Area of one tile

$= 24\times10 =240$ cm

$^2$

It is known that

$1$ sq. meter

$=10000$

$cm^2$

Required no. of tiles

$= \dfrac{\text{Total Area}}{\text{Area of one tile}} = \dfrac{1080\times 10000}{240} = 45000$ tiles

What is the area between the $2$ concentric circles?

Area of a circular ring

$= \pi(R^2-r^2)$

$= 3.14(4^2-2^2)$

$= 3.14 \times 12$

$= 37.68$

$cm^2$

In given figure depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is $60\ m$ and they are each $106\ m$ long. If the track is $10\ m$ wide, find:
(i) The distance around the track along its inner edge
(ii) The area of the track

The distance around the track along its inner edge

$= EF + HG$

$+$ circumference of semicircle with diameter

$EH+$ circumference of semicircle with diameter

$FG$

circumference of semicircle with diameter

$EH$

$+$ Circumference of semicircle with diameter

$FG$

$=2\times\pi\times30\,=60\pi \,m$

$\therefore$ The distance around the track along its inner edge

$=60\pi\,+\,106\,+\,106\,\,$

$=\,\dfrac{2804}{7}\,\,m$

$\therefore$ Area of track

$=$ Area of rectangle

$ABFE$

$+$ Area of rectangle

$HGCD$

$+$ (Area of semicircle with diameter

$AD$

$-$ Area of semicircle with diameter

$EH$ )

$+$ (Area of semicircle with diameter

$BC$

$-$ Area of semicircle with diameter

$FG$ )

$=(106\times10)$ +

$(106\times10)$ +

$(\dfrac{\pi}{2}\times40\times40-\dfrac{\pi}{2}\times30\times30)$ +

$(\dfrac{\pi}{2}\times40\times40-\dfrac{\pi}{2}\times30\times30)$

$=2120\,+\,2200\,=\,4320\,m^2$

In Fig., a square $OABC$ is inscribed in a quadrant $OPBQ$ . If $OA = 20 cm$ , find the area of the shaded region. (Use $\pi = 3.14$ )

From given figure

$OB\,=$ Diagonal of

$\square ABCO$

$=$ Radius of quadrant QOP

$\therefore$

$OB\,=OA \times\sqrt2$

$=20\sqrt2\,cm$

$\therefore$ Area of

$\square OABC\,=20\times 20\,\,=\,400\,cm^2$

Area of quadrant QOP

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

$=\dfrac{1}{4}\times3.14\times(20\sqrt2)^2$

$=\,628\,cm^2$

$\therefore$ Area of shaded region

$=628-400\,=\,228\,cm^2$

The circumference of a circular race track is $44\ m^{2}$ . Find its area

We know that

circumference of circular track

$=2\pi r$

$\begin{array}{l} \\ \Rightarrow 2\pi r=44 \\ \Rightarrow \dfrac { { 22 } }{ 7 } \times r=22 \\ \therefore r=7\ m \\ Now, \\ Area\, =\pi { r^{ 2 } } \\ =\dfrac { { 22 } }{ 7 } \times { 7^{ 2 } } \\ =22\times 7 \\ =154\, { m^{ 2 } } \end{array}$

Hence, this is the answer.

The radii of two circles are $8\ cm$ and $6\ cm$ respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles

Let

$r_1$ and

$A_1$ be the radius and area, respectively of first circle and

$r_2$ and

$A_2$ be the radius and area, respectively of second circle.

Let

$r_3$ and

$A_3$ be the radius and area, respectively of required circle.

For first circle,

$r_1 = 8$ cm ...Given

$A_1 = \pi {r_1}^2$

$\therefore A_1 = \pi \times 8^2\ cm^2 = 64\pi \ cm^2$ ...(1)

For second circle,

$r_2 = 6$ cm ...Given

$A_2 = \pi {r_2}^2$

$\therefore A_2 = \pi \times 6^2\ cm^2 = 36 \pi \ cm^2$ ...(2)

Now,for the required circle,

$A_3 = A_1+A_2$ ....Given

$\Rightarrow A_3 = 64 \pi + 36 \pi$ ....From (1) and (2)

$\Rightarrow A_3 = 100 \pi \ cm^2$

$\therefore \pi {r_3}^2 = 100 \pi$

$\Rightarrow r_3^{2} = 100$

$\Rightarrow r_3 = 10$

The radius of the required circle is

$10$ cm.

The radii of the two circles are $19\ \text{cm}$ and $9\ \text{cm}$ respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

Let

$r_1$ and

$C_1$ be the radius and circumference, respectively of 1st circle and

$r_2$ and

$C_2$ be the radius and circumference, respectively of 2nd circle.

Let

$r_3$ and

$C_3$ be the radius and circumference, respectively of the required circle.

$r_1 = 19 \text{ cm}$ ...Given

$\implies C_1 = 2\pi r_1$

$\therefore C_1 = 2 \pi \times 19 = 38\pi \text{ cm}$ ...(1)

$r_2 = 9\text{ cm}$ ...Given

$\implies C_2 = 2\pi r_2$

$\therefore C_2 = 2 \pi \times 9 = 18 \pi \text{ cm}$ ...(2)

$C_3 = C_1+C_2$ ....Given

$\therefore C_3 = 38 \pi + 18 \pi$ ....From (1) and (2)

$C_3 = 56 \pi \text{ cm}$

$\implies 2 \pi r_3 = 56 \pi$

$\therefore r_3 = 28\text{ cm}$

The radius of the required circle is

$28 \text{ cm}$

The inner and outer diameter of the circular ring is $6\ m$ and $12\ m$ . Find its area.

Area of a circular ring

$= \dfrac{\pi}{4}(D^2-d^2)$

$=\dfrac{3.14}{4}(12^2-6^2)$

$= 84.78$

$cm^2$

The circumference of a circle is equal to $72$ $\pi$ . Find the radius of this circle.

Given, Circumference of a circle

$=72 \pi$

We know that Circumference

$= \pi$

$d$

$d\pi=72\pi$

$d=72$

Since radius

$=$

$\dfrac {d}{2}$

$\therefore$ radius

$=\dfrac {72}{2}=36$

Observe the adjacent figure. It shows two circles with the same centre. The radius of the larger circle is $10cm$ and the radius of the smaller circle is $4cm$
Find the shaded area between the two circles. (take $\pi = 3.14$ )

Area of shaded region =

$\pi \left( { R }^{ 2 }-{ r }^{ 2 } \right)$

$3.14\left( { 10 }^{ 2 }-{ 4 }^{ 2 } \right)$

$3.14\times 84=263.76{ cm }^{ 2 }$

The radius of a circle is 21 cm. Find the perimeter and area of a quadrant of the circle.

Radius of a circle, r = 21 cm
Perimeter of a quadrant,

$P = \left( \dfrac{\pi}{2} + 2 \right ) r$ units

$= \left( \dfrac{22}{7 \times 2} + 2 \right ) \times 21 = \left( \dfrac{22}{14} + 2 \right ) \times 21$

$P= \left ( \dfrac{22 + 28}{14} \right ) \times 21 = \dfrac{50}{14} \times 21$

$= 75 cm$ .
Area of a quadrant,

$A = \dfrac{\pi r^2}{4}$ sq. units

$A = \dfrac{22}{7} \times \dfrac{21 \times 21}{4}$

$= 346.5 cm^2$ .

The area of a parallelogram is $153.6 {cm}^{2}$ . The base measures $19.2 cm$ . What is the measurement for the height of the parallelogram?

THEOREMS AND PROBLEMS ON PARALLELOGRAMS – EXERCISE 4.3.3- Class IX

Area of the parallelogram is

$A = bh$ where

$b$ is the base and

$h$ is the height of the parallelogram.

It is given that area of the parallelogram is

$A = 153.6$ cm

$^2$ and base

$b = 19.2cm$ , therefore,

$153.6=19.2 × h$

$\Rightarrow h=\dfrac { 153.6 }{ 19.2 } =8$

Hence, the height of the parallelogram is

$8$ cm.

Find the perimeter and area of a semicircle whose radius is 14 cm.

Radius of a semicircle,

$r = 14 cm$
Perimeter of a semicircle,

$P = (\pi + 2) r$ units

$\therefore P = \left( \dfrac{22}{7} + 2 \right ) \times 14$

$= \left( \dfrac{22+ 14}{7} \right ) \times 14 = \dfrac{36}{7} \times 14 = 72$
Perimeter of the semicircle = 72 cm.
Area of a semicircle,

$A = \dfrac{\pi r^2}{2}$ sq. units

$\therefore A = \dfrac{22}{7} \times \dfrac{14 \times 14}{2} = 308 cm^2$ .

In the figure, $PQSR$ represents a flower bed. If $OP = 21\ m$ and $OR = 14\ m$ , find the area of the shaded portion.

Given:

$OP = 21\ m$ and

$OR = 14\ m$
Area of the flower bed

$=$ Area of the quadrant OQP

$-$ Area of the quadrant

$OSR$

$= \dfrac{1}{4} \pi \times OP^2 - \dfrac{1}{4} \pi \times OR^2$

$= \dfrac{1}{4} \times \pi \times 21^2 - \dfrac{1}{4} \times \pi \times 14^2$

$=\dfrac{1}{4} \times \pi \times (21^2 - 14^2)$

$= \dfrac{1}{4} \times \dfrac{22}{7} \times (21 + 14) \times (21 - 14)$

$\therefore$ Area of the flower bed

$= \dfrac{1}{4} \times \dfrac{22}{7} \times 35 \times 7 = 192.5\ m^2$ .

The radii of two concentric circles are $19$ cm and $16$ cm, find the area of the ring enclosed between them.

Let,

$r_1=19 cm$ and

$r_2=16cm$

Therefore the area of the ring enclosed :

$=\pi (r_1^2-r_2^2)$

$=\pi (19^2-16^2)$

$=329.86$ cm

$^2$

Find the circumference of a circle whose radius is $7$ cm.

Fact: circumference of a circle whose radius is

$r$ is given by

$2\pi r$

Here

$r=7$ cm

Hence circumference of the circle

$=2\pi (7)=2\cdot \dfrac{22}{7}\cdot 7=2\cdot 22=44$ cm

Find the areaof a parallelogram $PQRS$ , if $PR = 24\ cm$ and $QU = ST = 8\ cm$ .

$\triangle{PQR}$

$PR=24$ cm

$QU=8$ cm

Area of

$\triangle{PQR}=\dfrac{1}{2}\times 24\times 8=96$ sq.cm

Now, in

$\triangle{PRS},\,ST=8\ cm,PR=24\ cm$

Area of

$\triangle{PRS}=\dfrac{1}{2}\times 24\times 8=96$ sq.cm

Area of parallelogram

$=$ area of

$\triangle{PQR} +$ area of

$\triangle{PRS}$

$=96+96=192$ sq.cm

In a school, exactly at the middle of the stage of the assembly hall, after drawing a circle of $2.8\ m$ radius, a beautiful picture is to draw in it. Cost for drawing is $Rs. 300$ per $1\ sq\ m$ . What is the total cost(in Rs) of drawing a picture in this circle?

Radius of circle

$(r)=2.8\space\mathrm{m}$

Area of circle

$=\pi r^2=\dfrac{22}{7}\times (2.8)^2=24.64\space\mathrm{m^2}$

Now, Cost of drawing

$=\mathrm{Rs.}\space300 \space\text{per}\space\mathrm{sq.m}$

$\therefore$ Cost of drawing

$24.64\space\mathrm{m^2}=\mathrm{Rs.} 300\times24.64=\mathrm{Rs.}\space7392$

The area of a rectangular park is $3100\text{ m}^2$ and its breadth is $50 \text{ m}$ . The value of 4 times its perimeter is _________.

Let the length be

$x \text{ m}$ .
Breadth

$= 50 \text{ m}$
Now, area =

$3100 \text{ m}^2$

$\displaystyle \Rightarrow x \times 50 = 3100$

$\Rightarrow x = \cfrac{3100}{50}$

$\Rightarrow x = 62$

$\therefore$ Length

$= 62 \text{ m}$

Perimeter of the rectangle

$=2(l+b)$

$= 2(62 + 50)$

$= 224 \text{ m}$

4 times the perimeter

$= (4 \times 224) \text{ m} = 896 \text{ m}$

A wire is bent to form a square enclosing an area of $484\ m^{2}$ . Using the same wire, a circle is formed. Find the area of the circle.

Area of square

$=484 m^2$

Side

$=\sqrt{484}\ m=22m$

Perimeter

$=22\times4\ m=88m$

So,

Circumference of circle

$=88\ m$

$\Rightarrow 2\pi r=88$

$\Rightarrow r=\cfrac{88\times7}{2\times22}=14m$

Area of circle

$=\pi r^2 =\cfrac{22}{7}\times14\times14\ m^2=616m^2$

Find the area of circle whose circumference is $44$ cm.

$\Rightarrow$ Circumference of circle

$=44\,cm$ [ Given ]

$\Rightarrow$ Circumference of circle

$=2\pi r$

$\Rightarrow$

$44=2\times \dfrac{22}{7}\times r$

$\Rightarrow$

$r=\dfrac{44\times 7}{44}=7\,cm$

$\Rightarrow$ Area of circle

$=\pi r^2$

$\Rightarrow$ Area of circle

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

$\Rightarrow$ Area of circle

$=22\times 7$

$\therefore$ Area of circle

$=154\,cm^2$

A park is in from of a rectangle $120$ m $\times$ $100$ m. At the centre of the park there is a circular lawn. The area of park excluding lawn is $8700$ $m^2$ . Find the radius of the circular lawn. (Use $\pi$ $= \dfrac{27}{7}$ ).

Let the radius of circular lawn be 'r'm

Area of shaded part

$=8700m^2$

Area of park-Area of circular lawn

$=8700m^2$

$\Rightarrow \left( 120\times 100 \right) -\pi { r }^{ 2 }=8700m^2$

$\Rightarrow \pi { r }^{ 2 }=12000-8700=3300\Rightarrow { r }^{ 2 }=\cfrac { 3300\times 7 }{ 22 } =150\times 7$

$\Rightarrow r=\sqrt { 150\times 7 } =32.04m$

The circumference of a circle exceeds the diameter by $16.8\ cm$ . Find the circumference of the circle.

Let the radius of the circle be

$r\,cm$ .

According to the given condition,

Circumference of a circle

$=$ Diameter of a circle +

$16.8$

$\therefore$

$2\pi r=2r+16.8$

$\Rightarrow$

$2\times \dfrac{22}{7}\times r-2r=16.8$

$\Rightarrow$

$\dfrac{44r-14r}{7}=16.8$

$\Rightarrow$

$\dfrac{30r}{7}=16.8$

$\therefore$

$r=\dfrac{16.8\times 7}{30}=3.92\,cm$

$\Rightarrow$ Circumference of a circle

$=2\pi r$

$\therefore$ Circumference of a circle

$=2\times \dfrac{22}{7}\times 3.92=24.64\,cm$

Find the circumference of circle whose radius is $14\ cm$ .

circumference of the circle with radius

$r$ is =

$2\pi r$

$\Rightarrow 2\pi r=14\Rightarrow r=\dfrac{14}{2\pi }=\dfrac{14\times 7}{2\times 22}=\dfrac{49}{22}$

$\Rightarrow r =2.227\ cm$

The fore wheel of a carriage makes $10$ revolutions more than the rear wheel in covering $120\ \text{m}$ . If the diameter of the fore wheel be increased by one-fourth, keeping the diameter of the rear wheel same, the fore wheel makes four revolutions more than the rear wheel in covering the same distance. Find the circumferences of the fore wheel and the rear wheel of the carriage.

Fore whel makes 110 revolutions more than are wheel distance converted

$=120\ \text{m}$

Let

$d_1$ be diameter of the fore wheel and

$d_2$ be diameter of the rear wheel then

$\dfrac {120}{\pi d_1} =\dfrac {120}{\pi d_2}+10......(1)$

later diameter of the fore wheel is increased by

$\dfrac{1}{4}$

$\implies \dfrac {120}{\pi \left (\dfrac {5}{4}d_1 \right)} =\dfrac {120}{\pi d_2}+6$

$\Rightarrow \ \dfrac {96}{\pi d_1}=\dfrac {120}{\pi d_2}+6........(2)$

substitude

$(1)$ in

$(2)$

$\Rightarrow \dfrac {96}{\pi d_1}=\dfrac {120}{\pi d_1}-10+6$

$\Rightarrow \ \dfrac {120}{\pi d_1}-\dfrac {96}{\pi d_1}=4$

$\Rightarrow \ \pi d_1=6$

$\therefore$ Circumference of the fore wheel is

$6\text{ m}$

Substuting

$\pi d_1$ in

$(1)$

$\implies \dfrac{120}{6}=\dfrac{120}{\pi d_2}+10$

$\implies \pi d_2=\dfrac{120}{10}=12$

$\therefore$ Circumference of the rear wheel is

$4\text{ m}$

A circle is circumscribing rectangle with side $16$ cm & $12$ cm. Find area of circle

In the above question we are given length of sides of the rectangle.

So according to the figure

$AC$ is the diagonal of the rectangle

$ABCD$ as well as the diamter of the circle.GIVEN

$AB=16CM$ AND

$BC=12CM$

By pythagoras theorem

$AC^{2}=12^{2}+16^{2}$

$\therefore$

$AC=\sqrt{144+256}$

$=20$

$AC=2r=20$

$\Rightarrow$

$r=10cm$

$area=\pi\times r^2$

$Area=100\times3.14$

$=314cm^{2}$

Find the circumference and area of circle having radius $4.2 cm.$

Given, radius

$r=4.2$ cmradius(r)=4.2cm

Circumference of the circle

$=2\pi r$

$=2\times \dfrac{22}{7}\times 4.2$

$=26.4 \, cm$

Area of the circle

$=\pi r^2$

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

$=55.44\, cm^2$

A copper wire when bent in the form of a square encloses an area of $121\ cm^{2}$ . If the same wire is bent into the form of a circle, find the area of the circle (Use $\pi = 22/7)$ .

Given

$a^{2}=121$

$\Rightarrow a=11$

$L=4a=44$

Given

$2\pi r=44$

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

$\therefore r=7$

Area

$=A=\pi \times 49=49\pi\ \ cm^{2}$

$=49\times \dfrac{22}{7}$

$A=154\ cm^{2}$

A wheel makes 1000 revolutions in covering a distance of 88 km.Find the radius of the wheel.

Distance covered in one revolution

$=\dfrac { 88\times 1000 }{ 1000 } =88m$

$2\pi R=88\Rightarrow 2\times \dfrac { 22 }{ 7 } \times R=88$

$\Rightarrow R=88\times \dfrac { 7 }{ 22 } \times \dfrac { 1 }{ 2 }$

$R=14m$ .

Prove that the area of a circular path of uniform width $h$ surrounding a circular region of radius $r$ is $\pi$ $h (2r + h)$ .

Let

$h$ be the width of the path. Let the radius of inner circle be

$r$ .

$\therefore$ Radius of outer circle

$=(r+h)$

$\Rightarrow$ Area of the path

$=$ Area of outer circle

$-$ Area of inner circle.

$\Rightarrow$ Area of the path

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

$\Rightarrow$ Area of the path

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

$\therefore$ Area of the path

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

A wheel has a diameter of $84$ cm. Find how many complete revolutions must it take to cover $792$ meters

Let r be the radius of the wheel , Then,
Diameter

$= 84 cm$

$\Rightarrow$

$2r = 84 cm$

$\Rightarrow$

$r = 42 cm$

$\therefore$ Circumference of the wheel

$=$

$2 \, \pi \, r \, \, = \, 2 \, \times \, \dfrac{22}{7} \, \times \, 42 \, cm$

$= 264 cm = 2.64m$

So, the wheel covers

$2.64$ meters in one complete revolution.

$\therefore$ Total number of revolutions in covering

$792$ meters

$=$

$\dfrac{792}{2.64}$

$= 300.$

Hence, the wheel takes

$300$ revolutions in covering

$792$ meters.

A steel wire when bent in the form of a square encloses an area of $121sq.cm$ , if the same wire is bent into the form of a circle, find the area of the circle.

Given that

$\Rightarrow$ Area of square

$=121$ sq.cm

$\Rightarrow$ Therefore, length of wire

$=\sqrt{121}$ cm

$\Rightarrow$ length of wire

$=11$ cm

This wire is bent in form of circle.

$\Rightarrow 2\pi r=11$

$\Rightarrow r=\dfrac{11}{2\pi}$

$\Rightarrow$ Area of circle

$=\pi r^2$

$\Rightarrow$ Area of circle

$=\pi (\dfrac{11}{2\pi})^2$

On solving we get

$\Rightarrow$ Area of circle

$=\dfrac{121}{4}cm^2$

A car travels 1 kilometre distance in which each wheel makes 450 complete revolutions. Find the radius of the its wheels.

35.35 cm
Let the radius of each wheel be r meters. Then,

Circumference of each wheel =

$2 \pi r \, = \, 2 \, \times \, \dfrac{22}{7} \, \times \, r \,\, meters$

$\Rightarrow$ Distance covered by the wheels in one revolution =

$2 \, \times \, \dfrac{22}{7} \, \times \, r \,\, meters$

$\Rightarrow$ Distance covered by the wheels in 450 revolutions =

$2 \, \times \, \dfrac{22}{7} \, \times \, r \, \times \, 450 \,\, meters$

It is given that the car travels 1 kilometer i.e. 1000 meter distance when its each wheel makes 450 revolutions.

$\Rightarrow 2 \, \times \, \dfrac{22}{7} \, \times \, r \, \times \, 450 \, = \, 1000 \,$

$\Rightarrow \, r \, = \, \dfrac{7 \, \times \, 1000}{2 \, \times \, 22 \, \times \, 450} \,$

$\Rightarrow r= \, 0.3535 \,\, meters \,$

$\therefore r= \, 35.35 \, cm$

Find the area of a quadrant of a circle whose circumference is 616 cm.

To find the quadrant of a circle we will have to first determine the circumference of the circle which is

$2\pi r$

Here, the circumference being

$44\text{ cm}$

$2\pi r=616$

$2(\dfrac{22}{7})r=616$

$r=\dfrac{616\times7}{22\times2}=\dfrac{4312}{44}=98 \text{ cm}$

Therefore, radius of circle,

$r=98 \ cm$

Now, Area of a circle

$= \pi r^2 = \dfrac{22}{7}(98^2) = 30184 \text{ cm}^2$

The area of the quadrant of a circle whose circumference is

$616 \text{ cm}$ is

$=\dfrac{Area}{4}=\dfrac{30184}{4}=7546\text{ cm}^2$

The diameter of a circular field is $\dfrac{56}{\pi}$ m. Find the circumference and hence find the cost of fencing it at the rate of $80$ per m.

Circumference of circular field

$=2\pi R=2\times {\pi} \times \dfrac { 56 }{ 2\pi} =56m$

Cost of fencing

$=Rs.\left( 56\times 80 \right) =Rs.4480$

The diameter of a wheel of a bus is $90$ cm which makes $315$ revolutions per minute. Determine its speed in kilometer per hour. (Use $\pi \, = \, \dfrac{22}{7}$ )

$\Rightarrow$ Diameter of wheel

$=90\,cm$

$\Rightarrow$ Radius of wheel

$(r)=\dfrac{90}{2}cm=45\,cm$

$\Rightarrow$ Distance covered in

$1$ revolution

$=2\pi r$

$\Rightarrow$ Distance covered in

$1$ revolution

$=2\times \dfrac{22}{7}\times 45=\dfrac{1980}{7}cm$

$\Rightarrow$ Distance covered in

$315$ revolutions

$=\dfrac{1980}{7}\times 315=89100\,cm=891\,m$

$\Rightarrow$ Hence, the speed is

$891\,m/min=\dfrac{891}{1000}\times 60km/hr=53.46\,km/hr$

In a circle of radius $6$ cm, a chord of length $10$ cm makes an angle of $110$ $^{\circ}$ at the centre of the circle. Find the area of the circle.

$\Rightarrow$ Radius of a circle

$=6\,cm$ [ Given ]

$\Rightarrow$ Area of a circle

$=\pi r^2$

$\Rightarrow$ Area of a circle

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

$\Rightarrow$ Area of a circle

$=\dfrac{22\times 36}{7}=113.14\,cm^2$

The outer circumference of a circular race-track is $528$ m. The track is everywhere $14$ m wide. Calculate the cost of levelling the track at the rate of $50$ paise per square metre (Use $\pi$ $= \dfrac{22}{7}$ ).

Outer circumference

$=528m$

$=2\pi r=528m\Rightarrow \Rightarrow R=\cfrac { 528\times 7 }{ 44 } =84m$

Width of track

$=14m$

Inner radius,

$r=84-14=70m$

$\therefore$ Area of track

$\pi \left( { R }^{ 2 }-{ r }^{ 2 } \right) =\cfrac { 22 }{ 7 } \left( { 84 }^{ 2 }-{ 70 }^{ 2 } \right)$

$=\cfrac { 22\times 2156 }{ 7 } =6776m^2$

Cost of leveling

$=$ Rs.

$\left( 6776\times 0.50 \right)$

$=$ Rs.

$3388$

In a circle of radius $6$ cm, a chord of length $10$ cm makes an angle of $110$ $^{\circ}$ at the center of the circle. Find the circumference of the circle.

$\Rightarrow$

$r=6\,cm$

$\Rightarrow$ Circumference of the circle

$=2\pi r$

$\therefore$ Circumference of the circle

$=2\times 3.14\times 6$

$\therefore$ Circumference of the circle

$=37.68\,cm$

From the circular sheet of radius $4\ cm$ , a circle of radius $3\ cm$ is removed. Find the area of the remaining sheet. $(Take\ \pi=3.14)$

The area of the circular sheet with $4\;{\rm{cm}}$ radius is,

$A = \pi {r^2}$

$= 16\pi$

The area of the circular sheet with $3\;{\rm{cm}}$ radius is,

$A = \pi {r^2}$

$= 9\pi$

Since the circle with radius $3\;{\rm{cm}}$ is removed, then the remaining area is,

$A = 16\pi - 9\pi$

$= 7\pi$

$= 7 \times 3.14$

$= 21.98\;{\rm{c}}{{\rm{m}}^2}$

The ratio of circumference of $2$ circle is $3:4$ then what is the ratio of the corresponding areas.

Suppose there are two circles

$A$ and

$B$ having radius

$r$ and

$R$ respectively then

$r:R=3:4$

$r=3x,R=4x$

Area of

$A=\pi r^2=\pi (3x)^2=9\pi x^2$

Area of

$B=\pi r^2=\pi (4x)^2=16\pi x^2$

Ratio of areas of circle

$A\ and\ B=\dfrac{9\pi x^2}{16\pi x^2}=\dfrac{9}{16}$

Hence, Ratio of areas of two circles is

$9:16$ .

To find area, any side of a parallelogram can be chosen as ____ of the parallelogram.

To find area, any side of a parallelogram can be chosen as the

$base$ of the parallelogram, we just have to take the height carefully w.r.t to the base chosen

Find the circumference of the circle with the following radius $\left(\text{Take }\pi=\dfrac {22}{7}\right)$

(a) $14\ cm$ (b) $28\ cm$ (c) $21\ cm$

(a) The radius is given as $14\ {cm}$ .

The circumference of the circle is,

$C = 2\pi r$

$= 2 \times \dfrac{{22}}{7} \times 14$

$= 88\ {cm}$

(b)

The radius is given as $28\ {cm}$

The circumference of the circle is,

$C = 2\pi r$

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

$= 176\ {cm}$

(c)

The radius is given as $21\ {cm}$

The circumference of the circle is,

$C = 2\pi r$

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

$= 132\ {cm}$

If the perimeter of a parallelogram is 140 m, the distance between a pair of opposite sides is 7 m and its area is 210 square m, find the length of two adjacent sides of the parallelogram.

Given that,

The perimeter of the parallelogram is

$140\ m$ .

Its area is

$210\ m^2$

Distance between a pair of opposite sides is

$7\ m$ .

To find out,

Length of the two adjacent sides of the parallelogram.

Let

$a,\ b$ be the two adjacent sides of the parallelogram, as shown in the figure above.

We know that, area of a parallelogram

$=side\times its\ corresponding\ height$

So, if

$a$ is the side,

$h$ will be its corresponding height.

Here,

$h=7\ m$

Hence,

$a\times h=210$

$\Rightarrow a=\dfrac{210}{7}$

$=30\ m$

We also know that, perimeter of a parallelogram

$=2(a+b)$

Here, perimeter

$=140\ m,\ a=30\ m$

Hence,

$2(30+b)=140$

$\Rightarrow 60+2b=140$

$\Rightarrow 2b=80$

$\Rightarrow b=40\ m$

Hence, the two adjacent sides of the parallelogram are

$30\ m$ and

$40\ m$ .

2113144_1038663_ans_cd6f0fc1fe57442dbe614d6adcdd5feb.png

A field is in the form of a circle. A fence is to be erected around the field. The cost of fencing would be Rs.5280 at the rate of Rs.24 per metre. Then, the field is to be throughly ploughed at the cost of $Rs. 0.50 per m^2$ . What is the amount required to plough the field?

Given cost of fence per

$=$ Rs.

$24$

Total cost of fence around the field

$=$ Rs.

$5280$

i.e.,

$2\pi r\times 24=5280$

$2\times \dfrac{22}{7}\times r=\dfrac{5280}{24}$

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

$r=\dfrac{7}{2}=3.5$

Area of the field

$=\pi r^2$

$=\dfrac{22}{7}\times \dfrac{7}{2}\times \dfrac{7}{2}=\dfrac{77}{2}$

cost of ploughing the field

$=0.50$ per

$m^2$

$\therefore$ Cost of ploughing the whole field

$=\dfrac{77}{2}\times \dfrac{1}{2}$

$=\dfrac{77}{4}$ .

Find the diameter of the circle whose circumference is $42\ cm$ .

Given: The circumference of is $42$ cm.

Let the radius and diameter of the circle $r$ and $d$ , respectively.

We know that,

Circumference $= 2\pi r$

$\therefore 2\pi r=42$

$\Rightarrow 2\left ( \dfrac{22}{7} \right ) r=42$

$\Rightarrow r = \dfrac{147}{22} \approx 6.68$ cm

Since, the diameter of the circle is twice of it radius.

$\therefore d=2r$

$\Rightarrow d =2\left ( \dfrac{147}{22} \right ) = \dfrac{147}{11} \approx 13.36$

Thus, the diameter of the given circle is $\dfrac{147}{11}$ cm or $13.36$ (approx) cm.

The value of the divisions on the outer rim of a graduated circle is $5'$ and the distance between successive graduations is $0.1$ . Find the radius of the circle.

Here, we know that

$1$ degree =

$60$ minutes

So,

$1$ minutes =

$\dfrac{1}{60}$ degree

$\Rightarrow5$ minutes =

$\dfrac{5}{60}$ degree

Also, we know that

$1$ degree =

$\dfrac{\pi }{180}$ radian

So,

$\dfrac{5}{60}$ degree =

$\dfrac{\pi }{180}\times \dfrac{5}{60}$ radian

To find the radius of the given circle, we have to use the formula

$\boldsymbol{\mathrm{Angle = \dfrac{Arc}{Radius}}}$

From the above formula we can write,

$\boldsymbol{\mathrm{Radius = \dfrac{Arc}{Angle}}}$

Value of

$1$ division on rim of circle is

$5'$ i.e

$5$ minutes,

and distance between two successive divisions is

$0.1$

So, we will substitute

$0.1$ for Arc and

$\dfrac{\pi }{180}\times \dfrac{5}{60}$ for Angle in the above formula and simplify for Radius.

$\boldsymbol{\mathrm{Radius = \dfrac{0.1}{\dfrac{\pi }{180}\times \dfrac{5}{60}}}}$

$\Rightarrow \boldsymbol{\mathrm{Radius=68.75}}$

Hence, the required radius of the circle is

$\boldsymbol{\mathrm{68.75}}$ cm.

In the adjoining figure, for the right angled $\triangle ABC$ and $\triangle BCD$ , $\dfrac{AB}{CD}=\dfrac{1}{2}$ . Find $\dfrac{A(\Delta ABC)}{A(\Delta BCD)}$ .

Area of triangle

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

$\times$ height

$\dfrac{A(\Delta ABC)}{A(\Delta BCD)}=\dfrac{\dfrac{1}{2}\times BC\times AB}{\dfrac{1}{2}\times BC\times CD}\\=\dfrac{AB}{CD}\\=\dfrac{1}{2}$ .

Fins the radius whose circumference is $448$

The circumference is only calculated in case of circle. So, the circumference of the circle is $448$ .

Let the radius of the circle is $r$ .

Write the formula to calculate circumference of circle.

Circumference $=2\pi r$

Substitute the values in the above equation.

$2\pi r=448$

$2\left ( \frac{22}{7} \right ) r=448$

Solve for $r$ .

$r=\dfrac{448\times 7}{2\times 22}$

$r=\dfrac{784}{11}$

$r\approx 71.27$

Thus, the radius of the given circle is $\dfrac{784}{11}$ units or $71.27$ (approx) units.

Find the area of the following circles, given that: $(Take \pi=\dfrac {22}{7})$
a)radius $=14\ mm$
b)diameter $=49\ m$
c)radius $=5\ cm$

(a)The radius of the circle is given as $14\;{\rm{mm}}$ .

The area of the circle is,

$A = \pi {r^2}$

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

$= 616\;{\rm{m}}{{\rm{m}}^2}$

(b) The diameter of the circle is given as $49\;{\rm{m}}$ .

The radius of the circle becomes $\dfrac{{49}}{2}\;{\rm{m}}$ .

The area of the circle is,

$A = \pi {r^2}$

$= \dfrac{{22}}{7} \times {\left( {\dfrac{{49}}{2}} \right)^2}$

$= 1886.5\;{{\rm{m}}^2}$

The area of the circle is,

$A = \pi {r^2}$

$= \dfrac{{22}}{7} \times {\left( 5 \right)^2}$

$= 78.5\;{\rm{c}}{{\rm{m}}^2}$

Find the area of a circle when: Radius $=5$ cm.

The area of circle in terms of radius

$r$ is

$A=\pi r^2$

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

$A=\pi r^2=\dfrac{22}{7}\times (5)^2=\dfrac{22\times 25}{7}=\dfrac{550}{7}\ cm^2$

$DL$ and $BM$ are the height on sides $AB$ and $AD$ respectively of parallelogram $ABCD$ figure. If the area of the parallelogram is $1470 \ cm^2$ , $AB=35\ cm$ and $AD=49\ cm$ , find the length of $BM$ and $DL$ .

Area of the parallelogram

$ABCD = AB \times DL = AD \times BM$

$\implies 1470 = 35 \times DL = 49 \times BM$

$\implies DL = \dfrac{1470}{35}$ and

$BM = \dfrac{1470}{49}$

$\therefore DL = 42$ cm and

$BM = 30$ cm

The radius of an air bubble is increasing at the rate of $0.5$ centimeter per second. At what rate is the volume of the bubble increasing when the radius is $1$ centimeter ?

The radius of a bubble is increasing at the Rate

$0.5\ cm/sec$ .

$\dfrac{dr}{dt}=0.5\ cm/sec.$

Rate of inc. of the volume

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

$\dfrac{dv}{dr}=\dfrac{4}{3}\pi3r^2\dfrac{dr}{dt}$

$=\dfrac{4}{3}\times \pi\times 3\times (1)^2\times 0.5$

$\boxed{\dfrac{dv}{dr}=2\pi\ cm^3/sec}$

This is the rate of increase in the volume of the bubble.

A metal wire, when bent in the form of an equilateral triangle of largest area, encloses an area of $484\sqrt 3 \,c{m^2}$ . If the same wire is bent into the form of a circle of largest area, find the area of this circle.

Area of equilateral triangle=

$\dfrac{\sqrt 3}{4}a^2$ where a is the side of the triangle.

Now given that

$\dfrac{\sqrt 3}{4}a^2=484\sqrt3$

$=>a^2=\dfrac{484\sqrt3 \times 4}{\sqrt 3}$

$=>a^2=\sqrt1936$

$a=44$

Now

$a=44$ ,

therefore length of wire=

$a+a+a+=3a$

Now, this wire is bent into a circle, so the circumference of the circle=

$2\pi r$ and length of wire =

$3\times 44=132$ then,

$=>2\pi r=132$

here put the value of

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

$=>r=\dfrac{21}{2}$

Now are of this circle=

$\pi r^2$

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

$= \dfrac{693}{2}=345.5cm^2$

If base and the corresponding altitude of a parallelogram are $9\cfrac{3}{4}cm$ and $12\cfrac{1}{4}cm$ respectively, then find the area of the parallelogram.

Area of parallelogram

$=base \times corresponding \: altitude$

$=9\dfrac{3}{4} \times 12\dfrac{1}{4}$

$=\dfrac{39}{4} \times \dfrac{49}{4}$

$=\dfrac{1911}{16}$

Area of parallelogram

$=119.4375\:cm^2$

Ram Lakhan has a field in the form of circle. The cost of ploughing the field at the rate of $RS 2.24$ per ${m^2}$ is $RS 45056$ . Find the cost of fencing the field at the rate of $RS 12.60$ per meter.

Rate of plowing

$=2.24 \ Rs \ per \ m^2$

Cost of plowing

$=45056 \ Rs$

To find the cost of fencing the field, if the rate of fencing is

$12.60 \ per \ m$

Area plowed for

$Rs \ 2.24 = 1 \ m^2$

Area plowed for

$Rs \ 45056 = \dfrac{1}{2.24} \times 45056 \ m^2=20114.2857 \ m^2$

Area of the circular field

$=20114.2857 \ m^2$

Let

$r$ be the radius of the field.

$So, \ \pi r^2=20114.2857$

$r^2=20114.2857\times \dfrac{7}{22}$

$r^2=6400 \ \implies r=80 m$

Perimeter of the field

$=2 \pi r=2\times\dfrac{22}{7}\times80=502.86 \ m$

So, total cost of fencing

$=502.86\times 12.60=6336 \ Rs$

If the area of the circular tin plate is $38.5 cm ^ { 2 }$ then, its circumference is

Aea

$=\pi r^2 =38.5cm^2\\r=3.50cm$

Circumference

$=2\pi r =21.989cm$

Find the area of a parallelogram $PQRS,$ if $PR = 24\ \text{ cm}$ and $QU = ST = 6\ \text{cm}$ .

Area of parallelogram

$=$ Area (PRS)

$+$ Area (PQR)

$=\frac{1}{2}(PR)(ST)+\frac{1}{2}(PR)(UQ)$

$=\frac{1}{2}(PR)(ST+UQ)$

$=\frac{1}{2}\times 24 \times 12= 144 \text{ cm}^{2}$

A circular garden whose diameter is 28 m needs to be front, what would be the length of the facing wire required if the gardener makes 3 rounds of fence. Also, find the cost of the wire if it cost Rs. 30 meters.

As the garden is in the shape of a circle.

Therefore,

Diameter of garden

$= 28 \; m \; \left( \text{Given} \right)$

$\Rightarrow$ Radius of garden

$\left( r \right) = \cfrac{28}{2} = 14 \; m$

As we know that circumference of the circle is given by-

Circumference

$= 2 \pi r$

where as,

$r$ is the radius of circle.

Therefore,

Circumference of garden

$= 2 \times \cfrac{22}{7} \times 14 = 88 \; m$

For

$3$ round of fence,

Length of wire required

$= 3 \times 88 = 264 \; m$

Given that the cost of wire is

$Rs. 30 \text{ per m}$

Therefore,

Cost of wire

$= 30 \times 264 = Rs. 7920$ .

Hence the length of wire required is

$264 \; m$ and the cost of required wire will be

$Rs. 7920$ .

The area of a circle is $38.5\ \text{m}^{2}$ , then its circumference will be___ $\text{m}$ .

Let radius of circle be

$r\ \text{m}$

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

$38.5=\pi {r}^{2}$

${r}^{2}=12.25$

$\Rightarrow \boxed {r=3.5\ \text{m}}$

$\therefore$ Circumference

$=2\pi r$

$=\left(2\times \dfrac{22}{7}\times 3.5\right)\ \text{m}$

$=21.99 \approx \boxed{22\ m}$ Ans.

Find the circumference of the circle with the following radius: (Take $\pi=\dfrac{22}{7}$ )
$28mm$

Let radius be =

$r$

Circumference of a circle

$=2\pi r$

$\Rightarrow$

$2\pi r=28mm$

$\Rightarrow$

$r=\dfrac{14}{\pi}$ mm

$\Rightarrow$

$r=4.46mm$

A path of width $8\ m$ runs around a circular park whose radius is $38\ m$ . Find the area of the path.

We have radius of circular park

$= 84/2 m$

$= 42 m$

area of circular park (A)

$= 22 / 7 × 42 ^ 2 m^2$

$= 5544 / 7 m^2$

radius of circle including the path and circular park = 42 + 8 m

$= 50 m$

area of the circle (a)

$= 22/7 × 50 ^ 2 m^2$

$= 55000 / 7 m^2$

therefore area of the path

$= a - A$

$= 5544 / 7 - 55000 / 7 m^2$

$= (5544 - 55000) / 7 m^2$

$= 2313.14286 m^2$

Find the circumference of the circle with the following radius: (Take $\pi=\dfrac{22}{7}$ )
$14\ cm$

Let radius be =

$r$

Circumference of a circle

$=2\pi r$

$\Rightarrow$

$2\pi r=14cm$

$\Rightarrow$

$r=\dfrac{7}{\pi}$ cm

$\Rightarrow$

$r=2.23cm$

Find the circumference of the circle with the following radius: (Take $\pi=\dfrac{22}{7}$ )
$21\ cm$

Let radius be =

$r$

Circumference of a circle

$=2\pi r$

$\Rightarrow$

$2\pi r=21cm$

$\Rightarrow$

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

$\Rightarrow$

$r=3.34cm$

A circle of radius $2\ cm$ is cut out from a square piece of an aluminum sheet of side $6\ cm$ . What is the area of the left over aluminium sheet? $\left (Take \pi =3.14\right)$

The area of the square piece of sheet

$=(side)^2=6^2\ cm^2=36 \ cm^2$

The area of the cutter out circle

$=\pi r^2 \ cm^2=\pi \times 2^2 \ cm^2=3.14\times 4\;cm^{2}=12.56\;cm^{2}$

$\therefore$ leftover aluminum sheet area

$=36-12.56$

$\therefore$ leftover aluminum sheet area

$=23.44\ cm^2$

A circle and a rectangle have the same perimeter. The sides of the rectangle are $18cm$ and $26cm$ . What is the area of the circle?

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

$r=\dfrac{l+b}{\pi }=\dfrac{18+26}{\pi }=\dfrac{44}{22}\times 7=14$

Area of circle =

$\pi r^{2}$

$= \frac{22}{7}\times 14\times 14$

$=44\times 14$

$= 616 cm^{2}$

1084289_1174018_ans_21dbdad388c84873a01a42e417b347a9.jpg

Find the area of the circles , given that: radius=14 mm $( Take \ \pi = \dfrac{{22}}{7})$

Radius of a circle

$=14\ m$

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

Area

$=\pi r^2 =\dfrac {22}{7}\times (14)^2$

$=\dfrac {22}{7}\times 14\times 14$

$=22\times 2\times 14$

$\boxed {Area\ =616\ m^2}$

A circular flower bed is surrounded by a path $4\ m$ wide bed is $66\ m$ . What is the area of this path? $(\pi=3.14)$

$D = 66 cm$

$w = 4 cm$

diameter of inner ring

$= 66 cm$

diameter of outer ring

$= 66+2 \times 4 = 74 cm$

$\therefore$ radius of innerring

$= \dfrac{d}{2} = 33cm(r_{1})$

radius of outer ring

$=\dfrac{d+2w}{2} = 37 cm (r_{2})$

Area of path

$= \pi(r_{2}^{2}-r_{1}^{2})$

$= \pi (37^{2}-33^{2})$

$= 280\pi$

$= 879.6 cm^{2}$

A rectangular plot is $112\ m$ long and $6\ m$ broad. It has $2\ m$ path all around it on the inside.Find the area of the path and the cost of constructing it at the rate of $Rs. 60$ per sq m.

Given that:

Length

$l= 112\ m$ ,

Breadth

$b= 6\ m$ ,

Width

$w=2\ m$ .

We know that Area of the path when it's inside

$= 2(w) \times (l+b - 2w)$

$= 2 \times 2 \times (112 + 6 - 2 \times 2.5)$

$= 452\ m^2.$

The cost of constructing it at

$Rs.60$ per sq.m

$= 452 \times60$

$=Rs. 27120$ .

This is the required answer.

$PQRC$ parallelogram, $QM$ is the height form $Q$ to $RC$ and $QN$ is the height from $Q$ to $PS$ . If $RC = 12 \mathrm { cm }$ and $QM = 7.6 \mathrm { cm }$ . Find .
(a) the area of the parallegram $PQRC$
( b ) $QN$ if $PS = 8 \mathrm { cm }$

Area of the parallelogram PQRS = Base× Height
$12cm\times$ $7.6cm=91.2cm^2$

Area = Base × Height

$=91.2cm^2=8cm\times$

$QN$

$QN=91.2\div8=11.2cm$

1087669_1179655_ans_f48597523dfa4d658b74e5cb4da3655d.PNG

The circumference of a given circular park is 55 m. It is surrounded by a path of uniform width 3.5 m. Find the area of the path.

1298829_1163204_ans_d30ccc0cdb9b47f0aeac227bae332e07.jpg

In the figure a circle is inscribe in a square of side $14\ cm$ . Find the area of the circle of the shaded region.

from figure,side of square

$=$ diameter of circle

$14=2r$

$2r=14$

$r=7\ cm$

Area of circle

$=\pi {(r)}^{2}$

$=\cfrac{22}{7}\times 7\times 7$

Area

$=154\ {cm}^{2}....(1)$

Also area of

$square={(14)}^{2}=196{cm}^{2}$

Area pf shaded portion

$=(196-154)\ {cm}^{2}$

$=42\ {cm}^{2}$

Hence, the area of shaded region

$=42\ {cm}^{2}$

Find the area of one fourth of a circle having radius $7 cm$ .

Area of the circle

$=\pi{r}^{2}=\dfrac{22}{7}\times {7}^{2}=154\ sq.cm$

Area of one fourth of a circle

$=\dfrac{1}{4}\times$ Area of the circle

$=\dfrac{1}{4}\times 154=38.5\ sq.cm$

The sum of the radii of two circle is $7$ cm, and the difference of their circumferences is $8$ cm. Find the circumferences of the circles.

1558959_1183657_ans_8949c4ad14134a5f985b5b88aaa0e836.jpg

A circular flower garden has an area of $314\ {m}^{2}$ . A sprinkler at the centre of the garden can cover an area that has a radius of $12\ m$ . Will the sprinkler water the entire garden? (Take $\pi=3.14$ )

${\textbf{Step -1: Finding the total area covered by the sprinkler.}}$

${\text{Radius of the area covered by the sprinkler = 12m}}$

$\therefore {\text{Total area covered by sprinkler is = }}\pi {{\text{r}}^2} = 3.14 \times {12^2}$

${\text{ = 452}}{\text{.16}}{{\text{m}}^2}$

${\text{Since the sprinkler covers more area than the area of garden}}$ ${314 \text{ m}^2}$

${\text{So, the sprinkler will water the entire garden.}}$

${\textbf{Hence, the sprinkler will water the entire garden.}}$

Saima wants to put a lace on the edge of a circular table cover of diameter $1.5\;m$ . Find the length of the lace required and also find its cost if one diameter of the lace costs $Rs\;15$ . (Take $\pi=3.14$ )

circumference

$=\pi d=1.5\pi m$

length of lace required = circumference of circle

$=1.5\times 3.14=4.71m$

length of lace required

$=4.71m$

one meter of lace

$=Rs. 15$

$4.71m$ lace

$=$ ?

$=\cfrac{4.71\times 15}{1}=Rs. 70.65$

It costs

$70.65Rs.$ to buy

$4.71m$ of lace

1358631_1193732_ans_466f9068aeb340ceb0fc2170cfd5d91f.PNG

Find the area of a flat circular ring formed by two concentric (circles with same centre) whose radii are 9 cm and 5 cm.

The area of the region enclosed by two concentric circles is given by:

$A=\pi(R^2-r^2)$

here

$R=9cm,\ r=5cm$

$A=\dfrac{22}{7}(9^2-5^2)$

$=\dfrac{22}{7}(81-25)$

$=176\ cm^2$

The diameter of circular park is 70 m, Calculate its surface area?

Given that,

Diameter $D=70\,m$

We know that,

The area of circular horizontal surface

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

$A=3.14\times 35\times 35$

$A=3847\,{{m}^{2}}$

Hence, this is the required solution

Find the area tabletop, which is in the shape parallelogram with base $4m$ and height $5m$ Find its area.

The base of parallelogram is

$4m$

The height of parallelogram is

$5m$

The area of parallelogram is

$b\times h$

$=4m\times 5m=20m^2$

If the area of a parallelogram is $83.2$ sq cm and its height is $6.4$ cm, find its find the length of its base.

1305563_1279098_ans_be3a31f4e14d462998f89b3a6872b6c5.JPG

Depicts an archery target marked with its five scoring regions from the center outwards as gold,red,blue,black and white. The diameter of the region representing gold score is $21\;cm$ and each of the other bands is $10.5\;cm$ wide. Find the area of each of the five scoring regions.

width of each circle

$l=150\ cm$

circle

$A$ :- diameter

$=k=21\ cm$

circle

$B\ \Rightarrow$ diametre

$=k+2l=42\ cm=2k$

circle

$C\ \Rightarrow$ diametre

$=k+4l=21\times 3\ cm=3k$

circle

$D\ \Rightarrow$ diametre

$=k+6l=21\times 4\ cm=4k$

circle

$3$ dimeter

$=k+8l=21\times 5\ cm =5k$

area of gold region = area of circle

$A=\dfrac {\pi k^2}{4}$

$=\dfrac {22}{4}\times \dfrac {21 \times 21}{4}=346.5\ cm^2$

are of red region = area of circle

$B-$ area of circle

$A$

$=\dfrac {\pi (2k)^2}{4}-\dfrac {\pi (k)^2}{4}=3x\dfrac {\pi k^2}{4}=3346.5\ cm^2$

$=1039.5\ cm^2$

area of blue region

$=area of circle$ C-

$are of circle$ B$$

$=\dfrac {\pi (3k)^2}{4}-\dfrac {\pi (2k)^2}{4}=5\times \left (\dfrac {\pi k^2}{4}\right)=5\times 346.5\ cm^2$

$=1732.5\ cm^2$

area of black region = are of circle

$D-$ area of circle

$C$

$=\dfrac {\pi (4k)^2}{4}-\dfrac {\pi (3k)^2}{4}=3dfrac {\pi k^2}{4}=7\times 346.5\ cm^2$

$=2425.5\ cm^2$

area of white region = area of circle

$E-$ area of circle

$D$

$=\dfrac {\pi (5k)^2}{4}-\dfrac {\pi (4k)^2}{4}=9\dfrac {\pi k^2}{4}$

$=9\times 346.5\ cm^2=3118.5\ cm^2$

$\therefore$ The area

$346.5,\ 1039.5,\ 1732.5,\ 2425.5$ and

$3118.5\ cm^2$ respectively

1358086_1193644_ans_94f6f9cd9744429b938b1aa84b2814b5.png

The circumference of a circle is $39.6cm$ . Find its area.
Use $\pi=\cfrac{22}{7}$

Circumference

$=39.6\ cm$

$2\pi r=39.6$

$r=\dfrac{39.6\times7}{2\times22}$

$r=6.3\ cm$

Area

$=\pi r^2$

$=\dfrac{22}{7}\times6.3^2$

$=124.74\ cm^2$

Find the breadth of a rectangular plot of land, if its area is $440$ $m^2$ and the length is $22 \ m$ . Also find its perimeter.

Given,

Length of the rectangular plot

$=22\:m$

and area of the rectangular plot

$=440\:m^2$

Let

$b$ be the breadth of the rectangular plot.

Area

$=l\times b$

$\Rightarrow 440=22\times b$

$\Rightarrow b=\dfrac{440}{22}$

$\Rightarrow b=20$

Thus, the breadth of the rectangular plot is

$20\:m$

Perimeter of the rectangular plot

$=2(l+b)$

$=2\times (22+20)\ m$

$=2\times42\ m$

$=84\ m$

Hence, the breadth of the plot is

$20\ m$ and the perimeter is

$84\ m$ .

(i) A circular pond of diameter $28\ m$ is surrounded by a path $7\ m$ wide. Find the area of the path.
(ii) Find the area of a $2\ m$ -wide circular path surrounding a circular plot of radius $20\ m$ .

$(i)$ Area of the path

$=\pi(R^2-r^2)$

Here,

$R=\dfrac{28}{2}+7=14+7=21\ m$

$r=21-7=14\ m$

Area of the path

$=\pi(21^2-14^2)$

$=\dfrac{22}{7}\times245$

$=770\ m^2$

$(ii)$ Area of circular path

$=\pi(R^2-r^2)$

Here,

$R=20+2=22$

$r=22-2=20\ m$

Area of circular path

$=\pi(22^2-20^2)$

$=\dfrac{22}{7}\times84$

$=264\ m^2$

Find the radius of the circle whose circumference is $44$ cm. $(\pi=22/7)$

Let

$r$ be the radius of the circle

Then the perimeter

$=2\pi r$

According to the problem

$2\pi r=44$

${r}=\dfrac{44}{2\pi}$

$r=7$

A flooring tile has the shape of a parallelogram whose base is $24\ cm$ and the corresponding height is $10\ cm$ . How many such tiles are required to cover a floor of area $1080\ {m}^{2}$ ? (If required you can split the tiles in whatever way you want to fill up the corners)

We know, Area of parellelograms

$=$ Base

$\times$ Corresponding height

Area of a tile

$=\dfrac{24}{100}\times\dfrac{10}{100}$

$=\dfrac{240}{10000}\\$

$=0.024\ m^2$

Area of the floor

$=1080\ m^2$

Now,

Number of tiles

$=\dfrac{Total\ area}{Area\ of\ one\ tile}$

$=\dfrac{1080}{0.024}$

$=45000\ tiles$

A wire in the form of a square with side $22cm$ . It is reshaped and bent into the form of a circle. Calculate the area of this circle. ( $\pi=\cfrac{22}{7}$ )

Given that side of the square is

$22$

$cm$

Perimeter of the square

$=4a$

$=4\times22\,cm$

$=88\ cm$

As the same wire is used to form a circle then both will have same perimeter.

Perimeter of circle

$=$ Perimeter of square

$\Rightarrow 2\pi r=88$

$\Rightarrow 2\times\dfrac{22}{7} r=88$

$\Rightarrow r=\dfrac{88\times7}{22\times2}$

$\Rightarrow r=14\ cm$

Hence radius of the circle is

$14\, cm$ .

then,

Area of the circle

$=\pi r^2$

$=\dfrac{22}{7}\times14\times14\,cm^2$

$=616\ cm^2$

Find the area of the given parallelogram.

$\textbf{Step -1: Find the area of given parallelogram.}$

$\text{Base}=7\text{ cm}$

$\text{Height}=4\text{ cm}$

${\because\text{Area of parallelogram}=\text{Base}\times\text{Height}}$

$\therefore\text{Area of given parallelogram}=7\times4$

$=28\mathrm{cm^2}$

$\textbf{Hence, the area of given parallelogram is }\mathbf{28cm^2.}$

A circle of circumference $88cm$ is cut out of a square sheet of paper of side $30cm$ . What is the area of the paper left?

Circumference

$=88\ cm$

$2\pi r=88$

$r=\dfrac{88\times7}{22\times2}$

$r=14\ cm$

Area of circle

$=\pi r^2$

$=\dfrac{22}{7}\times14^2$

$=616\ cm^2$

Area of sheet

$=30\times30$

$=900$

Remaining area

$=900-616=284\ cm^2$

On a square handkerchief, nine circular designs each of radius $7cm$ are made (see figure). Find the area of the remaining portion of the handkerchief.

Radius of a circle

$=7\ cm$

Length of side of the square

$=3(14)=42\ cm$

Area of the square

$=(side)^2=42^2=1764\ cm^2$

Area of one circle

$=\pi r^2=\dfrac{22}{7}\times7^2=154\ cm^2$

Area of nine circles

$=9\times154=1386\ cm^2$

Remaining portion (without design)

$=$ Area of handkerchief

$-$ Area of design

$=1764-1386$

$=378\ cm^2$

Hence, the area of the remaining portion is

$378\ cm^2$

How many times a wheel of radius $28\ cm$ must rotate to go $352\ cm$ . (Take $\pi=3.14$ )

Radius of the wheel

$=28$ cm

distance covered by the wheel in one rotation

$=2\times \pi \times r=2\times \cfrac { 22 }{ 7 } \times 28$ cm

$\therefore$ No. of times wheel must rotate to go

$352$ m

$=\cfrac { 352cm }{ 2\pi rcm } =\cfrac { 352\times 7 }{ 2\times 22\times 28 } =2$ times

The radii of two circles are $19cm$ and $9cm$ respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.

Let the radius of required circle

$=r\ cm$

Radius of 1st circle

$r_1=9\ cm$

Radius of 2nd circle

$r_2=19\ cm$

As per the question

Circumference of the required circle

$=$ Sum of circumference of two circles

Circumference of small circle

$=2\pi r_1$

$=2\pi\times9$

$=18\pi$

Circumference of small circle

$=2\pi r_2$

$=2\pi\times19$

$=38\pi$

Now,

Circumference of the required circle

$=$ Sum of circumference of two circles

$2\pi r=18\pi+38\pi$

$2\pi r=58\pi$

$r=\dfrac{56\pi}{2\pi}$

$r=28\ cm$

Hence, radius of new circle is 28 cm.

Find the area of the given parallelogram.

$\textbf{Step -1: Find the area of given parallelogram.}$

$\text{Base}=5\text{ cm}$

$\text{Height}=3\text{ cm}$

${\because\text{Area of parallelogram}=\text{Base}\times\text{Height}}$

$\therefore\text{Area of given parallelogram}=5\times3$

$=15\mathrm{cm^2}$

$\textbf{Hence, the area of given parallelogram is }\mathbf{15cm^2.}$

Find the area of a circle whose circumference is $484 \ cm$ .

$Circumference=484cm$

$2\pi r=484cm\Rightarrow r=\cfrac{484}{2\times22}\times 7cm\\area=\pi r^2cm^2=\cfrac{22}{7}\times77\times77 cm^2$

$=22\times11\times11\times77cm^2$

$=18634cm^2$

A road is $3.5$ m wide surrounding a circular plot whose circumference is $44$ m. Find the cost of paving the road at Rs. $10$ per $\text{m}^{2}$ .

Radius of plot

$=\cfrac { 44 }{ 2\pi } =\cfrac { 44 }{ 44 } \times 7=7$ m.

Area of road

$=\pi \left[ { \left( 10.5 \right) }^{ 2 }-{ 7 }^{ 2 } \right] =61.25\times \pi$

$=192.42255$ m

$^2$ .

$\therefore$ Cost Rs.

$=1924.22$

Find the area of the given parallelogram.

$\textbf{Step -1: Find the area of given parallelogram.}$

$\text{Base}=2.5\text{ cm}$

$\text{Height}=3.5\text{ cm}$

${\because\text{Area of parallelogram}=\text{Base}\times\text{Height}}$

$\therefore\text{Area of given parallelogram}=2.5\times3.5$

$=8.75\mathrm{cm^2}$

$\textbf{Hence, the area of given parallelogram is }\mathbf{8.75cm^2.}$

Find the area of the Parallelogram ABCD and hence find the value of $x$ .

As opposite sides of a parallelogram are equal.

Also, we know that area of parallelogram

$=$ base

$\times$ height

$=AP\times DC=QC\times BC$

$\Rightarrow$ Area

$=8\times x=10\times 4=40cm^2$

$\Rightarrow x=\dfrac{40}{5}=8$ cm.

What will be the perimeter of a quadrant of a circle of radius $R$ .

Perimeter of quadrant

$=R+R+\cfrac { 2\pi R }{ 4 }$

$=2R+\cfrac { \pi R }{ 2 }$

Find the diameter of a circle whose circumference is:
i) $4.4$ m
ii) $6.6$ m
iii) $11$ m

Circumference

$=\pi D$

$\Rightarrow Dia=\dfrac{circumference}{\pi}$

$i). Dia=\dfrac{4.4}{22}\times 7=1.4m$

$ii). Dia=\dfrac{6.6}{22}\times 7=2.1m$

$iii).Dia=\dfrac{11}{22}\times 7=3.5m$

Hence, solved

X and Y are points on the side LN of the triangle LMN such that LX=XY=YN.Through X, a line is drawn parallel to LM to meet MN at Z. Prove that ar(LZY)=ar(MZYX).

$\begin{array}{l} LX=XY=YN\left( { given } \right) \\ \because LM\parallel XZ \\ \Rightarrow Draw\, \, ZL{ \bot _{ r } }LY \\ \Rightarrow In\, \, \Delta LZY\, \, XT\parallel YZ \\ \Rightarrow ar\Delta LZY=\frac { 1 }{ 2 } \left( { LY } \right) XZL=\left( { 2XY } \right) \times \frac { 1 }{ 2 } \times \left( { ZL } \right) =\left( { ZL } \right) \times \left( { XY } \right) \\ \Rightarrow In\, \, quadrilateral\, \, \left( { MZYX } \right) =\left( { base\times height } \right) \\ \Rightarrow ar\left( { MZYX } \right) =\left( { XY } \right) \times \left( { ZL } \right) \\ \Rightarrow Hence,\, \, ar\left( { \Delta LZY } \right) =arMZYX\, \, proved. \end{array}$
1216565_1213743_ans_edc56df3b5b148fc9348a2ad55b8c9cc.png

1216565_1213743_ans_edc56df3b5b148fc9348a2ad55b8c9cc.png

Find the area of the given parallelogram.

$\textbf{Step -1: Find the area of given parallelogram.}$

$\text{Base}=5\text{ cm}$

$\text{Height}=4.8\text{ cm}$

${\because\text{Area of parallelogram}=\text{Base}\times\text{Height}}$

$\therefore\text{Area of given parallelogram}=5\times4.8$

$=24\mathrm{cm^2}$

$\textbf{Hence, the area of given parallelogram is }\mathbf{24cm^2.}$

Find the area of the given parallelogram.

$\textbf{Step -1: Find the area of given parallelogram.}$

$\text{Base}=2\text{ cm}$

$\text{Height}=4.4\text{ cm}$

${\because\text{Area of parallelogram}=\text{Base}\times\text{Height}}$

$\therefore\text{Area of given parallelogram}=2\times4.4$

$=8.8\mathrm{ cm^2}$

$\textbf{Hence, the area of given parallelogram is }\mathbf{8.8cm^2.}$

Find the area of parallelogram (in square m) whose base and altitude are given base $=15$ dm, altitude $=4.5$ dm.

Area of parallelogram

$=B\times H$

$=15$ dm

$\times 4.5$ dm

$=.675m^2$ .

A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to from a platform 22 m by 14 m. Find the height of the platform.

1306798_1242305_ans_c1216641ca69426a9af3de8c2f25a931.jpg

Find the area of a circle whose radius is
$87.5\ km$

Area of circle

$=\pi r^2$

Given

$r=87.5\ cm$

$=\pi (87.5)^2$

$=24052.82\ cm^2$

Find the number of revolutions made by a circular wheel of area $1.54m^2$ in rolling a distance of $176 m$ .

Distance covered by wheel in n revolutions with radius

$r=2\pi rn.$

$\therefore 2\pi r\,n=176m$ .........(i)

Area of wheel (circular)=

$1.54m^2$

$\Rightarrow \pi r^2=1.54$

$\Rightarrow r^2=\dfrac{1.54}{\pi}=\dfrac{154\times 7}{22\times 100}=\dfrac{7\times 7}{10\times 10}$

$\Rightarrow r=0.7m$

Now,

$2\pi r\,n=176$

$\Rightarrow 2\times \dfrac{22}{7}\times 0.7\times n=176$ [From (I)]

$\Rightarrow n=\dfrac{176\times 7\times 10}{2\times 22\times 7}=40$

Hence ,

$n=40$ revolutions.

Find the area of a circle whose radius is
$28\ cm$

Area of circle

$=\pi r^2$

Given

$r=28\ cm$

$=\pi (28)^2=\dfrac{22}{7}\times 784$

$=2464\ cm^2$

The radii of two circles are $8\ cm$ and $6\ cm$ respectively. Find the radius of the circle having area equal to the sum of the area of the two circles.

Given radii of two circles are

$8$ cm and

$6$ cm. We have to find the radius of the circle having area equal to sum of area of the two circles.

$\therefore$ Sum of area of two circles

$=\pi \times 8^2+\pi\times 62$

$=64\pi+36\pi$

$=100\pi$

$\therefore \pi R^2=100r$

$R=10$ cm

$\therefore$ Radius of the circle is

$R=10$ cm.

1188101_1242779_ans_7bcc8c18c3584273bea1b6e2a2ff120e.jpg

Find the area of a circle whose radius is
$5.6\ cm$

Area of circle

$=\pi r^2$

Given

$r=5.6\ cm$

$=\pi (5.6)^2=\dfrac{22}{7}\times 31.36$

$=98.56\ cm^2$

The circumfrence of a circle is $264cm$ . Find its area.

We have,

Circle

$=264$

radius of circle

$=r$

Circumference

$=2\pi r$

$2\pi r=264$

$\dfrac {22}{7}\times r=132$

$r=132\times \dfrac {7}{22}$

$r=42\ cm$

Area

$=\pi r^{2}$

$=\dfrac {22}{7}\times 42 \times 42$

$=5544\ cm^{2}$

Hence, this is the answer.

The circumfrence of a circle is $35.2m$ . Find its area.

Circumference of circle

$=2\pi r$

$2 \pi r=35.2$

$2\times \dfrac {22}{7}\times r=35.2$

$\dfrac {44}{7}\times r=35.2$

$r=35.2\times \dfrac {7}{44}$

$r=\dfrac {246.4}{44}$

$r=5.6$

Are of a circle

$=\pi r^{2}$

$=\dfrac {22}{7}\times (5.6)^2$

$=\dfrac {22}{7}\times 31.36$

$= 98.56\ cm^2$

The circumference of of a circle is $616\ \text{cm}$ . Find its area

Given. Circumference

$=616\ cm$

$2\pi r=616$

$2\times \dfrac{22}{7}\times r=616$

$r=\dfrac{616\times 7}{2\times 22}$

$r=\dfrac{616\times 7}{44}$

$r=14\times 7=98\ cm$

Area of circle

$=\pi r^2$

$=\dfrac{22}{7}\times 98\times 98$

$=22\times 14\times 98=30184\ cm^2$

If two angles of a triangle are ${30}^{\circ}$ and ${45}^{\circ}$ , then find its third angle.

1371914_1265983_ans_94d23278c9064e0ab4e8975a7f42669d.PNG

The diameter of a wheel is $1.4\ m$ , find its circumference.

b'radius=diameter/2

radius of a wheel=1.4/2

=0.7

$\begin{array}{l}circumference = 2\pi r\ = 2\left( {\frac{{22}}{7}} \right)\left( {0.7} \right)\\\ = 4.4m\end{array}$

Thus circumference is 4.4m

Find the angle in radian through which a pendulum swings if its length is 75 and the tip describes an arc of length 21 cm .

We know that

$l=r\theta$

Here,

$l=$ length of arc

$=21$ cm

$r=$ radius

$=$ length of pendulum

$=75$ cm

We have

$l=r\theta$

$\Rightarrow 21=75\times \theta$

$\Rightarrow \theta=\dfrac{21}{75}=\dfrac{7}{25}$ radian

The adjacent sides of a parallelogram are $7.5\ cm$ and $4\ cm.$ If the distance between the longer vides is $2\ cm,$ find the distance between the shorter sides.

The distance between longer sides,

$EC = 2\ cm$

Let, the distance between shorter sides,

$FC = x$

Area of the parallelogram

$=b \times h$

$A = 7.5 \times 2$

$= 15\ cm^2$

Now, considering

$AD$ as the base and

$FC$ as the height.

But the area of parallelogram should remains same,

$A = b \times h$

$15 = 4 \times h$

$h = 3.75\ cm$

$\therefore FC = 3.75\ cm$

1342588_1262327_ans_bec16184ef134666b89fefd175f739c7.png

The sum of circumference and the radius of a circle is $51\ cm$ , find the radius of the circle.

Sum of circumference and radius = 51 cm.

Find radius = (?)

$\rightarrow$ circumference + radius = 51

$\therefore 2\pi r+r=51$

$\therefore r(2\pi +1)=51$

$\therefore r=\frac{51}{2\pi +1}$

$=\frac{51}{2\times \frac{22}{7}+1}$

$=\frac{51}{\frac{44+7}{7}}$

$= \frac{51\times 7}{51}$

$\therefore r= 7 cm$

1129641_1249132_ans_d3f2afefb1204c21a188ce7e9a861bd1.jpg

D is the mid point of side ${\text{BC}}$ of $\Delta {\text{ABC}}$ and ${\text{E}}$ is the mid point of ${\text{BD}}$ . If ${\text{O}}$ is the mid point of ${\text{AE}}$ , prove that ${\text{ar}}\Delta {\text{BOE = }}\;\frac{{\text{1}}}{{\text{8}}}{\text{ar}}\Delta {\text{ABC}}$

Since

$AD$ and

$AE$ are median of

$\triangle ABC$ and

$\triangle ABD$ respectively.

$\therefore \ ar (\triangle ABD)=\dfrac {1}{2}ar (\triangle ABC).....(i)$
and

$ar (\triangle ABE)=\dfrac {1}{2}ar (\triangle ABD)....(ii)$

$OB$ is a median of

$\triangle ABE$

$\therefore \ ar(\triangle BOA)=\dfrac {1}{2}ar (\triangle ABE)...(iii)$
From

$(i), (ii)$ , we have

$ar (\triangle BOE)=\dfrac {1}{8} ar (\triangle ABC)$
1406673_1264980_ans_b922b556233249e892b970b635fef8a5.png

The ratio of the outer and inner circumference of a circular path as shown in the figure is $36:25$ . If the path is $5\ m$ wide, then find :
(i) the area enclosed by the path
(ii) the diameter of the inner circle

Let the radius of outer and inner circle be

$R$ and

$r$ respectievly.

According to the given condition,

$\cfrac { 2\pi R }{ 2\pi r } =\cfrac { 36 }{ 25 }$

$25R=36r$

$\implies R=\cfrac { 36r }{ 25 }$

Width of the path is given as

$5$ m which is equal to

$R-r$ .

$R-r=5$

So,

$\cfrac { 36r }{ 25 } -r=5\quad \Rightarrow \cfrac { 11r }{ 25 } =5\quad \Rightarrow r=\cfrac { 125 }{ 11 }$ m

And

$R=\cfrac { 36r }{ 25 } =\dfrac{36\times 125}{25\times 11}=\cfrac { 180 }{ 11 }$ m

Area of path

$=\pi ({ R }^{ 2 }-{ r }^{ 2 })=\cfrac { 22 }{ 7 } \times \cfrac { 1 }{ 121 } ({ (180) }^{ 2 }-{ (125) }^{ 2 })$

Area of path

$=435.71{ m }^{ 2 }$

Diameter of inner circle

$=2r=\cfrac { 250 }{ 11 }$ m

If the circumference of a circle is $176\ cm$ , find its radius.

The circumference of the circle

$=176$

i.e.,

$2\pi r=176$

$2\times \cfrac{22}{7}\times r=176$

$r=28cm$

Radius

$=28cm$

Find the circumference of circle whose area is 16 times the area of the circle with diameter 1.4 m.

Given that, area of a circle is $16$ times the area of another circle with diameter $1.4 \ m$ .

To find out: Circumference of the first circle.

Now, radius of the second circle $=\frac{1.4}{2}=0.7\ m$

We know that, area of a circle $=\pi r^2$

$\therefore \$ Area of the second circle $=\frac{22}{7}\times 0.7\times 0.7$

$\Rightarrow 1.54\ m^2$

Then,

Area of first circle $=16\times 1.54=24.64 \ m^2$

$\therefore \ \pi {{r}^{2}}=24.64$

${{r}^{2}}=\dfrac{24.64}{\pi }$

${{r}^{2}}=\dfrac{24.64}{\dfrac{22}{7}}$

${{r}^{2}}=\dfrac{24.64\times 7}{22}$

${{r}^{2}}=7.84\,$

$r=\sqrt{7.84}$

$r=2.8\ m$

Now, Circumference of circle $=2\pi r$

$=2\times \dfrac{22}{7}\times 2.8$

$=\dfrac{123}{7}$

$=17.6\ m$

Hence, the circumference of the circle is $17.6 \ m$ .

The sum of circumferences of two circles is $132 cm.$ If the radius of one circle is $14 cm,$ find the radius of the second circle.

$132=2\pi r_{1}+2\pi r_{2}$

$\Rightarrow \frac{132}{2\pi }=r_{1}+r_{2}$

$\Rightarrow \frac{132\times 7}{2\times 22}=14+r_{2}$

$\Rightarrow 21= 14+r_{2}$

$\Rightarrow 7=r_{2}$

$r_{2}= 7\,cm$

1195853_1280269_ans_f7fbc85c6ab040e191ec7e7bdb0597d5.jpg

What is the area of the circle whose circumference is $170\ m$ .

Given circumference

$=170$ m

$=$ c

Area of circle

$=\dfrac{c^2}{4\pi}$

$=\dfrac{170^2}{4\pi}=2299.79$ .

1202605_1282309_ans_aedc0f9161894b7ebb2fdaf2ef433431.jpg

Find the area of a sector of a circle with radius $14\ cm$ , if the length of the corresponding arc is $22\ cm$ .

We have,

Radius of circle $=14\,cm.$

Length of arc $=22\,cm.$

Using formula,

$Angle=\dfrac{radius}{arc}$

Now,

$Angle=\dfrac{22}{14}$

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

$Angle=\dfrac{\pi }{2}\,\,\,\,\left( \pi =\dfrac{22}{7} \right)$

$Angle=\theta =\dfrac{\pi }{2}$

So,

Area of sector $=\dfrac{\theta }{{{360}^{o}}}\times \pi {{r}^{2}}$

$=\dfrac{{{90}^{o}}}{{{360}^{o}}}\times \dfrac{22}{7}\times 14\times 14$

$=\dfrac{1}{4}\times 22\times 2\times 14$

$=22\times 7$

$= 154\,sq.\,cm.$

Hence, this is the answer.

If the area of a parallelogram is $29.6$ sq cm and its base is $8$ cm, find its height.

1305562_1279089_ans_001c489158954ebca8d7b08296bb42c3.JPG

How many time will the wheel of a can rotation in a iorrney of 88 km, given that the diameter of the wheel is 56 cm?

Circumference =

$2\pi r=1$ revolution

D= 56 cm,

$r=\frac{56}{2}= 28\, cm$

$\therefore 2\pi r=2\times \frac{22}{7}\times 28= 44\times 4 = 176\, cm$

For 176 cm, wheel rotates 1 time

$\therefore$ for

$88 km = 88\times 1000\times 100\, cm$ , wheel rotates =

$\frac{1}{176}\times 8800000$

= 50000 times.

1170798_1272255_ans_d9285265dd044c64a96fba5fdbc873b0.jpg

The area of circle is $25 \pi \text{ cm}^2.$ Find its circumference.

Let the radius of the circle be

$r.$

The area of the circle is given as,

$\begin{aligned}{}\pi {r^2} &= 25\pi \\{r^2} &= 25\\r &= 5\text{ cm}\end{aligned}$

Hence, the circumference of circle will be,

$\begin{aligned}{}2\pi r& = 2 \times \frac{{22}}{7} \times 5\\ &= \frac{{220}}{7}\\ &= 31.428\text{ cm}\end{aligned}$

Hence, the circumference of the circle is equal to

$31.428\text{ cm}.$

A round table cover has six equal designs as shown in the figure. If the radius of the cover is 28 cm,then find the cost of making the design at the rate of rs/. 0.35 per $cm^{2}$ .

$OB = OC = 28 cm$

Area of circle

$= \pi r^2 = \dfrac{22}{7} \times 28 \times 28$

$= 88 \times 28$

$= 2464 cm^2$

Since, ABCDEF is a regular hexagon

$\therefore \Delta OBC$ is an equilateral

$\Delta$

$ar(\Delta OBC) = \dfrac{\sqrt{3}}{4} \times 28 \times 28 = 339.08 cm^2$

Area of 6 triangles

$= 6 \times 339.08 = 2034.48 cm^2$

Area of shaded region = Area of circle - Area of 6.

$= 2464 - 2034.48 = 429.52 cm^2$

Rate

$= 0.35 cm^2$

$\therefore$ Rate

$= 0.35 \times 429.52 = Rs. 150.33$

1170795_1274017_ans_d72bdf7b641c4376825951bd8be1653a.png

Find the area of the mirror segment of a circle of radius $42$ cm, if length of the correspondence are is $44$ cm.

1313055_1271512_ans_002687e277fe47b2b2d7c4ffd1e16e42.JPG

Find the total cost of wooden fencing around a circular garden of diameter 28 m, if 1 m of fencing costs 2300.

Given,

$D = 28 \text{ m}$

So, fencing will be equal to the Perimeter of circle.

$\therefore$ Length

$= \pi D = \cfrac{22}{7}\times 28 \text{ m}$

Also, given that cost

$= 2300 \text{ Rs/m}$

$\therefore$ Total cost

$= 2300\times \cfrac{22\times 28}{7}$

Cost

$= 28914.3 \text{ Rs}$

If the circumference of a circle is $\pi$ units more than the diameter of circle then find the diameter of the circle .

Let the diameter of the circle be $=D$

Now, Circumference of the circle $=\pi \times Diameter$

According to given question,

$\pi \times D=\pi +D$

$\dfrac{22}{7}D=\dfrac{22}{7}+D$

$\dfrac{22}{7}D=\dfrac{22+7D}{7}$

$22D=22+7D$

$22D-7D=22$

$15D=22$

$D=\dfrac{22}{15}$

Then,

$D=x=\dfrac{22}{15}$

Hence, this is the answer.

Find the distance covered by the wheel of a car in $500$ revolution if the diameter of the wheel is $77\ cm$

1306687_1280694_ans_697796d04c2f4774a9bbc320bc48bf3c.jpeg

The diameter of a wheel of a car is 1.12 m. Find the distance covered by the car by completing one revolution. $\left(take \pi = \dfrac{22}{7}\right)$

We know that, Radius of a circle

$r=\dfrac{diameter}{2}=\dfrac{1.12}{2}=0.56m$

In one complete revolution, car covers a distance equal to perimeter of wheel

$=2\pi r=2\times \dfrac{22}{7}\times 0.56=3.52 m$

1181952_1293239_ans_e61209fdf52f4e3b91b533905bfe1e2f.jpg

Two concentric circles are of radii $7\mathrm { cm }$ and $5\mathrm { cm } .$ Find the area of the portion between two circles.

Given radii of concentric circles are

$7\mathrm { cm }$ and

$5\mathrm {cm }$

Area of two circles will be

$\pi(7)^2,\pi(5)^2\implies 49\pi,25\pi$

Area of the portion between two circles is

$49\pi-25\pi=24\pi$

If the circumference of two circle are in the ratio $3 : 4.$ Find the ratio of their areas.

Circumference of two circles

$=3:4$

$\dfrac { { 2\pi r }_{ 1 } }{ { 2\pi r }_{ 2 } } =\dfrac { 3 }{ 4 }$

$\dfrac { { r }_{ 1 } }{ { r }_{ 2 } } =\dfrac { 3 }{ 4 }$

Ratio of radius

$=3:4$

Area

$=\dfrac { { \pi r }_{ 1 }^{ 2 } }{ { \pi r }_{ 2 }^{ 2 } } =\dfrac { { 3 }^{ 2 } }{ { 4 }^{ 2 } } =\dfrac { 9 }{ 16 }$

Ratio of areas

$=9:16$

Find the area of the triangle whose vertices are
$(2,3) (-1,0),(2,-4)$

We have,

$\left( {{x}_{1}},{{y}_{1}} \right)=\left( 2,3 \right)$

$\left( x,{{y}_{2}} \right)=\left( -1,0 \right)$

$\left( {{x}_{3}},{{y}_{3}} \right)=\left( 2,-4 \right)$

We know that,

Area of triangle

$=\dfrac{1}{2}\left| \left[ {{x}_{1}}\left( {{y}_{2}}-{{y}_{3}} \right)+{{x}_{2}}\left( {{y}_{3}}-{{y}_{1}} \right)+{{x}_{3}}\left( {{y}_{1}}-{{y}_{2}} \right) \right] \right|$

$=\dfrac{1}{2}\left| \left[ 2\left( 0+4 \right)-1\left( -4-3 \right)+2\left( 3-0 \right) \right] \right|$

$=\dfrac{1}{2}\left| \left[ 8+7+6 \right] \right|$

$=\dfrac{1}{2}\left|21 \right|$

$= \dfrac{21}{2}\,\text{sq}\text{.}\,\text{unit}$

Find the area of the circle, whose radius is $4.2\ cm$ .

We know that,

Area of circle=

$\pi {r^2}$

$\begin{array}{l} =\frac { { 22 } }{ 7 } \times 4.2\times 4.2 \\ =\frac { { 22\times 6\times 42 } }{ { 100 } } \\ =\frac { { 5544 } }{ { 100 } } \\ =55.44c{ m^{ 2 } } \end{array}$

Hence,

We get

$55.44cm^2$ .

Find circumference of a circle, if its area is $154\ cm^{2}$ .

$\begin{array}{l} Area\, \, of\, \, circle=154 \\ \pi { r^{ 2 } }=154 \\ \dfrac { { 22 } }{ 7 } \times { r^{ 2 } }=154 \\ { r^{ 2 } }=\dfrac { { 154\times 7 } }{ { 22 } } \\ r=7\, \, cm \\ circumference\, \, =2\pi r \\ =2\times \dfrac { { 22 } }{ 7 } \times 7=44\, \, cm. \end{array}$

Find the diameter of a circle whose circumference is equal to the some of circumference of circle with radii $10\ cm,12\ cm$ ,and $18\ cm$

Let d be the required diameter, then radius of the circle

$\dfrac{d}{2}$

It is given that,

$2\pi\dfrac{d}{2}=2\pi(10)+2\pi(12)+2\pi(18)$

$\Rightarrow 2\pi\left(\dfrac{d}{2}\right)=2\pi(10+12+18)$

$\Rightarrow \dfrac{d}{2}=40$

$\Rightarrow d=80\ cm$ .

PQRS is a parallelogram. If PQ=12 cm, QR=8 cm and AR=9 cm, find the area of parallelogram ABCD and the length of RB.

1240173_1324649_ans_0300767049f7414f986d538a67e7b2d4.jpg

Find the area of the parallelogram.

Area of parallelogram

$= DC\times AE=7\times 4=28 { cm }^{ 2 }$

1173974_1308788_ans_1eafdec8950e4c94b0d66003ebe0b431.png

Two circular pieces of equal radii and maximum area, touching each other are cut out from a rectangular cardboard of dimension $14 \mathrm { cm } \times 7 \mathrm { cm } .$ Find the area of the remaining cardboard.

Length of the rectangular cardboard =

$\displaystyle 14 \ cm$ and

Breadth of the rectangular cardboard =

$\displaystyle 7 \ cm$

Area of cardboard = Area of rectangle =

$\displaystyle length \times breadth = 14 \times 7 = 98 \ cm^2$

Let the two circles with equal radii and maximum area have a radius

$\displaystyle r$ each.

$\displaystyle 2r = 7$ or

$\displaystyle r = \frac{7}{2} \ cm$

Again,Area of two circular cut outs =

$\displaystyle 2 \times \pi r^2 = 2 \times \frac{\pi}{7} \times (\frac{7}{2})^2 = 77 \ cm^2$

Now,The area of remaining cardboard = Area of cardboard - Area of two circular cut outs =

$\displaystyle 98 – 77= 21$

Therefore, area of remaining cardboard is

$\displaystyle 21 \ cm^2$ .

1557533_1303128_ans_79b082db45e04ec7acb4c73353debcd5.JPG

Find the circumference of the circle, whose radius is $17.5\ cm$ .

$\begin{array}{l} r=17.5\, \, cm \\ circumference=2\pi r \\ =2\times \dfrac { { 22 } }{ 7 } \times 17.5 \\ =110\, \, cm \end{array}$

The locus of $z$ satisfying the equation $arg$ $\left( \dfrac { z - 2 } { z + 2 } \right) = \dfrac { \pi } { 3 }$ is a portion of circle, then the length of that portion of circumference of the circle is equal to

$\begin{array}{l} z=x+iy \\ z-2=\left( { x-2 } \right) +iy \\ z+2=\left( { x+2 } \right) +iy \\ \frac { { z-2 } }{ { z+2 } } =\frac { { \left[ { \left( { x-2 } \right) +iy } \right] \left[ { \left( { x-2 } \right) -iy } \right] } }{ { \left[ { \left( { x+2 } \right) +iy } \right] \left[ { \left( { x+2 } \right) -iy } \right] } } \\ \frac { { z-2 } }{ { z+2 } } =\frac { { \left( { { x^{ 2 } }-4 } \right) +{ y^{ 2 } }+i\left( { yx+2y } \right) +i\left( { -yx+2y } \right) } }{ { { { \left( { x+2 } \right) }^{ 2 } }+{ y^{ 2 } } } } \\ \frac { { z-2 } }{ { z+2 } } =\frac { { \left( { { x^{ 2 } }+{ y^{ 2 } }-4 } \right) +i\left( { 4y } \right) } }{ { { { \left( { x+2 } \right) }^{ 2 } }+{ y^{ 2 } } } } \\ \left( { \frac { { z-2 } }{ { z+2 } } } \right) =\frac { { \frac { { 4y } }{ { { { \left( { x+2 } \right) }^{ 2 } }+{ y^{ 2 } } } } } }{ { \frac { { { x^{ 2 } }+{ y^{ 2 } }-4 } }{ { { { \left( { x+2 } \right) }^{ 2 } }+{ y^{ 2 } } } } } } =\frac { { 4y } }{ { { x^{ 2 } }+{ y^{ 2 } }-4 } } =\sqrt { 3 } \\ \left( { { x^{ 2 } }+{ y^{ 2 } }-4 } \right) =\frac { { 4y } }{ { \sqrt { 3 } } } \\ { x^{ 2 } }+{ y^{ 2 } }-\frac { { 4y } }{ { \sqrt { 3 } } } -4=0 \\ { x^{ 2 } }{ \left( { y-\frac { 2 }{ { \sqrt { 3 } } } } \right) ^{ 2 } }=4+\frac { 4 }{ 3 } \\ { x^{ 2 } }+{ \left( { \frac { { y-2 } }{ { \sqrt { 3 } } } } \right) ^{ 2 } }=\frac { { 16 } }{ 3 } \\ { x^{ 2 } }{ \left( { \frac { { y-2 } }{ { \sqrt { 3 } } } } \right) ^{ 2 } }={ \left( { \frac { 4 }{ { \sqrt { 3 } } } } \right) ^{ 2 } } \\ Length\, \, of\, \, circumference\, \, =2\pi \left( { \frac { 4 }{ { \sqrt { 3 } } } } \right) \\ =\frac { { 8\pi } }{ { \sqrt { 3 } } } \end{array}$

A circular flower bed is surrounded by a path $4 \,m$ wide. The diameter of the flower bed is $66 \,m$ . What is the area of this path? $(\pi = 3.14)$

1171128_1323358_ans_a09fb9480d35428194cfa8ca884019d3.jpg

Find the circumference of the circle with the following radius. (Take $\pi=22/7$ )
$28\ cm$

Circumference

$=2\pi r$

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

$=176cm^2.$

Hence, the answer is

$176cm^2.$

The sum of radii of two circles is $4.2\ \text{m}$ and the difference of their circumference is $13.4\ \text{m}$ , then find the circumference of both the circles.

1202653_1347264_ans_e52180dd3a1b434cb6875f303a95552e.jpg

Find the radius of the circle whose area is $38.5\ m^{2}$ .

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

$\Rightarrow 38.5=\dfrac { { 22 } }{ 7 } \times r\times r \\$

$\Rightarrow 38.5\times 7=22\times r\times r \\$

$\Rightarrow 269.5=22\times r\times r \\$

$\Rightarrow \dfrac { { 269.5 } }{ { 22 } } =r\times r \\$

$\Rightarrow 12.25=r\times r \\$

$Hence, \\ r=3.5$

A wheel has a diameter of $12$ cm. How many revolutions will it take to cover a distance of $26.4$ m?

Diameter of the wheel

$=12$ cm

Circumference of the wheel

$=2\pi r$

$=2\pi\cdot \dfrac{12}{2}$

$=12\pi$

$=12\times \dfrac{22}{7}$

$=\dfrac{264}{7}$ cm.

$\therefore$ Distance covered in

$1$ revolution

$=\dfrac{264}{7}$ cm

$\therefore$ No. of revolutions to cover a distance of

$26.4$ m

$=\dfrac{26.4m}{\dfrac{264cm}{7}}$

$=\dfrac{2640cm}{\dfrac{264}{7}cm}$

$=\dfrac{2640\times 7}{264}$

$=70$ revolutions.

If the circumference of the circular sheet is $154 cm$ , find its radius. Also find the area of the sheet. (Take $\pi = \frac { 22 } { 7 } )$

Circumference

$2\pi r=154$ cm

$\Rightarrow r=\dfrac{154}{2\times\dfrac{22}{7}}$

$=\dfrac{154\times7}{2\times 22}=24.5$ cm

Area of the circle

$\pi{r}^{2}=\dfrac{22}{7}\times24.5\times 24.5=1886.50{m}^{2}$

If the circumference of a circle is $264\ cm$ , then find its radius. Take $\pi=\dfrac{22}{7}$

Circumference of a circle

$= 2\pi r=264cm$

$\Rightarrow r=\dfrac { 264 }{ 2\pi } =\dfrac { 264\times 7 }{ 2\times 22 }$

$=\dfrac { 132\times 7 }{ 22 } =\dfrac { 66\times 7 }{ 11 }$

$=6\times 7=42cm$

Two crossroads, each $2$ m wide run at right angles through the center of a rectangular park of length $64$ m and breadth $38$ m ,such that each is parallel to one of the sides of the rectangle. Find the area of the roads. Also, find the area of the remaining part of the park.

Area of a rectangular park

$= l\times b = 64 \times 38 = 2432\ \text{m}^{2}$

Area of a horizontal road

$= l \times b = 64 \times 2 = 128\ \text{m}^{2}$

Area of a vertical road

$= l \times b = 38 \times 2 = 76\ \text{m}^{2}$

Common area

$= 2 \times 2 = 4\ \text{m}^{2}$

Area of the roads

$= 204 - 4 = 200\ \text{m}^{2}$

Area of the remaining part of the park

$= (2432 - 200)\ \text{m}^2 = 2232 \ \text{m}^{2}$

The sum of circumference and diameter of a circle is $145\text{ cm}$ , find the diameter of the circle.

Let

$r$ be the radius of the circle. So, diameter

$d$ will be equal to

$2r$ .

$\begin{aligned}{}2\pi r + 2r &= 145\quad\quad\quad\quad[\text{Given}]\\2r\left( {\frac{{22}}{7} + 1} \right) &= 145\\d \times \frac{{29}}{7} &= 145\\d &= 35\text{ cm}\end{aligned}$

Hence, the diameter of the circle is

$35\text{ cm}$ .

The minute hand of a circular clock is $15\ cm$ long. How far does the tip of the minute hand move in $1$ hour. (Take $\pi=3.14$ )

Distance moved by minute hand in

$1$ hour

$=$ Circumference of the circle made by minute hand.

Radius

$=r=15$ cm

Circumference

$=2\pi r=2\times 3.14\times 15$

$=\dfrac{314}{100}\times 30$

$=\dfrac{942}{10}=94.2$ cm

$\therefore$ Minute hand moves

$94.2$ cm in one hour.

The diameter of a bicycle wheel is 28 cm. What distance will it cover in 100 revolutions?

Circumference of the wheel is given as

$\pi d$ where

$d$ is the diameter of the wheel.

In one revolution, total distance travelled by the wheel is equal to its circumference.

So in

$100$ revolutions, total distance travelled by the wheel will be equal to

$100\pi d$ .

Let

$Z$ be the required distance.

$Z=100\times \dfrac{22}{7}\times 28$ cm

$=100\times 88$ cm

$=8800$ cm

$=8.8$ m

The diameters of the front and the rear wheels of a tractor are $63\ cm$ and $1.54\ m$ respectively. Calculate their areas and circumference.

Solution :-

Diameter of front wheel = 63 cm

Radius = 63/2 cm

Circumference of front wheel = 2πr

⇒ 2*22/7*63/2

⇒ 2772/14

Circumference of front wheel = 198 cm

Diameter of rear wheel = 1.54 m or 154 cm

Radius = 154/2

= 77 cm

Circumference of rear wheel = 2*22/7*77

⇒ 3388/7

Circumference of rear wheel = 484 cm

Area of front wheel = πr²

⇒ 22/7*63/2*63/2

⇒ 87318/28

Area of front wheel = 3118.5 cm²

Area of rear wheel = 22/7*77*77

⇒ 130438/7

Area of rear wheel = 18634 cm²

Find the area of each of the following parallelograms:

Area of

$11$ gm

$=2$ (area of triangle)

$=2\times \left(\dfrac{1}{2}\times 3\times 5\right)$

$=15cm^{2}$ .

1215296_1352362_ans_2cc22d6e50444446a681cb7de081eb71.jpg

The circumference of a circle is 39.6 cm . Find its area .

Given : Circumference of a circle

$=2\pi r=39.6$ cm

$\Rightarrow 2\times \dfrac{22}{7}r=39.6$

$\Rightarrow r=\dfrac{39.6\times 7}{44}=6.3$ cm

Area of a circle

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

$=\dfrac{22}{7}\times 6.3\times 6.3=124.74‬$ sq.cm

Two circles touch each other externally. The sum of their areas is $74\pi cm^{2}$ and the distance between their centre is $12 cm$ . Find the diameter of the circle.

R.E.F image

$Let A_{1}$ be the area of circle

$c_{1}$

$Let A_{1}$ be the area of circle

$c_{1}$

$Let r_{1}$ be the radius of circle

$c_{1}$

$Let r_{2}$ be the radius of circle

$c_{2}$

$r_{1}+_{2}=12cm.----(1)$

$A_{1}+A_{2}=74\pi cm^{2}$

$\pi r_{1}^{2}+\pi r_{2}^{2}=74\pi$

$r_{1}^{2}+r_{2}^{2}=74-----2$

Form

$(1),r_{1}=12-r_{2}----(3)$

Put (3) in (2)

$(12-r_{2})^{2}+r_{2}=74$

$144-24r_{2}+r_{2}^{2}=74$

$144-24r_{2}+2r_{2}^{2}=74$

$2r_{2}^{2}-24r^{2}+144-74=0$

$2r_{2}^{2}-24r_{2}+70=0$

$r_{2}^{2}-12r_{2}+35=0$

$r_{2}^{2}-5r_{2}-7r_{2}+35=0$

$r_{2}(r_{2}-5)-7(r_{2}-5)=0$

$(r_{2}-7)(r_{2}-5)=0$

$1f=r_{2}-7=0 r_{2}=-5=0$

$1fr_{2}=7 1fr_{2}=5$

$n=12-7 r_{1}=12-5$

$n=5cm r_{1}=7cm.$

$d_{1}=2r_{1},d_{2}=2r_{2} d_{1}r_{1},d_{2}=2r_{2}$

$d_{1}=10cm d_{1}=14cm$

$d_{2}=14cm d_{2}=10cm$

1208418_1348049_ans_dbeee06692164097854d3a94da0c2456.jpg

The inner circumference of a circular track [given figure] is $220$ m. The track is $7$ cm wide everywhere . Calculate the cost putting up a fence along the outer circle at the rate of Rs. $2$ per meter.

Let the outer and inner radii of the track be

$r\ m$ and

$R\ m$ respectively. Then,

Inner circumference

$2 \pi r=220\ m$

$\Rightarrow 2 \times \dfrac{22}{7} \times r =220$ and

$\Rightarrow r = 2 \dfrac{222 \times 7}{2 \times 22} =35 \ m$

Since the track is

$7\ m$ wide everywhere. Therefore,

$R=$ Outer radius

$=r+7 = (35+7)m =42\ m$

$\therefore$ Outer circumference

$=2 \pi R= 2 \times \dfrac{22}{7} \times 42\ m=264\ m$

Rate of fencing

$= Rs.\ 2$ per metre

$\therefore$ Total cost of fencing

$=(Circumference \times rate )= Rs. (264 \times 2) =Rs. 528$

From a circular sheet of radius $4\ cm$ , a circle of radius $3\ cm$ is removed. Find the area of the remaining sheet. (Take $\pi =3.14$ )

Area of the remaining part

$=$ Area of bigger circle

$-$ Area of the smaller circle.

Consider the bigger circle: Radius

$r=4\ cm$

Area

$=\pi{r}^{2}=3.14\times{4}^{2}=\dfrac{314}{100}\times 16=\dfrac{5024}{100}=50.24\ {cm}^{2}$

Consider the smaller circle: Radius

$r=3\ cm$

Area

$=\pi{r}^{2}=3.14\times{3}^{2}=\dfrac{314}{100}\times 9=\dfrac{2826}{100}=28.26\ {cm}^{2}$

Area of the remaining part

$=$ Area of bigger circle

$-$ Area of the smaller circle.

$=50.24\ {cm}^{2}-28.26\ {cm}^{2}=21.98\ {cm}^{2}$

$\therefore$ Area of remaining sheet is

$21.98\ {cm}^{2}$

The centre of a circle is $C\left(4,5\right)$ and one end of a diameter is $A\left(3,2\right)$ , find the coordinates of the other end.

Let the other end of the diameter of the given circle be

$B\left(\alpha,\beta\right)$ whose one end is

$A\left(3,2\right)$

Midpoint of

$AB=\left(\dfrac{\alpha+3}{2},\,\dfrac{\beta+2}{2}\right)$

Since the centre

$C\left(4,5\right)$ of the circle is the mid-point of

$AB$ ,

$\dfrac{\alpha+3}{2}=4$ and

$\dfrac{\beta+2}{2}=5$

$\Rightarrow\,\alpha+3=8$ and

$\beta+2=10$

$\Rightarrow\,\alpha=8-3=5$ and

$\beta=10+2=8$

$\therefore\,$ The coordinates of the other end of the diameter are

$\left(5,8\right)$

Find the area of the parallelogram shaped field having length of the base $21\ m$ and height $15\ m$ .

Area of parallelogram =

$b\times h$

$= 21\times 15$

$= 315 m^{2}$ .

The outer circumference of a circle track is $196.68\ m$ and the track is everywhere $6\ m$ wide. Find the cost of levelling the track at the rate of $25$ paise per $m^{2}$ .[Take $\pi=3.14$ ]

Let

$R$ be the radius of outer track and

$r$ be the radius of inner track

Given:

$2\pi R= 196.68$

$\implies R =\dfrac{196.68}{2\times 3.14}= 31~m$

Now,

$r=R-6=25~m$

Area of Track

$=$ Area of outer circle

$-$ Area of inner circle

$=\pi [R^{2}-r^{2}]$

$=\pi [(31)^2-(25)^2]$

$=\pi (961-625)$

$=3.14(336)$

$=1055.04\text{ m}^2$

Now, the cost of levelling

$1\text{ m}^2$ is

$25$ paise

$\therefore$ Total cost

$= 1055.04\times \dfrac{25}{100}$

$=\text{Rs }263.76$

1194463_1353696_ans_8c15ce452ec04c7b9a143db5ee0d7f6d.png

The circumference of the front wheel of a cart is 3 m and that of the back is 4 m. What is the distance travelled by the cart, when the front wheel makes five more revolutions than the rear wheel?

Given that, the circumference of the front wheel of a cart is

$3\ m$ and that of the back wheel is

$4\ m$ .

Also, the front wheel makes five more revolutions than the back.

To find out: The distance travelled by the cart.

Let the distance be

$x \ m$ .

$\therefore \ \dfrac{x}{3}-\dfrac{x}{4}=5$

$\Rightarrow \dfrac{4x-3x}{12}=5$

$\Rightarrow x=5\times 12=60 \ m$

Hence, the distance travelled by the cart is

$60\ m$ .

The inner circumference of a circular race track is $44 m$ less than the outer circumference. If the outer circumference is $396 m$ , then find the width of the track.

Outer circumference

$= 396 m$

Inner circumference

$= (396-44)m = 352 m$

Let outer radius be

$R$ and inner radius be

$r$ .

$\therefore2 \pi R = 396$

$\Rightarrow 2\times \dfrac{22}{7} \times R = 396$

$\Rightarrow R=\dfrac{396\times7}{2\times22}$

$\Rightarrow R=63$

and,

$2 \pi r = 352$

$\Rightarrow 2\times \dfrac{22}{7} \times r = 352$

$\Rightarrow r=\dfrac{352\times7}{2\times22}$

$\Rightarrow r=56$

Therefore, width of track =

$(63-56)m=7$

$m$ .

A chord of a circle of radius $10\ cm$ subtends a right angle at the centre. Find the area of corresponding minor segment $(\pi=3.14)$

Area of sector OAPB

$=\,\dfrac{\theta}{360}\times \pi\times r^2$

$=\dfrac{90}{360}\times \pi r^{2}$

$=\dfrac{1}{4}\times 3.14\times 10\times 10$

$=78.5cm^{2}$

$Area of \bigtriangleup OAB=\dfrac{1}{2} r^2sin\theta =\dfrac{1}{2}\times 10\times 10$

$= 50cm^{2}$

Area of minor segment = Area of sector

$OAPQ$ - area of

$\bigtriangleup (AOB)=78.5 cm^{2}-50 cm^{2}$

$=28.5cm^{2}$

1210764_1378418_ans_5e5293df38c84a6597c4df901c123220.png

A path $3\ m$ wide runs around the inside of a square park of side $60\ m$ . Find the area of the path.

REF.image

Given ABCD is a park of [side = 6cm ]

now

Portion outside (EFGH) is a running path of width = 3m

area of running path = area of ABCD - area of EFGH

length of side EFGH

$= 60-6=54 cm$

Then

area of running path

$= (60)^{2} - (54)^{2}$

$Area = 3600 - 2916$

$Area = 684cm^{2}$

1215301_1368380_ans_63b6820997b448d89f1043a661fb3127.jpg

What will be the width of the following circular race track with inner diameter 10 cm and outer diameter 14 cm ?

2040856_1363259_ans_2f257f7b905245148c8f924de491579c.jpg

The adjacent sides of a parallelogram $36\ cm$ and $27\ cm$ in length. If the perpendicular distance between the shorter sides is $12\ cm$ , find the distance between the longer sides.

REF.Image

Area of parallelogram

= base

$\times$ height

$= 27 \times 12 cm$

$324 cm^{2}$

Area =

$b\times h$

$324=36\times h$

$\frac{324}{36}=h$ h= 9 cm

1210748_1370913_ans_9b8e496621864e03bfd833f1b198b60c.jpg

The side of a square is 10 cm. Find the area of inscribed and circumscribed circles.

Side of square

$= 10\text{ cm}$

Radius of inscribed circle =

$\dfrac{r}{2}=5\text{ cm}$

Area of inscribed circle =

$\pi r^{2}=\dfrac{22}{7}\times 25=78.5\text{ cm}^{2}$

Radius of circumscribed circle

$=$ Diagonal of square / 2

$=\dfrac{10\sqrt{2}}{2}=5\sqrt{2}$

Area of circumscribed circle

$=\pi r^{2} = \dfrac{22}{7} \times 50=157$

$\text{cm}^{2}$

1282664_1374199_ans_b64c3f012e914c3b99c04cf184c47853.png

Find the height of a parallelogram whose area is 54 c ${ m }^{ 2 }$ and the base is 15 cm.

Area of a parallelogram

$=$ base

$\times$ height.

$\Rightarrow 54=15\times$ height

$\Rightarrow$ height

$=\dfrac{54}{15}=\dfrac{18}{5}=3.6$ cm

$\therefore$ height of the parallelogram

$=3.6$ cm

Area of circle is given below find their diameters.
$12474\ sq\ cm$

Area of a circle

$\pi{r}^{2}$

$\Rightarrow \pi{r}^{2}=12474$ sq.cm

$\Rightarrow \dfrac{22}{7}\times {r}^{2}=12474$ sq.cm

$\Rightarrow {r}^{2}=12474\times \dfrac{7}{22}=3,969‬$ cm

$\Rightarrow r=\sqrt{3,969}=63‬$ cm.

$\therefore$ Diameter

$=2r=2\times 63=126$ cm.

The diameter of a wheel is $28\ cm.$ How many revolutions will it make to travel $352\ m$ ?

Diameter of a wheel

$=28$ cm

Radius of a wheel

$=\dfrac{28}{2}=14$ cm

Distance travelled to make one revolution

$=$ Circumference of the wheel

$=2\pi r$

$2\pi\times r=2\times\dfrac{22}{7}\times 14=88$ cm

To make one revolution the wheel has to travel

$=88$ cm

Number of revolutions to travel

$352$ m is

$=\dfrac{Total \,\,distance \,\, travelled}{Distance\,\, travelled\,\, to\,\, make\,\, one\,\, revolution}=\dfrac{352}{88}=4$

The area of circle if the circumference is $20\pi$ is ____

Given:

The circumference of circle is

$20\pi$

$\Rightarrow 2\pi r=20\pi$

$\Rightarrow r=10\ cm$

Area of circle

$=\pi r^2$

$=\pi (10)^2$

$=314.28\ cm^2$

What is the diameter of a circle whose area is equal to the sum of the areas two circles of radii $40\ cm$ and $9\ cm$ ?

Area of the circle

$=$ sum of area of two circles

$\pi{R}^{2}=\pi\times{40}^{2}+\pi\times {9}^{2}$

$\Rightarrow {R}^{2}=1600+81=1681$

$\therefore R=\sqrt{1681}=41$ cm

$\therefore$ Diameter of given circle

$=41\times 2=82$ cm

Find the radius of a circle whose circumference is $39.6\ cm$

Circumference of a circle

$=2\pi r=39.6$ cm

$\Rightarrow 2\times \dfrac{22}{7} r=39.6$

$\Rightarrow r=\dfrac{39.6\times 7}{2\times 22}= 6.3‬$ cm

The circumference of a circle exceeds the diameter by $16.8$ cm. Find the radius of the circle.

Let the radius of the circle be

$r$ cm.

Diameter

$= 2r$ cm and Circumference =

$2\pi r$ cm

Therefore,

$2\pi r=2r+16.8\\$

$2\times \dfrac{22}{7} \times r=2r+16.8\\$

$44r=14r+16.8\times 7\\$

$30r=117.6\\$

$r=3.92\\$

Hence, radius

$= 3.92$ cm

If the area of a circle is $154 \mathrm{cm}^{2}$ , what is the perimeter of the circle?

Let the radius of a circle be

$r\ \mathrm{cm}$

Given Area of the circle is

$154\ \mathrm{cm^2}$

$\implies \pi r^2=154$

$\implies r^2=154\times \dfrac{7}{22}=7^2$

$\implies r=7$

Perimeter of the circle is

$2\pi r=2\times \dfrac{22}{7}\times 7=44\ \mathrm{cm}$

In the given figure ABC is a right triangle and right angled at B such that $\angle BCA=2\angle BAC$ .
Show that hypotenuse AC = 2BC.
(Hint : Produce CB to a point D that BC = BD)

$\triangle{ABD}$ and

$\triangle{ABC}$ we have

$BD = BC$

$AB = AB$ [Common]

$\angle{ABD}=\angle{ABC}={90}^{\circ}$

$\therefore$ By SAS criterion of congruence we get

$\triangle{ABD}\cong \triangle{ABC}$

$\Rightarrow AD = AC$ and

$\angle{DAB}=\angle{CAB}$ [By CPCT]

$\Rightarrow AD = AC$ and

$\angle{DAB}=x$ [

$\therefore\angle{CAB}=x$ ]

Now,

$\angle{DAC}=\angle{DAB}+\angle{CAB}= x + x = 2x$

$\therefore \angle{DAC}=\angle{ACD}$

$\Rightarrow DC = AD$ [Side Opposite to equal angles]

$\Rightarrow 2BC = AD$ since

$DC = 2BC$

$\Rightarrow 2BC = AC$ Since

$AD = AC$

Hence proved.

1303696_1378622_ans_58b27270328f4a508e0cbbf5e662a8e0.png

Find the diameter of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters $36$ m and $20$ m.

Given, two circles with diameters

$36m$ &

$20m$ .

So, radius of first circle

$=\dfrac{36}{2}\ m=18\ m$

Circumference

$=2\pi r=2\pi\times 18\ m=36\pi \ m$

And, radius of second circle

$=\dfrac{20}{2}\ m=10\ m$

Circumference

$=2\pi r=2\pi\times 10\ m=20\pi \ m$

Sum of circumferences of two given circles

$=(36\pi +20\pi)\ m =56\pi\ m$

Let

$R$ be radius of the required circle

Circumference

$=2\pi R$

ATQ,

$2\pi R=56\pi$

$\Rightarrow R=28$

So, diameter

$D=2\times28\ m=56\ m$

$\therefore$ Diameter of required circle is

$56\ m$

Find the radius of a circle whose circumference is $44$ cm

Circumference of the circle

$=2\pi r$

$2\times\dfrac{22}{7}\times r=44$

$\Rightarrow r=\dfrac{44\times 7}{2\times 22}=7$ cm

$\therefore$ radius of the circle is

$7$ cm

A circular grass plot of land $42\ m$ is diameter has a path $3.5\ m$ wide running round its outside. Find the cost of gravelling the path at $Rs. 4$ per square metre.

Area of inner circle

$=\pi{r}^{2}=\dfrac{22}{7}\times 21\times 21=22\times 3\times 21=1386$ sq.m

Area of outer circle

$=\pi{r}^{2}=\dfrac{22}{7}\times 24.5\times 24.5=22\times 24.5\times 2.5=1886$ sq.m

Area of path

$=1886-1386=500.5$ sq.m

Cost of gravelling

$=500.5\times 4=$ Rs.

$2002$

1304264_1378676_ans_ea7fb39e53a3428bb718b1dc57c33310.PNG

Measure of an arc of a circle is ${80^ \circ }$ and its radius is $18$ cm. Find the length of the arc. $(\pi = 3.14)$

Length of an arc

$=\dfrac{\theta }{360}\times 2\pi r$

$=\dfrac{80^o}{360^o}\times 2\times 3.14\times 18$

$=8\times 3.14\ cm$

$= 25.12 \ cm\$

Find the area of the circle in which a square of area $64\ cm^{2}$ is inscribed. (use $\pi=3.14$ ).

Given the area of the square is

$64$ cm

$^2$ .

Then its side will be of length

$8$ cm.

Then its diagonal will be of length is

$8\sqrt{2}$ cm.

Since the square is inscribed in a circle then the diagonal of the square will be the diameter of the circle.

Let

$r$ be the radius of the circle.

Then

$2r=8\sqrt{2}$

or,

$r=4\sqrt{2}$ .

Then the area of the circle

$=\pi r^2=3.14\times (4\sqrt 2)^2$

$=3.14\times 32$

$=100.48$ cm

$^2$ .

Find the area of a parallelogram with base $32$ $cm$ and height $16.5$ $\mathrm{cm}$

Base of parallelogram

$=32cm$

Height of parallelogram

$=16.5cm$

Area of parallelogram is

$= 32\times 16.5=528cm^2$

The area of triangle with sides $5,12,13$ is

Sides of triangle are

$5,12,13$

Semi perimeter is

$s=\dfrac{a+b+c}2=\dfrac{5+12+13}2=15$

Area of triangle is

$\sqrt {s(s-a)(s-b)(s-c)}\\\sqrt {15(10)(3)(2)}=\sqrt {900}=30$

The area of sector of circle of radius $14cm$ subtending an angle of $30^{\circ}$ is

The angle subtended at center of circle is

$30^{\circ}$

The radius of circle is

$14cm$

$\theta$ is the angle subtended at the center of circle having radius

$r$ then area of the sector is given as,

$A=\dfrac{\theta}{360^o}\times \pi r^2$

According to the given condition,the Area of sector is given as

$A=\dfrac{30}{360}\times \dfrac{22}7 \times 14\times 14\\=\dfrac 1{12}\times 22\times 2\times 14\\=\dfrac 13\times 11\times 14\\=\dfrac{154}3\\=51.3\ cm^2$

$PQRS$ is a parallelogram. $QM$ is the height from $Q$ to $SR$ and $QN$ is the height from $Q$ to $PS$ . If $SR=12 cm$ and $QM=7.6 cm$ . Find:
the area of the parallelogram PQRS
$QN$ ,if $PS=8cm$

Given

$SR=12cm$ ,

$QM=7.6cm$

Area of parallelogram is

$SR\times QM=12\times 7.6=91.2cm^2$

Area is also given as

$=QN\times PS\\QN=\dfrac{91.2}{8}=11.4cm$

1301583_1401136_ans_f3ffd4a327b94357b605794787d96005.jpg

Find area of parallelogram whose base is 24cm and height is 5cm

Base of parallelogram

$b=24\ cm$

Height of parallelogram

$h=5\ cm$

Area of parallelogram is

$A= bh$

$=24\times 5=120\ cm^2$

Find area of triangle If base is $4cm$ an height is $5cm$

Base is

$4cm$

Height is

$5cm$

Area of triangle is

$\dfrac 12\times 4\times 5=10cm^2$

Find the ring formed by two concentric circles of radius $14cm,7cm$

Area of ring is given by

$\pi(R^2-r^2)\\\dfrac{2}{7}\times (14^2-7^2)\\\dfrac{22}7\times (196-49)\\\dfrac{22}{7}\times 147\\22\times 21=462cm^2$

The circumference of a circle is $396$ cm. Find its area.

circumference of a circle

$=2\pi r=396\ cm$

$r=\dfrac{396}{2\pi}$

Area of the circle

$=\pi{r}^{2}=\pi\times\dfrac{396}{2\pi}\times \dfrac{396}{2\pi}$

$=198\times \dfrac{198}{\dfrac{22}{7}}$

$=198\times 198\times\dfrac{7}{22}=12,474$ sq.cm

Establish the formula for area and circumference of circle.

Circumference of circle is

$2\pi r$

Area of circle is

$\pi r^2$

Find the diameter of the circle if its. Circumference is $62.8\ cm\ (\pi=3.14)$

Let the diameter be

$d$

As we know that

Circumference of the circle with diameter

$d$ is

$\pi d$

Given circumference is

$62.8\ \mathrm{cm}$

$\implies \pi d=62.8$

$\implies d=\dfrac{62.8}{3.14}=20\ \mathrm{cm}$

So diameter of the circle is

$20\ \mathrm{cm}$

The diameter of the wheel of a cart is $1.05\ m$ . How much distance will cart cover in $1000$ rotations of the wheel?

$1$ rotation is equal to the circumference

Now, circumference of a circle =

$2\pi r$ or

$\pi d$ .

Circumference

$=\pi \times 1.05$

$=3.29867$ m

Now, as cart covered

$1000$ rotations, the distance is

$3.29867\times 1000=3298.67$ m

(approximately

$= 3299$ m)

Area of circular park of radius $7m$ is

Radius is

$7m$

Area is

$\pi r^2=\dfrac {22}7\times 7^2\\22\times 7=154cm^2$

The diameter of a wheel is $70$ cm. find the distance covered by wheel in $5$ revolution.

Distance covered will be

$=$ Number of revolutions

$\times$ Circumference

Radius of wheel

$=\dfrac{70}{2}= 35$ cm

Distance covered in

$5$ revolutions will be

$D=5 \times 2\pi \times 5=50\pi=50\times\dfrac{22}{7}=157.14$ m

Find the circumference of a circle whose radius is 35 cm.

Given

Radius

$(r)=35 cm$

$\pi = 22/7.$

circumference of a circle

$=2\pi r$

$=2.\dfrac{22}{7}.35$

$= 44\times5$

$= 220.cm$

$\therefore$ circumference of a circle is

$220 cm.$

Work out the area and the circumference of each circle. Give your answers correct to one decimal place (1 d.p)
Radius = 8 cm

Circumference

$=2\pi r$

$=2\times 3.14\times 8$

$=50.24\cong 50.2$ cm

Area

$=\pi r^2=3.14\times (8)^2$

$=200.96cm^2$ .

1199916_1445247_ans_1fad626ff7e44767bd30aedc259f71a6.jpg

The area of triangle with base $5cm$ and Height $6cm$ is

The height of triangle is

$6cm$

The base of triangle is

$5cm$

The area of triangle is

$\dfrac 12 \times b\times h\\$

$\dfrac 12\times 6\times 5\\15cm^2$

If the circumference of a circle is $176cm$ , find its radius.

Circumference of a circle

$=2\pi r$

$\Rightarrow 176=2\times\dfrac{22}{7}r$

$\Rightarrow r=\dfrac{176\times 7}{2\times 22}=28$ cm

$\therefore$ Radius of the circle

$=28$ cm

Find the circumference of a circle whose radius is 15.4 cm.

The radius of the circle

$= 15.4 cm$

we know that

circumference of the circle

$=2\pi r$

$=2.\pi \times15\times4$

$=2\times\dfrac{22}{7}\times15\times4$

$= 96.8 cm$

$\therefore$ circumference of the circle

$= 96.8 cm$

There is a circular pond and foot-path runs along its boundary. A man walks around it, exactly once, keeping close to the edge. If his step is $55$ cm long and he takes exactly $400$ steps to go round the pond. What is the diameter of the pond?

Let the radius of the pond be 'r'. since man takes exactly 400 step of 55 cm to go around the pond.

Hence, the circumference of the pond =

$55\times 400$

$=22,000 cm$

$=\frac{22000}{100}$

$= 220 m.$

circumference of the pond is

$220m.$

circumference of the circular pond

$=2\pi r$

$220=2\pi r$

$\Rightarrow 220=2\times \dfrac{22}{7}\times r$

$\Rightarrow r=220\times \dfrac{7}{44}$

$\Rightarrow r=35m$

$\therefore$ Diameter of the pond

$=2\times r$

$=2\times 35$

$= 70 m.$

The diameter of a bicycle wheel is $28$ cm. What distance will it cover in $100$ revolutions?

Diameter

$= 28$ cm.

radius

$= \dfrac {d} {2} = \dfrac {28}{2} = 14$ cm

circumference = 1 revolution

$\therefore$ circumference =

$2\pi r$

$=2.\dfrac{22}{7}.14$

$= 88$ cm

The distance will cover in 100 revolutions =

$100\times 88$

$= 8800$ cm

$= 88$ m.

From circular card sheet of radius $14$ cm, a square of $4$ cm is removed as shown in the adjoining figure.Find the area of the remaining sheet.
( $\pi = \cfrac{{22}}{7}$ )

Area of shaded regon

$=\pi(14)^2-(side)^2\\=\cfrac{22}{7}\times14\times14-(4)^2\\=600cm^2$

1380520_1525840_ans_33f3e85bb7b74626900cfeb5d6dd9fd3.png

If the circumference of the two circles be $396$ cm and $374$ cm, find the area of the in-between region.

Since

$396>374$

Outer circle has circumference

$=396$ cm and inner circle has circumference

$=374$ cm

Circumference of outer circle

$=2\pi r=396$ cm

$\Rightarrow 2\times \dfrac{22}{7}r=396$

$\therefore r=\dfrac{396}{2}\times\dfrac{7}{22}=63$ cm

Area of outer circle

$=\pi{r}^{2}=\dfrac{22}{7}\times 63\times 63=12,474{cm}^{2}$

Circumference of inner circle

$=2\pi r=374$ cm

$\Rightarrow 2\times \dfrac{22}{7}r=374$

$\therefore r=\dfrac{374}{2}\times\dfrac{7}{22}=59.5‬$ cm

Area of inner circle

$=\pi{r}^{2}=\dfrac{22}{7}\times 59.5\times 59.5=3,540.25‬{cm}^{2}$

$\therefore$ Area of the shaded region

$=$ Area of outer circle

$-$ Area of inner circle

$=12,474{cm}^{2}-3,540.25‬{cm}^{2}=8,933.75‬‬{cm}^{2}$

Find the circumference of circle whose area is 16 times the area of the circle with diameter 7 cm.

$\textbf{Hint: Use area of circle as}$ $\pi {{r}^{2}}$ and circumference as $2\pi r$

$\textbf{Step-1: Finding radius of required circle.}$

$\text{Diameter of given circle = 7cm, Radius = 7/2 cm}$

$\text{Area of required circle is }$

$\Rightarrow \dfrac{49}{4}\pi \times 16$

$\Rightarrow 196\pi c{{m}^{2}}$

$\textbf{Step-2: Radius of required circle is -}$

$Radius=\sqrt{\dfrac{Area}{\pi }}=\sqrt{\dfrac{196\pi }{\pi }}$

$\Rightarrow \sqrt{196}=14cm$

$Radius=\sqrt{\dfrac{Area}{\pi }}=\sqrt{\dfrac{196\pi }{\pi }}$

$\Rightarrow \sqrt{196}=14cm$

$\textbf{Step-3: Circumference of required circle is –}$

$\therefore Circumference=2\pi r=2\times \dfrac{22}{7}\times 14$

$\Rightarrow 88cm$

$\textbf{Final Answer: Therefore, the circumference of cicle whose area is 16 times the area of}$

$\textbf{circle with diameter as 7 cm is 88 cm.}$

The inner circumference of a circular track is $264\ m$ and the width of the track is $7\ m$ . Find :
the cost of fencing the outer circumference at the rate of $Rs\ 50\ per\ m$ .

Let

$r$ be the radius of the inner track.

Inner circumference of track

$= 2 \pi r = 264 \; m \quad \left( \text{Given} \right)$

$\Rightarrow 2 \times \cfrac{22}{7} \times r = 264$

$\Rightarrow r = 264 \times \cfrac{7}{2 \times 22}$

$\Rightarrow r = 42 \; m$

Width of track

$\left( w \right) = 7 \; m \quad \left( \text{Given} \right)$

$\therefore$ Outer radius

$\left( R \right) = r + w$

$= 42 + 7 = 49 \; m$

Therefore,

Outer circumeference of track

$= 2 \pi R$

$= 2 \times \cfrac{22}{7} \times 49 \\ = 308 \; m$

Cost of fencing

$= Rs. 50$ per meter

Therefore,

Cost of fencing the outer circumference of track

$= 50 \times 308 = Rs. 15400$

The inner circumference of a circular track is $264\ m$ and the width of the track is $7\ m$ . Find :
the radius of the inner track.

Let

$r$ be the radius of inner track.

Inner circumference of track

$= 2 \pi r = 264 \; m \quad \left( \text{Given} \right)$

$\Rightarrow 2 \times \cfrac{22}{7} \times r = 264$

$\Rightarrow r = 264 \times \cfrac{7}{2 \times 22}$

$\Rightarrow r = 42 \; m$

From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed (as shown in the adjoining figure). Find the area of the remaining sheet.

Area of sheet

$=\pi r^2$

$=\dfrac{22}{7}\times 14\times 14$

$=616 \text{ cm}^2$

Area of

$2$ small circles

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

$=2 \pi r^2$

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

$=77\text{ cm}^2$

Area of the rectangle

$=l \times b$

$=3\times 1$

$=3 \text{ cm}^2$

So, remaining area

$=616-(77+3)$

$=616-80$

$=536 \text{ cm}^2$

A copper wire when bent in the form of a square encloses an area of $484\ cm^2$ . If the same wire is bent into the form of a circle, find the area enclosed by it.

We have, Area of the square

$=a^2=484\ cm^2$

$\therefore$ Side of the square

$=\sqrt{484}\ cm=22\ cm$

So, Perimeter of the square

$=4(side)=(4\times 22)\ cm=88\ cm$

Let

$r$ be the radius of the circle. Then, according to equation,

Circumference of the circle = Perimeter of the square

$\Rightarrow 2\pi r=88 \Rightarrow 2\times \dfrac{22}{7}\times r=88$

$\Rightarrow r=\dfrac{88\times 7}{2\times 22}\Rightarrow r=14\ cm$

$\therefore$ Area of the circle

$=\pi r^2=\left\{ \dfrac{22}{7}\times (14)^2 \right\} \ cm^2=616\ cm^2$

The length of two adjacent sides of a parallelogram are $17$ cm and $12$ cm. one of its diagonals is $25$ cm long. Find the area of the parallelogram. Also find the length of altitude from vertex to the side of length $12$ cm.

Area of triangle

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

Area of parallelogram

$=2\times$ Area of the triangle formed by the adjacent sides and the diagonal

Sides are

$17cm,12cm,25cm$

Semi perimeter(s)

$=\dfrac{12+17+25}{2}=\dfrac{54}{2}=27$ cm

Area

$=\sqrt{27(27-12)(27-17)(27-25)}$

$\Rightarrow A=\sqrt{27\cdot 15\cdot 10\cdot 2}$

$\Rightarrow A=\sqrt{3\times 3\times 3\times 3\times 5\times 2\times 5\times 2}$

$\Rightarrow A=3\times 3\times 5\times 2$

$\Rightarrow A=90 \,\,\mathrm {cm}^2$

$\therefore$ Area of parallelogram

$=2(90)=180cm^2$ .

Let the height of the parallelogram be

$h$

Given base

$b=12 \,\,\mathrm {cm}$

$A=b\times h$

$\Rightarrow 180=12\times h$

$\Rightarrow \dfrac{180}{12}=h$

$\therefore h=15\,\,\mathrm {cm}$

$\therefore$ Length of altitude

$=h=15 cm$ .

Find the area in square meter of a parallelogram , whose base is $140\ cm$ and altitude is $20\ cm$ .

$Base =140\ cm=1.4\ m$

$Altitude =20\ cm=0.2\ m$

Area of parallelogram

$=b\times h$

$=1.4\times 0.2$

$=0.28\ m^2$

Hence, this is the answer.

Find the appropriate radius of a circle whose circumference is 78mm

1311165_1536165_ans_1550ab8847cd40f093d5e08e6445a169.jpeg

The inner circumference of a circular track is $264\ m$ and the width of the track is $7\ m$ . Find the length of the outer circumference.

Let

$r$ be the radius of the inner track.

Inner circumference of track

$= 2 \pi r = 264 \; m \quad \left( \text{Given} \right)$

$\Rightarrow 2 \times \cfrac{22}{7} \times r = 264$

$\Rightarrow r = 264 \times \cfrac{7}{2 \times 22}$

$\Rightarrow r = 42 \; m$

Width of track

$\left( w \right) = 7 \; m \quad \left( \text{Given} \right)$

$\therefore$ Outer radius

$\left( R \right) = r + w$

$= 42 + 7 = 49 \; m$

Therefore,

Length of outer circumference of track

$= 2 \pi R$

$= 2 \times \cfrac{22}{7} \times 49 \\ = 308 \; m$

$PQRS$ is a parallelogram with altitude $PM$ on base $SR$ and $SN$ is the altitude on base $QR.$ If $SR=8.2\ cm,\ PM=4.8\ cm$ and $SN=6\ cm.$ What is $PS$ ?

Given:

$SR=8.2\ cm$

$PM=4.\ cm$

$SN=6\ cm$

Let us denote area of parallelogram

$PQRS$ by

$A.$

By considering base

$SR$ and altitude

$PM,$

$A=PM\times SR$

$=8.2\times 4.8$

$=39.36\ cm^2$

Now, by considering base

$QR$ and altitude

$SN,$

$A=SN\times QR$

$\Rightarrow 39.36=6\times QR$

$\Rightarrow QR=6.56\ cm$

As we know that the opposite sides of a parallelogram are equal so,

$QR=PS=6.56\ cm$

The inner circumference of a circular track is $264\ m$ and the width of the track is $7\ m$ . Find :
the radius of the outer circumference.

Let

$r$ be the radius of the inner track.

Inner circumference of track

$= 2 \pi r = 264 \; m \quad \left( \text{Given} \right)$

$\Rightarrow 2 \times \cfrac{22}{7} \times r = 264$

$\Rightarrow r = 264 \times \cfrac{7}{2 \times 22}$

$\Rightarrow r = 42 \; m$

Width of track

$\left( w \right) = 7 \; m \quad \left( \text{Given} \right)$

$\therefore$ Outer radius

$\left( R \right) = r + w$

$= 42 + 7 = 49 \; m$

The adjacent sides of a parallelogram are $10\ m$ and $8\ m$ . If the distance between the longer sides is $4\ m$ , find the distance between the shorter sides.

Area of the parallelogram

$=$ base

$\times$ height

$= 10 \times 4= 40$ sq.m

Now the shorter side

$= 8$ m

Area of the parallelogram

$=8\times h$

$\Rightarrow 40 = 8h$

$\Rightarrow h =\dfrac{40}{8}=5$ m

$\therefore$ the distance between shorter side is

$5$ m

1287747_1447833_ans_8c6677124fe44bfbb50ea99f583acbea.png

The diameter of the moon is approximately on fourth of the diameter of the earth. Find the ratio of their surface areas.

In the adjoining figure, $\Delta ABC$ is an isosceles triangle in which $AB=AC$ . if $BM \bot AC\,\,and\,\,\,CN \bot AB$ , prove that:
(i) $\Delta BMC \cong \Delta CNB$
(ii) $BM = CN$

1230336_1462470_ans_527e36e2280741fba91a5a2557c4f772.jpg

In fig. $APB$ and $AQP$ are semi-circles, and $A O = \mathrm { OB } .$ area of the shaded region. [Use $\pi = \dfrac { 22 } { 7 } ]$

1313301_1465627_ans_21e9c53305684b998a6a519880d6a777.JPG

The base of a parallelogram measures $1$ m $60$ cm and its height is $75$ cm . Find its area in $\text{m}^2$ .

1311585_1466407_ans_3f25fda2c14642b5abc84d3dc522c49e.jpeg

Find the perimeter of a square whose area is $196\ \text{ cm }^{ 2 }$ .

Area of square,

$A$ of side

$a$ is given by:

$A=a^2$

$a^2=196\ \text{cm}^2$ (Given)

$a=\sqrt{196}\ \text{cm}$

$a=14\ \text{cm}$

Perimeter of square

$P$ of side

$a$ is given by:

$P=4a$

$P=4\times 14\ \text{cm}$

$P=56\ \text{cm}$

Hence, the perimeter is

$56\ \text{cm}$ .

Find the circumference of a circle whose radius is -
$35 cm$

1311286_1504974_ans_b606c7ff825047e8b0e60aef4609e96c.jpeg

Find the area of circle $x^{2}+y^{2}=25$ .

1313515_1470801_ans_f47f39b40c2248809fba814c5a629a66.JPG

Find the area of a parallelogram with base $5\ cm$ and altitude $4.2\ cm$ .

As we know that,

Area of parallelogram

$=$ Base

$\times$ altitude

Given:-

Base of parallelogram

$= 5 \; cm$

Altitude

$= 4.2 \; cm$

Therefore,

Area of parallelogram

$= 5 \times 4.2 = 21 \; {cm}^{2}$

Find the area of a triangle with vertices A(3,0), B(7,0) and C(8,4)

1297850_1482462_ans_90146569766a456e89e7febf76335dfb.jpg

A wire is in the shape of a square with each side as 16.5 cm is bent in the shape of a circle. FInd the circumference and diameter of the circle.

1319048_1586226_ans_80ec734551d748aa97ac83642ee933e6.jpg

The given figure is formed by there semi-circle,in which OA=OB=14cm, then area of the figure is

1300266_1608044_ans_5bd4087e74be43eabf18be7b71a5178f.jpg

In the given figure, the side of square is $28\mathrm { cm }$ and radius of each circle is half of the length of the side of the square, where $O$ and $O ^ { \prime }$ are centres of the circles. Find the area of shaded region.

Side of square

$28cm$

The area of square

$28\times 28=784cm^2$

The radius of circle is

$14cm$

The area of circle is

$=\dfrac {22}7\times 14\times 14$

$\Rightarrow 28\times 14=392cm^2$

The overlapped area is

$\dfrac {90}{360}\times 392=\dfrac 14\times 392=98cm^2$

The resultant area of circle is

$392-98=294cm^2$

The total area is

$784+294+294=1372cm^2$

Prove that the area of a circular path of uniform width h surrounding a circular region of r is $\pi h (2r+h)$ .

Area of circular region

=area of outer-area of inner circle

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

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

$=\pi \left( { h }^{ 2 }+2rh \right)$

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

1236283_1525677_ans_daab9e6f84044ac498e06c3abc312ef8.png

Find the circumference of a circle whose radius is $4.25\ cm$ .

Given,

Radius of the circle

$=4.25\ cm$

Circumference

$=2\pi r$

$=2\times \dfrac{22}{7}\times 4.25\ cm$

$=2\times \dfrac{22}{7}\times \dfrac{425}{100}\ cm$

$=\dfrac{11}{7}\times 17\ cm$

$=\dfrac{187}{7}\ cm$

$=26.71\ cm$

Hence, the circumference is

$26.71\ cm$ .

Find the area of circle of a diameter is $4.2$ .

Given

Diameter of circle

$= 4.2$

radius

$= \dfrac{4.2}{2}=2.1$

Area of circle =

$\pi r^{2}$

$=\dfrac{22}{7}\times (2.1)^{2}$

$=\dfrac{22}{7}\times 2.1\times 2.1$

$=13.86$

If 1 cm represent $5$ m, what would be an area of $6$ square cm, represent ?

Given

$1\ cm$ represents

$5\ m$

$\implies 1\ cm=5\ m$

given area

$=6\ cm^2=6\times (5\ m)^2=6\times 25 \ m^2=150\ m^2$

area of

$6\ cm^2$ represents

$150\ m^2$

There are two circles of different radii such that radius of the smaller circle is three seventh that of the larger circle. A square whose area equals 3969 square cm has its side as thrice the radius of the larger circle. What is the circumference of the smaller circle?

Given that, there are two circles of different radii such that radius of the smaller circle is three seventh that of the larger circle.

Also, a square whose area is

$3969 \ cm^2$ has its side as thrice the radius of the larger circle.

To find out: The circumference of the smaller circle.

We know that, area of a square

$=side\times side$

$\therefore \$ Side of the square

$=\sqrt{3969}=63\ cm$

Hence, radius of the larger circle

$=63\times \dfrac{1}{3}$

$=21\ cm$

And radius of smaller circle

$=21\times \dfrac{3}{7}=9\ cm$

We know that, the circumference of a circle is

$2\pi r$

$\therefore \$ The circumference of the smaller circle

$=2\times\dfrac{22}{7}\times 9\quad \quad \left[\because \ \pi=\dfrac{22}{7}\right]$

$\Rightarrow 56.5714\ cm$

Hence, the circumference of the smaller circle is

$56.5714\ cm$ .

Consider two concentric circles. The radius of the big circle is $42$ cm and the radius of the small circle is $35$ cm. Find the area between the circles.

Given

${ R }_{ 1 }=4㎝,{ R }_{ 2 }=35cm$

Area between circles

$=$ area of shaded region

$=\pi { R }_{ 2 }^{ 2 }-\pi { R }_{ 1 }^{ 2 }$

$=\pi \left( { 42 }^{ 2 }-{ 35 }^{ 2 } \right)$

$=1693.32 \, cm^2$

1234733_1525478_ans_4d9e115fb16d420d89f1975baa9c9415.png

In a parallelogram it is being given that base = 14 dm and height = 6.5 dm. Find its area in ${ m }^{ 2 }$ .

Area =

$base\times height$

Area=

$14dm \times6.5dm$

Area=

$91dm^{2}$

Area=

$9100cm^{2}$

Area=

$0.91m^{2}$

Find the circumference of a circle whose radius is -
$15.4 cm$

1311291_1504996_ans_0fc59c60023a443bbb6babea627130ae.jpeg

Sudha wants to polish a circular table table top of diameter $2.8$ meter.Find the cost of polishing if the rate of polishing of $Rs25$ per square meter.

Diameter

$=2.8m$

Radius

$=1.4m$

Area of circular table

$=\pi { r }^{ 2 }=\pi \left( 1.4 \right) \left( 1.4 \right) =6.15㎡$

Rate of polishing=Rs.

$25$

Cost of polishing

$=25\times 6.15$

$=153.75$ Rs.

Find the circumference of a circle whose radius is
(i) $28\text{ cm}$
(ii) $10.5\text{ cm}$

Circumference of a circle of radius

$r$ is

$2\pi r$

(i)

$r=28\text{ cm}$

$\implies$ Circumference

$=2\times \dfrac{22}{7} \times 28=176 \text{ cm}^2$

(ii)

$r=10.5\text{ cm}$

$\implies$ Circumference

$=2\times \dfrac{22}{7} \times 10.5=66 \text{ cm}^2$

Find the area of the following parallelogram.

We know that the area of parallelogram

$=$ base

$\times$ height
Here base

$=5$ cm and height

$=3$ cm

$\therefore$ Area of parallelogram

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

The area of a circle inscribed in an equilateral triangle is $154cm^2$ . Find the perimeter of the triangle. Take $\sqrt 3 = 1.73$

Area of circle =

$\displaystyle \pi r^2$ (where r be the radius of circle)

$\displaystyle \frac{22}{7} \times r^2$

Given: area of a circle =

$\displaystyle 154 \ cm^2$

$\displaystyle \frac{22}{7} \times r^2 = 154$

$\displaystyle r^2 = 154 \times \frac{7}{22}$

$\displaystyle r = 7$

Radius of the circle is

$\displaystyle 7\ cm$ .

Let us consider each side of the equilateral triangle is

$\displaystyle “a”$ and height is

$\displaystyle h$ , then

$\displaystyle r = \frac{h}{3}$

$\displaystyle h = 3r = 3 \times 7 = 21$

Height of the triangle is

$\displaystyle 21 \ cm$

From right triangle APC:

$\displaystyle h^2 = a^2 - (\frac{a}{2})^2 = \frac{3a^2}{4h}$

$\displaystyle h = \frac{\sqrt{3}}{2} (a)$

$\displaystyle 21 = \frac{\sqrt{3}}{2} \times a$ (putting value of

$\displaystyle h$ )

$\displaystyle a = 14 \sqrt{3} \ cm$

Perimeter of the triangle = Sum of all the sides

$\displaystyle = 3a = 3 \times 14 \sqrt{3}= 42 \times 1.73= 72.66 \ cm$

1557461_1543161_ans_ba919485871c4918a19f88cfd8fe6132.png

Find the cost of the polishing a circular table top of diameter 4 m, if the rate of polishing it is Rs. 25 per metre square. $[\text{use} \ \pi=3.14]$

Given that the diameter of the circular table top is

$4\ m$ .

To find out: Cost of polishing the top of the table at

$Rs.\ 25\ per\ m^2$ .

Radius of the table

$(r)=\dfrac{d}{2}$

$=\dfrac{4}{2}=2\ m$ .

We know that, area of a circle

$=\pi r^2$

$\therefore \$ area of the table top

$=3.14\times 2\times 2\quad \quad \left[\because \ \pi=3.14\right]$

$\therefore \$ Area

$=12.56\ m^2$

Now, cost of polishing

$1\ m^2$ area

$=Rs.\ 25$ .

$\therefore \$ Cost of polishing

$12.56\ m^2$ area will be:

$12.56\times 25=Rs.\ 314$

Hence, the cost of polishing the circular table top of diameter

$4\ m$ is

$Rs.\ 314$ .

Find the area of the circle whose circumference is $77\ cm$ . $\left( {take\;\pi = \dfrac{{22}}{7}} \right)$

$2\pi r = 77\ cm$ = circumference

$= 2\times\dfrac{22}{7}\times r = 77$

$r = \dfrac{77\times 7}{22\times 2} = \dfrac{49}{4}$ cm

Area

$=\pi r^2 = \dfrac{22}{7}\times\dfrac{49}{4}\times\dfrac{49}{4}$

$= \dfrac{11\times 7\times 49}{8} = \dfrac{3775}{8} = 471.625\ cm^2$

Of the two concentric circles, the circumference of the outer circle is 132 cm and the circumference of the inner circle is 88 cm. Find the difference between their radii.

1313404_1532870_ans_ed721f3c3d844271abba72a3dd851886.jpeg

The ratio of diameter of two circle is $4 : 5$ . Find out ratio of their areas.

Let the diameters of the two circles are

$d_1$ and

$d_2$ .

Then the ratio of their areas

$=\dfrac{r_1^2}{r_2^2}$

$=\dfrac{4^2}{5^2}$

$=\dfrac{16}{25}$ .

So the ratio is

$16:25$ .

The radius of the circle is 3.5 cm. Find its circumference and area.

Given radius of circle is

$r=3.5\ cm$

as we know that

Perimeter of circle with radius

$r$ is

$2\pi r$ and area of circle is

$\pi r^2$

So perimeter of circle is

$2\pi (3.5)=\dfrac{22}{7}\times 7=22\ cm$

Area of circle is

$\pi (3.5)^2=\dfrac{22\times 7\times 7}{7\times 4}=38.5\ cm^2$

The circumference of a circle is $31.4 {cm}$ Find the radius and the area of the circle
(Take $\pi=3.14$ )

1317018_1538439_ans_ce6cdbbabfdb440182635714feab7241.jpeg

Find the area of the following parallelogram.

We know that the area of parallelogram

$=$ base

$\times$ height

Here base

$=7$ cm and height

$=4$ cm

$\therefore$ Area of parallelogram

$=7\times 4=28cm^2$ .

Circumference of a circle is 88 cm. Find the diameter of the circle.

Let the diameter be

$d\ cm$

As we know that

Circumference of circle with diameter

$d$ is

$\pi d$

Given Circumference of a circle is

$88\ cm$

$\implies \pi d=88$

$\implies \dfrac{22}{7}\times d=88$

$\implies d=\dfrac{88\times 7}{22}=28$

So the diameter is

$28\ cm$

Find the area of the parallelogram whose base and height are given as shown in the figure.

We know that the area of parallelogram

$=$ base

$\times$ height
Here base

$=2$ cm and height

$=4.4$ cm

$\therefore$ Area of parallelogram

$=2\times 4.4=8.8cm^2$ .

Find the area of the following parallelogram.

We know that the area of parallelogram

$=$ base

$\times$ height
Here base

$=5$ cm and height

$=4.8$ cm

$\therefore$ Area of parallelogram

$=5\times 4.8=24cm^2$ .

How many times will the wheel of a car rotate in a journey of $88$ km if the diameter of the wheel is $56$ cm $\left(\pi =\dfrac{22}{7}\right)$ .

${\textbf{Hint: Circumference of wheel = 2}}\pi {\text{r}}$

${\textbf{Step -1: Find circumference}}$

${\text{Journey = 88km}}$

${\text{Diameter of wheel is = 56 cm}}$

${\text{Circumference of the wheel}} = 2\pi r$

$= {\text{ }}\left( {\dfrac{{22}}{7}} \right) \times 56$

$= {\text{ }}176\text{ cm.}$

${\textbf{Step -2: Find number of rotation}}$

${\text{Number of rotation}} = {\text{ }}\dfrac{{{\text{88}}\;{\text{km}}}}{{{\text{176}}\;{\text{cm}}}}$

$= {\text{ }}\dfrac{{{{88 \times 1000 \times 100}}}}{{{\text{176}}\;}}$

${\text{ = 50000}}\;$

${\textbf{Final Answer: So the number of rotations is 50000}}$

PQRS is a parallelogram. QM is the height from Q to SR and QN is the height from Q to PS. If SR $=12\ cm$ and QM $=7.6\ cm$ Find QN, if PS $=8\ cm.$

Given: SR

$=12$ cm, QM

$=7.6$ cm, PS

$=8$ cm.

Area of parallelogram

$=$ base

$\times$ height

$=12\times 7.6=91.2cm^2$ .

Area of parallelogram

$=$ base

$\times$ height

$\Rightarrow 91.2=8\times$ QN

$\Rightarrow QN=\dfrac{91.2}{8}=11.4$ cm.

In figure $CD||AE$ and $CY||BA$
Name a triangle equal in area of $\triangle CBX$

1660548_1670479_ans_e70e81fc582f4feebf370cbc70ce7b68.jpeg

The ratio of the radii of two circles is $3:2$ . What is the ratio of their circumference?

If the radius of circle is

$r$ then its circumference is given as

$2\pi r$

Thus it can be seen that circumference is directly proportional to the radius of the circle.

Hence ratio of circumference of two circle will be equal to the ratio of their respective radii which is equal to

$3:2$

Fill in the blanks:
the diameter of a circle passes through __________

The radius of a circle is the distance from the centre of a circle to any point on the circumference. A diameter of a circle is any line passing through the centre of the circle which connects two points on the circle.

Hence, the diameter of a circle passes through the centre of the circle.

From a circular sheet of radius $4\text{ cm,}$ a circle of radius $3\text{ cm}$ is removed. Find the area of the remaining sheet. (Take $\pi =3.14$ )

Radius of circular sheet(R)

$=4\text{ cm}$ and radius of removed circle(r)

$=3\text{ cm}$
Area of remaining sheet

$=$ Area of circular sheet

$-$ Area of removed circle

$=\pi R^2-\pi r^2=\pi(R^2-r^2)$

$=\pi(4^2-3^2)=\pi(16-9)$

$=3.14\times 7=21.98 \text{ cm}^2$
Thus, the area of the remaining sheet is

$21.98\text{ cm}^2$

A point $D$ is taken on the side $BC$ of a $\triangle ABC$ such that $BD=2DC$ . Prove that $ar(\triangle ABD)=2\ ar(\triangle ADC)$ .

1657221_1670358_ans_7ce9b96cba144ba084a7a483a95896d8.jpeg

Fill in the blanks:
All the diameters of any given circle are _________

We know that two or more lines are said to be concurrent if they pass through a common point i.e., they intersect at a point. The point

$O$ is called the point of concurrence. We know that diameter of circle always passes through the centre of the circle. Hence all the diameters of a circle passes through the centre

$O$ .

Hence, the diameter of a circle are concurrent.

From a circular card sheet of radius $14\ cm,$ two circles of radius $3.5\ cm$ and a rectangle of length $3\ cm$ and breadth $1\ cm$ are removed (as shown in the adjoining figure). Find the area of the remaining sheet. $\bigg($ Take $\pi =\dfrac{22}{7}\bigg)$

Radius of circular sheet(R)

$=14$ cm and Radius of smaller circle(r)

$=3.5$ cm
Length of rectangle(l)

$=3$ cm and breadth of rectangle(b)

$=1$ cm

According to question,
Area of remaining sheet

$=$ Area of circular sheet

$-[$ Area of two smaller circle

$+$ Area of rectangle

$]$

$=\pi R^2-[2(\pi r^2)+(l\times b)]$

$=\dfrac{22}{7}\times 14\times 14-\left[\left(2\times \dfrac{22}{7}\times 3.5\times 3.5\right)-(3\times 1)\right]$

$=22\times 14\times 2-[44\times 0.5\times 3.5+3]$

$=616-80$

$=536 cm^2$

Therefore the area of remaining sheet is

$536 cm^2$ .

A circular flower garden has an area of $314$ $m^2$ . A sprinkler at the centre of the garden can cover an area that has a radius of $12\ m.$ Will the sprinkler water the entire garden? (Take $\pi =3.14$ )

Circular area by the sprinkler

$=\pi r^2=3.14\times 12\times 12$

$=3.14\times 144=452.16m^2$
Area of the circular flower garden

$=314m^2$
Since Area of circular flower garden is smaller than area by sprinkler.
Therefore the sprinkler will water the entire garden.

Pragya wrapped a cord around a circular pipe of radius $4$ cm (adjoining figure) and cut off the length required of the cost. Then she wrapped it around a square box of side $4$ cm (also shown). Did she have any cord left? (Take $\pi =3.14$ )

Radius of pipe

$=4$ cm
Wrapping cord around circular pipe

$=2\pi r$

$=2\times 3.14\times 4=25.12$ cm
Again, wrapping cord around a square

$=4\times$ side

$=4\times 4=16$ cm
Remaining cord

$=$ Cord wrapped on pipe

$-$ Cord wrapped on square

$=25.12-16=9.12$ cm
Thus, she has left

$9.12$ cm cord.

The circumference of a circle is $31.4\text{ cm}$ . Find the radius and the area of the circle. (Take $\pi =3.14$ )

The circumference of the circle is,

$2\pi r=31.4$

$\Rightarrow 2\times 3.14\times r=31.4$

$\Rightarrow r=\dfrac{31.4}{2\times 3.14}$

$\Rightarrow r=5\text{ cm}$

Then area of the circle is,

$\pi r^2=3.14\times 5\times 5$

$=78.5\text{ cm}^2$

Therefore, the radius and the area of the circle are

$5\text{ cm}$ and

$78.5\text{ cm}^2$ respectively.

Find the circumference of a circle whose radius is $1.4\,m$

Here, radius

$r = 1.4 \,m = \dfrac{14}{10} \,m$
Circumference of a circle

$= (2 \pi r) units$
Where,

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

$= 2 \times \dfrac{22}{7} \times \dfrac{14}{10}$

$= 1 \times 22 \times \dfrac{2}{5}$

$= 8.8 \,m$

A road which is $7$ m wide surrounds a circular park whose circumference is $88 m$ . Find the area of the road.

Given

$w=7m$

$2\pi r=88m$

$\Rightarrow 2.\dfrac{22}{7} r=88$

$\Rightarrow 2.\dfrac{1}{7}.r=4$

$\Rightarrow r=2\times 7$

$r=14$

So,

$r=14$

$R=r+w=14+7=21$

$\therefore$ Area of road

$=\pi R^2-\pi r^2$

$=\pi (R+r)(R-r)$

$=\dfrac{22}{7}(21+14)(21-14)$

$=\dfrac{22}{7}(35)(7)$

$=770m^2$

1501132_1700916_ans_f6bb931b82484ed3a2d059263481c3d8.png

Each side of a square flower bed is $2\ m\ 80\ cm$ long. It is extended by digging a strip $30\ cm$ wide all around it. Find the area of the enlarged flower bed and also the increase in the area of the flower bed.

$side=2.8m$

extended by digging strip of 30cm on all sides

So final Side=

$2.8+0.6=3.4m$

Area of Enlarged flower bed=

$3.4\times 3.4=11.56m^{2}$

Initial Area=

$2.8\times2.8=7.84m^{2}$

Increased Area=

$11.56-7.84=3.72m^{2}$

Rakesh has a rectangular field of length $80\ \text{m}$ and breadth $60\ \text{m}$ . In it, he wants to make a garden $10\ \text{m}$ long and $4\ \text{m}$ broad at one of the corners and at another corner, he wants to grow flowers in two floor-beds each of size $4\ \text{m}$ by $1.5\ \text{m}$ . In the remaining part of the field, he wants to apply manures. Find the cost of applying the manure at the rate of $\text{Rs}\ 300/\text{m}^2$ .

Area of rectangular field

$=80 \times 60=4800 \text{ m}^2$

Area of garden

$=10\times 4=40\text{ m}^2$

Area of two flower-beds

$=2(4\times 1.5)=12\text{ m}^2$

Area of the remaining field

$=4800-40-12=4748 \text{ m}^2$

$\because$ Cost of applying manure in

$1\text{ m}^2$

$=\text{Rs }30$

$\therefore$ Cost of applying manure in

$4748\text{ m}^2=4748\times 30=\text{Rs }1,42,440$

A wire when bent in the form of an equilateral triangle encloses an area of $\displaystyle 121\sqrt{3} \ cm^2$ . The same wire is bent to form a circle. Find the area enclosed by the circle.

Let the square be of side

$\displaystyle 'a'$ cm and radius of the circle be

$\displaystyle 'r'$

Area enclosed by the square

$\displaystyle = 484 \ cm^2$

Also, we know that Area of squre

$\displaystyle = side \times side$

Aera of the squre

$\displaystyle = a^2$

$\displaystyle \Rightarrow a^2 = 484$

$\displaystyle \Rightarrow a = \sqrt484$

$\displaystyle \Rightarrow a = 22$

Therefore, side of square is

$\displaystyle 22$ cm.

Perimeter of square

$\displaystyle = 4 \times side = 4 \times 22 = 88$

From the statement, circumference of the circle = Perimeter of square

$\Rightarrow \displaystyle 2 \pi r = 88$

$\displaystyle r = (88\times 7)/(2\times 22) = 14$

Radius of the circle

$\displaystyle = 14$ cm

Area of cirle

$\displaystyle = \pi r^2 = \dfrac{22}{7} \times 14 \times 14 = 616 cm^2$

The circumference of a circle exceeds its diameter by $\displaystyle 45$ cm. Find the circumference of the circle.

Given : circumference of a circle exceeds its diameter by

$\displaystyle 45$ cm.

$\displaystyle \Rightarrow$ Circumference of circle

$\displaystyle = 45 +$ Diameter of circle

$\Rightarrow\displaystyle 2 \pi r = 45 + 2 r$

$\Rightarrow \displaystyle 2r(\pi - 1) = 45$

$\Rightarrow \displaystyle 2r\left(\frac{22}{7} - 1 \right) = 45$

$\Rightarrow \displaystyle \frac{15}{7}r = \frac{45}{2}$

$\therefore \displaystyle r = 10.5$

Circumference of a circle

$\displaystyle = 2 \pi r = 2\times \frac{22}{7} \times 10.5 = 66 cm$

The area of a circle is $\displaystyle 98.56 \ cm^2$ . Find its circumference.

Area of a circle

$\displaystyle = 98.56 \ cm^2$

We know, Area of the circle

$\displaystyle = \pi r^2$

Where,

$\displaystyle r =$ radius of the circle

So,

$\displaystyle \pi r^2 = 98.56$

Put

$\displaystyle \pi = \frac{22}{7}$

$\displaystyle r^2 = \frac{98.56}{\frac{22}{7}} = 31.36$

$\displaystyle r = 5.6$

Circumference of circle

$=2\times\pi\times\displaystyle 5.6=35.2$ cm

Find the area of a circle whose circumference is $\displaystyle 8\pi$ .

Circumference of circle

$\displaystyle = 8 \pi$

$\displaystyle 2 \pi r = 8 \pi$

$\displaystyle r = 4$

Area of circle

$\displaystyle = \pi r^2 = \pi (4)^2 = 16 \pi$

The radius of the wheel of a vehicle is $\displaystyle 42 \ cm$ . How many revolutions will it complete in a $\displaystyle 19.8 \,km$ long journey?

Radius of the wheel =

$\displaystyle r = 42 \ cm$ (given)

Circumference of wheel = Circumference of circle =

$\displaystyle 2 \pi r = 2 \times \frac{22}{7} \times 42= 264$

Circumference of wheel is

$\displaystyle 264 \ cm$ .

Again, distance travelled by wheel =

$\displaystyle 19.8 \ km = 1980000 \ cm$ (Convert km into cm)

Now, number of revolutions taken by a wheel =

$\displaystyle 1980000/264 = 7500$

A rope by which cow is tethered is increased from $\displaystyle 16 \ m$ to $\displaystyle 23 \ m$ . How much additional ground dows it have now to graze?

At radius =

$\displaystyle 16 \ m$ , cow can graze the area of plot is

$\displaystyle \pi r^2 = \frac{22}{7} \times 16 \times 16 = 804.5 \ m^2$

At radius =

$\displaystyle 23 \ m$ , cow can graze the area of plot is

$\displaystyle \pi r^2 =\frac{22}{7} \times 23 \times 23= 1662.57 \ m^2$

Additional ground area =

$\displaystyle 1662.57 - 804.5 = 858 \ m^2$

Find the area of a circle whose radius is $21\,cm.$

Given, radius

$r = 21 \,cm$ and

$\pi = \dfrac{22}{7}$
Then,
Area of the circle =

$\pi r^2$ sq units

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

$= 22 \times 3 \times 21$

$= 1386 \,cm^2$

The short and long hands of a clock are $\displaystyle 4 \ cm$ and $\displaystyle 6 \ cm$ long respectively. Find the sum of distances travelled by their tips in $\displaystyle 2$ days. [ take $\displaystyle \pi = 3.14$ ].

The short and long hands of a clock are

$\displaystyle 4 \ cm$ and

$\displaystyle 6 \ cm$ long respectively.

In an hour the hour hand completes one rotation in

$\displaystyle 12$ hours therefore in

$\displaystyle 24$ hours it will complete

$\displaystyle 2$ rotations.

Similarly, the minute hand completes one rotation therefore in

$\displaystyle 24$ hours the minute hand will complete

$\displaystyle 24$ rotations.

Distance travelled by its tip in

$\displaystyle 2$ days =

$\displaystyle 4$ (circumference of the circle with radius

$\displaystyle 4 \ cm$ )=

$\displaystyle ( 4 \times 2\pi \times 4) = 32\pi \ cm$ (As, in

$\displaystyle 2$ days, the short hand will complete

$\displaystyle 4$ rounds)In

$\displaystyle 2$ days, the long hand will complete

$\displaystyle 48$ rounds length moved by its tip =

$\displaystyle 48$ (circumference of the circle with radius

$\displaystyle 6 \ cm$ )

$\displaystyle (48 \times 2\pi \times 6) \ cm = 576 \pi \ cm$

Now,

Sum of the lengths moved=

$\displaystyle (32 \pi + 576 \pi) = 608 \pi \ cm = (608 \times 3.14) = 1909.12 \ cm$

The hypotenuse of a right-angled triangle is 6.5 cm and its base is 6.0 cm. Find the length of the perpendicular and the area of the triangle.

We are given with a right-angled triangle whose measures are:
Base

$=6\,cm$
Hypotenuse

$= 6.5\,cm$

Using Pythagoras Theorem:
Hypotenuse

$^2 =$ Base

$^2 +$ Perpendicular

$^2$

$(6.5)^{2} = (6)^{2} +$ Perpendicular

$^{2}$
or Perpendicular

$^{2}= 42.25 - 36 =6.25$
or Perpendicular

$= 2.5\,cm$

Now,
Area of a triangle

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

$= \dfrac{1}{2} \times 6\,cm \times 2.5\,cm$

$= 7.5\,cm^{2}$

A park is in the shape of a circle of diameter $\displaystyle 7 \ m$ . It is surrounded by a path of width $\displaystyle 0.7 \ m$ . Find the expenditure of cementing the path at the rate of $\displaystyle Rs. \ 110$ per sq. m.

Diameter of park =

$\displaystyle 7 \ m$

so, radius of park =

$\displaystyle r_1 = 3.5 \ m$

Width of park =

$\displaystyle 0.7 \ m$

Bigger radius of park =

$\displaystyle r_2 = 0.7 + 3.4 = 4.2 \ m$

Now, Area of path = area of bigger circle -area of smaller circle=

$\displaystyle \pi r_2^2 - \pi r_1^2 = \pi (r_2^2 - r_1^2) = \frac{22}{7} (4.2^2 -3.5^2) = \frac{22}{7} (17.64 - 12.25) = 16.94 \ m^2$

Also,Cost of expenditure

$= rate × area=$

$\displaystyle 110 \times 16.94 = 1863.40$

Cost of Expenditure of cementing the path is

$\displaystyle Rs. \ 1863.40$ .

There are $20$ cylindrical pillars in a building, each having a diameter of $50\ cm$ and height $4\ m$ . Find the cost of cleaning them at $Rs\ 14\ per\ m^{2}$

It is given that
Diameter of one pillar

$=50 cm=0.5 m$
So the radius of one pillar

$=0.5/2 =0.25 m$
Height of one pillar

$=4 m$

We know that

Lateral surface area of one pillar

$=2\ \pi rh$

By substituting the values

Lateral surface area of one pillar

$=2\times (22/7)\times 0.25\times 4$

So we get

Lateral surface area of one pillar

$=6.85\ m^2$

So the lateral surface area of

$20$ pillars

$=20\times 6.285=125.714\ m^2$

It is given that the cost of cleaning

$=Rs. 14$ per

$m^2$

So the cost of cleaning

$125.714\ m^2 = Rs. (14\times 125.714)$

We get

Cost of cleaning

$125.714\ m^2 =1760$

Therefore, the cost of cleaning

$125.714\ m^2$ is

$Rs. 1760$ .

A racetrack is in the form of a ring whose inner circumference is $\displaystyle 352 \ m$ and outer circumference is $\displaystyle 396 \ m$ . Find the width and the area of the track.

Let r and R be the radii of inner circle and outer boundaries respectively.

Then,

Circumference of inner circle =

$\displaystyle 2\pi r = 352$

$\displaystyle r = \frac{352}{2\pi}$

and Circumference of outer circle =

$\displaystyle 2\pi R = 396$

$\Rightarrow \displaystyle R =\frac{396}{2\pi}$

Width of the track =

$\displaystyle ( R - r) = \frac{396}{2\pi} - \frac{352}{2\pi} = \frac{44}{2\pi} = \frac{44}{2} \times \frac{7}{22} = 7$

Width of the track is

$\displaystyle 7 \ m$ .

Again,

$\displaystyle R + r = \frac{396}{2\pi} + \frac{352}{2\pi} = \frac{748}{2\pi}$

Area of the track =

$\displaystyle \pi (R^2 - r^2) = \pi (R - r)(R + r) = \pi \times 7 \times \frac{748}{2\pi} = 2618$

Therefore, the area of the track is

$\displaystyle 2618 \ m^2$ .

A path of $\displaystyle 8 \ m$ width runs around the outsider of a circular park whose radius is $\displaystyle 17 \ m$ . Find the area of the path.

Width of the path =

$\displaystyle 8 \ m$

Inner radius of the circular park =

$\displaystyle 17 \ m$

Outer radius of the circular park =

$\displaystyle (17 + 8) = 25 \ m$

Now,Area of the path =

$\displaystyle \pi [(25^2 - 17^2) ] =\pi(25+17)(25-17)= \frac{22}{7} \times 42 \times 8 = 1056$

Area of the path

$=\displaystyle 1056 \ m^2$

The difference between the circumference and radius of a circle is $\displaystyle 37 \ \text{cm}$ . Using $\displaystyle \pi = \dfrac{22}{7}$ , find the circumference of the circle.

Circumference of a circle - Radius

$\displaystyle = 37\ \text{cm}$ ...(Given)

We know that, Circumference of a circle

$\displaystyle = 2 \pi r$ (Where

$r$ = radius)

According to the statement:

$\displaystyle 2 \pi r - r = 37$

$\displaystyle 2 \times \frac{22}{7} \times r - r = 37$

$\displaystyle \left(\frac{37}{7}\right)r = 37$

$\therefore \displaystyle r = 7\ \text{cm}$

Therefore, circumference of a circle

$\displaystyle = 2 \pi r = 2 \times \frac{22}{7} \times 7 = 44\ \text{cm}$

Find the area of a circle whose diameter is $1.4\,m$ .

Given, Diameter of a circle

$= 1.4 \,m$
So, radius of the circle

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

Then,
Area of the circle

$= \pi r^2$ sq units

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

$= 22 \times 0.1 \times 0.7\ m^2$

$= 1.54 \,m^2$

Hence, the area of the circle is

$1.54 \,m^2$

Find the circumference of a circle whose radius is $28\ cm$

Given radius

$=28\ cm$

We know that, Circumference of the circle i.e,

$C=2\pi r$

$=(2\times 22/7\times 28)$

$=(56\times 22/7)$

$=176\ cm$

Thus, the circumference of the circle is

$176\ cm$

Find the circumference of a circle whose diametre is $35\ cm$

Given

Diameter

$=35\ cm=2\times radius=2r$

Circumference of the circle,

$C=2\pi r$

$=\pi \times (2r)$

$=\pi \times$ diameter of the circle

$=22/7\times 35$

$=110\ cm$

Hence, the circumference of the circle is

$110\ cm$

Find the area of a circle whose diameter is $28\,cm$

Given, Diameter of a circle

$= 28 \,cm$
WKT, Radius of a circle

$r = \dfrac{d}{2} = \dfrac{28} {2} = 14 \,cm$
Then,
Area of the circle

$= \pi r^2$ sq units

$= \dfrac{22} {7} \times 14 \times 14$

$= 22 \times 2 \times 14$

$= 616 \,cm^2$

Find the area of a circle whose radius is $3.5\,m$

Given, radius

$r = 3.5 \,m$ and

$\pi = \dfrac{22} {7}$
Then,
Area of the circle

$= \pi r^2$ sq units

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

$= 22 \times 0.5 \times 3.5$

$= 38.5 \,m^2$

Find the ratio of the area of a circle and an equilateral triangle whose diameter and a side are respectively equal.

Let radius be

$r$ and side be

$a$

Given,

$2r=a\\\Rightarrow \dfrac{r}{a}=\dfrac{1}{2}$

$\dfrac{\text{Area of circle}}{\text{Area of equilateral triangle}}=\dfrac{\pi r^2}{\dfrac{\sqrt 3}{4}a^2}\\=\dfrac{4\pi }{\sqrt 3}\left( \dfrac{r}{a}\right)^2\\=\dfrac{4\pi}{\sqrt 3}\times \dfrac{1}{4}\\=\dfrac{\pi }{\sqrt 3}$

If circumference and the area of a circle are numerically equal, find the diameter of the circle.

1975268_1785612_ans_e2f2ee2f8e9c4a8c8639cffe49e95255.jpg

Find the area of a circle whose circumference is $22\ cm$ .

Let

$r$ be the radius of the circle. Then,

Circumference

$=22\ cm$

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

$\therefore$ Area of the circle

$=\pi r^{2}=\dfrac{22}{7}\times \dfrac{7}{2}\times \dfrac{7}{2}\ cm^{2}=38.5\ cm^{2}$

Two sides of a triangular field are 85m and 154m in length and its perimeter is 324m. Find
(i) the area of the field and
(ii) the length of the perpendicular from the opposite vertex on the side measuring 154m.

It is given that two sides of a triangular field are 85m and 154m

Consider the third side as x m

It is given that the perimeter

$= 324m$

We can write it as

$85 + 154 + x = 324$

On further calculation we get

$x = 324 - 85 - 154$

By subtraction

$x = 85m$

Consider

$a = 85m, b = 154m and c = 85m$

We know that

$S=\dfrac{a+b+c}{2}$

$S=\dfrac{85+154+85}{2}$

By division

$S=162m$

We know that

Area

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

By substituting the values

Area=

$\sqrt{162(162-85)(162-154)(162-85)}$

So we get

Area=

$\sqrt{162\times77\times8\times77}$

It can be written as

Area=

$\sqrt{2\times9\times9\times7\times11\times2\times2\times2\times7\times11}$

On further calculation

Area=

$\sqrt{11\times11\times9\times9\times7\times7\times2\times2\times2\times2}$

We get

Area=

$11\times9\times7\times2\times2$

By multiplication

Area=

$2772m^2$

We know that

Area of the triangle =

$1/2 \times b \times h$

By substituting the values

$1/2 \times 154 \times h = 2772$

On further calculation

$77 \times h = 2772$

So we get

$h = 2772/77$

By division

$h = 36 m$

Therefore, area of the field is

$2772 m^2$ and the length of the perpendicular from the opposite vertex on the side measuring

$154m$ is

$36m.$

Find the area of the circle inscribed in a square of side $a\ cm$ .

Diameter of the circle

$=a$

$\Rightarrow Radius=\dfrac{a}{2}\Rightarrow Area=\pi \left( \dfrac{a}{2}\right)^2=\dfrac{\pi a^2}{4}\ cm^2$

A bicycle wheel makes $5000$ revolutions in moving $11\ km$ . Find the diameter of the wheel.

Let the radius of the wheel be

$r\ cm$ .

Distance covered by the wheel in one revolution

$=\dfrac{Distance\ moved}{Number\ of\ revolutions}=\dfrac{11}{5000}\ km$

$=\dfrac{11}{5000}\times 1000\times 100\ cm=220\ cm$

$\therefore$ Circumference of the wheel

$=220\ cm$

$\Rightarrow 2\pi r=22\ cm \Rightarrow 2\times \dfrac{22}{7}\times r =220$

$\Rightarrow r=\dfrac{220\times 7}{2\times 22}\Rightarrow r=35\ cm$

$\therefore$ Diameter

$=2r\ cm=(2\times 35)\ cm=70\ cm$

Hence, the diameter of the wheel is

$70\ cm$ .

Difference between the circumference and radius of $37\ cm$ . Find the area of circle.

Given

$2\pi r-r=37$

$r(2\pi-1)=37$

$r=\dfrac{37}{2\pi-1}=\dfrac{37}{2\times \dfrac{22}{7}-1}=\dfrac{37\times 7}{37}=7$

So area of circle

$=\pi r^{2}$

$=\dfrac{22}{7}\times 7\times 7=154\ cm^{2}$

Find the area of a triangular field whose sides are 91m, 98m and 105m in length. Find the height corresponding to the longest sides.

Consider a = 91m, b = 98m and c = 105m
So we get

$S=\dfrac{a+b+c}{2}$

$S=\dfrac{91+98+105}{2}$
By division
s=147
We know that
Area=

$\sqrt{s(s-a)(s-b)(s-c)}$
By substituting the values
Area=

$\sqrt {147(147-91)(147-98)(147-105)}$
So we get
Area=

$\sqrt {147\times 56 \times 49 \times 42}$
It can be written as
Area=

$\sqrt {49\times3\times7\times2\times2\times2\times49\times7\times3\times2}$
On further calculation
Area =

$49\times3\times2\times2\times7$
So we get
Area =

$4116m^2$
It is given as
b = Longest side =

$105 m$
Consider h as the height corresponding to the longest side
We know that
Area of the triangle =

$1/2 \times b \times h$
By substituting the values

$1/2 \times b \times h = 4116$
On further calculation

$105 \times h = 4116\times2$
So we get

$h = (4116\times2)/105$
By division
h =

$78.4 m$

The difference between the semiperimeter and the sides of a $\Delta$ ABC are 8cm, 7cm and 5cm respectively.Find the area of the triangle.

Consider a, b and c as the sides of a triangle and s as the semi perimeter

So we get

s - a = 8cm

s - b = 7cm

s - c = 5cm

We know that(s - a) + (s - b) + (s - c) = 8 + 7 + 5

On further calculation

3s - (a + b + c) = 20

We know that a + b + c = 2s

3s - 2s = 20

So we gets = 20

By substituting the value of s

a = s - 8 = 20 - 8 = 12cm

b = s - 7 = 20 - 7 = 13cm

c = s - 5 = 20 - 5 = 15cm

We know that

Area=

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

By substituting the values

Area=

$\sqrt{20(20-12(20-13)(20-15)}$

So we get

Area=

$\sqrt{20\times8\times7\times5}$

It can be written as

Area=

$\sqrt{4\times5\times4\times2\times7\times5}$

By multiplication

Area=

$20\sqrt{14}$

Therefore, the area of the triangle is

$20\sqrt{14}cm^2$

The radius of a wheel is $0.25\ m$ . Find the number of revolutions it will make to travel a distance of $11\ km$ .

1711113_1785613_ans_a0eb49be055042baaeb133629f5ab76f.PNG

$1\ sq\ m = 1\ m\times$ $m = 100\ cm \times$ $cm$ .

$1\ sq\ m = 1\ m\times 1\ m = 100\ cm \times 100\ cm$ .

Tahir measured the distance around a square field as $200$ rods (lathi). Later he found the length of this rod was $140\ cm$ . Find the side of this field in meters.

$\textbf{Step 1: Find the total distance}$

$\textbf{Total Number of rods = 200}$

$\textbf{Length of each rod = 140 cm}$
Total distance = Total number of rods

$\times$ Length of each rod

$= 200 \times 140$

$= 28000\ cm$

$\textbf{Step 2: Calculate the perimetre of square field}$

Perimeter of a square = 4

$\times$ side length

$\therefore$ length of side of given square field

$= \dfrac{28000}{4}$

$= 7000\ cm$
We know that, 100 cm = 1m

$\therefore$ length of side of square field

$= \dfrac{7000}{100}$

$= 70\ m$

$\therefore \textbf{Length of side of square field is 70 m}$

Mukesh walks around a circular track of radius 14 m with a speed of 4 km/hr. If he takes 20 rounds of the track for how long does he walk

Radius of the circular track = 14 m
Circumference of the circular track =

$2\pi r$ =

$2 \times 22/7 \times 14$ =

$44\times 2$ = 88 m
Total distance cover in 20 rounds =

$88\times 20$ = 1760 m
speed of Mukesh on the circular track = 4Km/h
=

$\frac{4\times 1000}{60}=\frac{2\times 100}{3}=\frac{200}{3}$ m/min
Time taken by Mukesh =

$\frac{1760}{\dfrac{200}{3}}=\frac{1760\times 3}{200}=\frac{176\times 3}{20}$ = 26.4 min
26.4 min = 26 min and 24 sec.

There is a circular pond and a foothpath runs along its boundary. A person walks around it, exactly once keeping close to the edge. If his step is 66 cm long and he takes exactly 400 steps to go around the pound, find the diameter of the pond.

Let the radius of the pond be r.
Then, diameter of the pond d=

$2 \times r$
Since, a person takes exactly 400 steps with 66 cm long each step to go round the pond.
Hence, the circumference of the pond =

$66 \times 400$ = 26400 cm
=26400/100 = 264 m
As the circumference of a circle is

$2\pi r$

$\Rightarrow$

$2\pi r$ = 264

$\Rightarrow$ r= 264/2 \times 7/22 = $$ \dfrac{264\times 7}{44} = 42 m
Hence, the radius of the pond = 42 m.
So, diameter of the pond =

$2\times 42$ = 84 m.

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

Area of shaded region = are of the parallelogram ABCD-Area of ABE
Area of parallelogram =

$Side \times height$
Area of the parallelogram ABCD =

$40 \times 30$ = $$ 1200 cm^{2}
Area of ABE =

$\frac{1}{2} \times AB \times EF$ =

$\frac{1}{2} \times 40 \times 30$ =

$600 cm^{2}$
Thus, the area of shaded region = 1200-600 =

$600 cm ^{2}$

In Fig., area of AFB is equal to the area of parallelogram ABCD. If altitude EF is $16\, cm$ long. find the altitude of the parallelogram to the base AB of length $10\, cm$ . What is the area of triangle DAO, where O is the mid-point of DC ?

Given, Area of triangle

$AFB$

$=$ Area of parallelogram

$ABCD$

$\Rightarrow \dfrac {1}{2} \times AB \times EF = CD\times EG$ , where

$EG$ is height of parallelogram.

Let

$h=EG$ be corresponding height

$\Rightarrow \cfrac {1}{2} \times 10 \times 16 =10 \times h \cdots\cdots\{\because CD=AB= 10\ cm\}$

$\Rightarrow h =8\, cm$

Now, In triangle

$DAO$ ,

$O$ is the mid-point of

$DC$

$\therefore DO$

$= 5\,cm$

And from figure it is clear that, the height of triangle

$DAO$ is equal to height of parallelogram.

So, Area of triangle

$DAO$

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

$=\dfrac {1}{2} \times 5 \times 8 = 20\, cm^2$

1674219_1793735_ans_8f57c8ab0cd2443ba978b9aadd538627.png

$ABCD$ is a parallelogram in which $AE$ is perpendicular to $CD$ (Fig.). Also $AC =5\, cm, DE =4\, cm$ , and the area of triangle $AED = 6\, cm^2$ . Find the perimeter and area of $ABCD$ .

Given,

Area of triangle

$AED = 6\, sq. cm$

$AC= 5\,cm, DE=4\,cm$

Area of triangle

$AED$

$\Rightarrow 6=\dfrac {1}{2}\times DE \times AE=\dfrac {1}{2}\times 4 \times AE$

$\Rightarrow AE =\dfrac {6\times 2}{4}= 3\, cm$

Now, in right-angled triangle

$AEC$ ,

$\Rightarrow EC^2 = AC^2 - AE^2$

$\Rightarrow EC^2 = 5^2 - 3^2$

$\Rightarrow EC^2 = 16$

$\Rightarrow EC = 4\, cm$

$\Rightarrow DE+ EC= DC$

$\Rightarrow DC= 4+4 = 8\, cm$

Since,

$ABCD$ is a parallelogram,

$AB=DC = 8\, cm$

Now, by Pythagoras theorem in triangle

$AED$

$\Rightarrow AD^2 = AE^2 +ED^2$

$\Rightarrow AD^2 =3^2 +4^2$

$\Rightarrow AD^2 =25$

$\Rightarrow AD = 5\, cm$

Thus,

$AD=BD= 5\,cm$

Perimeter of parallelogram

$ABCD$ ,

$=2(l +b)$

$= 2(DC +AD) = 2(8+5)$

$= 2\times 13=26\,cm$

Area of parallelogram

$ABCD$

$= DC \times AE = 8 \times 3=24\, cm^2$

Find the radius of a circle whose circumference is equal to the sum of the circumference of two circles of radii $15 \,cm$ and $18 \,cm.$

Circle I

$(C_1)$ Circle II

$(C_2)$ Circle III

$(C)$

$r_1 = 15 \,cm$

$r_2 = 18 \,cm$

$r = ?$
According to the question,

$C_1 + C_2 = C$

$\Rightarrow 2\pi r_1 + 2\pi r_2 = 2\pi r$

$\Rightarrow 2 \pi[ r_1 + r_2] = 2\pi r$

$\Rightarrow r_1 + r_2 = r$

$\Rightarrow 15 + 18 = r$

$\Rightarrow r = 33 \,cm$ is the required radius of the circle.

Find the area of the shaded region.

Let

$R$ be radius of bigger hole and

$r$ be radius of smaller circle

From the figure,

$\Rightarrow r=\cfrac {7}{2}\, cm$

$\Rightarrow R=r+7=\dfrac {7}{2} +7 =\cfrac{21}{2} cm$

Now, area of shaded region

$=\text{ area of bigger circle-area or smaller circle}$

$= \pi (R^2-r^2)$

$=\dfrac {22}{7}\times \left ( \dfrac {21}{2} \times \dfrac {21}{2} - \dfrac {7}{2} \times \dfrac {7}{2} \right )=\cfrac{22}7\times \left(\cfrac{441}4-\cfrac{49}4\right)$

$= \dfrac {22}{7}\times \dfrac {392}{4}$

$= 11\times \cfrac{56}2=308\, cm^2$

Therefore,

$308\, cm^2$ is the area, shaded Portion.

Find the total cost of wooden fencing around a circular garden of diameter $28\, m$ . if $1\,m$ of fencing costs $Rs.\ 300$ .

Given,

Diameter of circular garden

$= 28\,m$

So,

$\text{Length of the fencing = circumference of circle}$

$=\pi d =\dfrac {22}{7}\times 28 = 88\,m$

$1\,m$ of fencing costs

$Rs.\ 300$ .

Hence, total cost of fencing

$88 \times 300 = Rs.\, 26400$

A circular pond is surrounded by a $2\, m$ wide circular path. If outer circumference of circular path is $44\, m$ , find the inner circumference of the circular path. Also find area of the path.

Let

$R$ be radius of outer circle and

$r$ be radius of inner circle.

It has been given that

$44\, m$ is the circumference of outer circle,

$\Rightarrow 2\pi R= 44\, m$

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

$\Rightarrow R = \cfrac {44}{2\times\cfrac {22}{7}} = \dfrac{7\times 44}{2 \times 22}= 7\, m$

It is given that circular path is

$2\ m$ wide.

So, radius of inner circle

$r=R-2=7-2=5\ m$

So, inner circumference of path

$=2\pi r=2\times\cfrac{22}7\times 5=31.42\ m$

Therefore,

$\text{Area of the path = area of outer circle - area of inner circle}$

$\Rightarrow \pi (R^2-r^2)$

$= \cfrac {22}{7} (7^2-5^2)= \cfrac{22}{7}\times 24$

$= 75.43\, m^2$

1674704_1795190_ans_e0a8fcd4635f4b90b84e240d7c03d26a.png

Find the areas of the shaded region.

Diameter of the complete circle

$= 14\,cm$

So, Radius

$= 7\,cm$

From the figure, Area of the complete circle

$=\dfrac {22}{7}\times 7\times 7 = 154\,cm^2$

Diameter of the smaller circle

$=\cfrac {7}{4}$

$\Rightarrow \text{radius} (r) =\cfrac {7}{8} \,cm$

There, are two such smaller circles.

So, Area of two smaller circles

$= 2\times \pi r^2 = 2\times \dfrac {22}{7}\times \dfrac {7}{8}\times \dfrac {7}{8} = \dfrac {77}{16}\, cm^2$

Now, area of shaded portion

$=$

$\text{area of complete circle}$

$-$

$\text{area of two smaller circles}$

$= 154 -\cfrac {77}{16} =\cfrac {2464 -77}{16}$

$= \dfrac {2387}{16} = 149 .19\ cm^2$

Therefore,

$149.19\ cm^2$ is the area of shaded portion.

How much distance, in meters, a wheel of $25\, cm$ radius will cover if it rotates $350$ times?

Given,

Radius of wheel

$=25\, cm= 25\left(\cfrac1{100}\, m\right)=0.25\ m$

So,

$\text{Distance traveled in one rotation = circumference of circle}$

$=2\pi r= 2\times \dfrac {22}{7} \times 0.25$

$=\dfrac{11}{7} \, m$

Hence, distance traveled in

$350$ rotations

$=\dfrac{11}{7} \times 350 =550\,m$

Therefore, the wheel covers

$550\, m$ distance.

Find the cost of digging a cuboidal pit $8m$ long, $6 m$ board and $3 m$ deep at the rate of $Rs.30 per m^3.$

length of cuboidal pit

$l = 8 m$
Breadth

$b = 6 m$ height

$h = 3 m$

$\therefore$ Volume of pit

$V= l \times b \times h$

$= 8 \times 6 \times 3$

$= 143 m^3$

$\therefore$ cost of digging a cuboidal

$1 m^3$ is

$Rs. 30$
Cost of digging a cuboidal pit for

$144 m^3$ ....?....

$=144 \times 30$

$=Rs 4320$

A rectangular room is 9 m long and 6 m wide. Find the cost of covering the floor with carpet 2 m wide at Rs 35 per metre.

Given:

Length of room

$l = 9m$
Width of room

$b = 6m$

$\Rightarrow$ Area of room

$=l \times b=9 \times 6$

$\Rightarrow$ Area of room

$= 54\ m^2$

Width of carpet

$= 2\ m$
Area of carpet

$= 54\ m^2$
Length

$= ?$

Length of carpet =

$\dfrac{\text {Area of carpet}}{\text{width of carpet}}=\dfrac{54}{2}$

Length of carpet

$= 27 \ m$

$\Rightarrow$ Cost of covering

$1\ m = Rs.\ 35$

$\Rightarrow$ Cost of covering

$35\ m=$

$Rs. 35 \times 27 m = Rs. 945$

The base of a right angled triangle is 12 cm, its hypotenuse is 13 cm long . Find its area and the perimeter .

It is given that

$ABC$ is a right angled triangle

$BC = 12$ cm and

$AB= 13$ cm
Using the Pythagoras theorem

$AB^{2} = AC^{2}+ BC^{2}$
Substituting the values

$13^{2} = AC^{2}+ 12^{2}$
so we get

$AC^{2} = 169 - 144 \\= 25$

$\Rightarrow AC=5$

$Area =\dfrac12\times 12\times 5\\=30 cm^2$

Perimeter

$=12+13+5$

$=30cm$

A B C is a triangle in which AB = AC=4 cm and $\angle A = 90^{0}$ . Calculate the area of $\Delta ABC$ Also fine the length of perpendicular from A to BC,

1801359_1811902_ans_0e71c3453b104ff28de80fd726f21dee.PNG

A dinner plate is in the form of a circle. A circular region encloses a beautiful design as shown in Fig. The inner circumference is $352\, mm$ and outer is $396\, mm$ . Find the width of circular design.

Let

$R$ be radius of outer circle and

$r$ be radius of inner circle,

It has been given that

$352\,mm$ is the circumference of inner circle,

$\Rightarrow 2\pi R = 352\, mm$

$\Rightarrow 2\times\cfrac{22}7\times r = 352\, mm$

$\Rightarrow r=\cfrac{352\times 7}{2\times 22}=\dfrac{2464}{44}= 56\, mm$

In addition

$396\,mm$ is the circumference of outer circle,

$\Rightarrow 2\pi R = 396\, mm$

$\Rightarrow 2\times\cfrac{22}7\times R = 396\, mm$

$\Rightarrow R= \dfrac{396\times 7}{2\times 22}= 63\, mm$

Therefore, widih of circular design

$=R-r= 63-56=7\,mm$

A room is 4 m long and 3 m 25 cm wide. How many square metres of carpet s needed to cover the floor of the room?

Length of the room

$= 4 m$
Breadth of the room

$= 3 m 25 cm = 3.25 m$

$\therefore$ Area of the room = Length

$\times$ Breadth

$= 4\times 3.25 m =13 sq. m$
Hence,

$13\,\, square \,\,metres$ of carpet is needed to cover the floor of the room.

The moon is about $384000\, km$ from earth and its path around the earth is nearly circular. Find the length of path described by moon in one complete revolution (Take, $\pi= 3.14$ )

Length of path described by moon in one complete revolution is circumference of circle with radius

$384000\ km$

$= 2\pi r$

$= 2\times 3.14 \times 384000$

$= 2411520\,km$

Find the area of a triangle whose base is 6 cm and corresponding height 4 cm.

We know that
Area of triangle

$=\dfrac12 \times base \times height$
Substituting the values

$=\dfrac12 \times 6 \times 4$
By further calculation

$= 6 \times 2$
=

$12 cm^{2}$

In the adjoining figure, the area enclosed between the concentric circles is $770{ cm }^{ 2 }$ . If the radius of the outer circle is $21cm$ , calculate the radius of the inner circle.

Given,

Radius of outer circle

$(R)=21cm$

Radius of inner inner circle

$(r)=r\ cm$

Area of shaded portion

$=770{ cm }^{ 2 }$

$\Rightarrow \pi \left( { R }^{ 2 }-{ r }^{ 2 } \right) =770$

$\Rightarrow \pi \left( {21 }^{ 2 }-{ r }^{ 2 } \right)=770$

$\Rightarrow 441-{r}^{2}=770\times \dfrac7{22}=35\times 7=245$

$\Rightarrow {r}^{2}=441-245=196$

$\Rightarrow r=\sqrt{196}=14\ cm$

In Fig, triangle AEC is right-angled at E, B is a point on EC, BD is the altitude of triangle ABC, AC $= 25\, cm, \,BC =7\, cm$ and $AE = 15\, cm$ .
Find the area of triangle ABC and the length of DB.

Given,

$AC=25\,cm$ ,

$AE= 15\,cm$ ,

$BC=7\,cm$

Applying Pythagoras theorem in triangle

$AEC$ ,

$\Rightarrow AC^2 = AE^2+ EC^2$

$\Rightarrow EC^2 = 25^2 -15^2$

$\Rightarrow EC^2= 625- 225=400$

$\Rightarrow EC =20\,cm$

Now,

$EB=EC-BC=20-7=13\,cm$

Area of triangle

$AEC$ ,

$=\cfrac {1}{2} \times AE \times EC$

$=\cfrac {1}{2} \times 15\times 20= 150\, cm^2$

Now, Area of triangle

$AEB$ ,

$=\cfrac {1}{2} \times AE \times EB$

$=\cfrac {1}{2} \times 15\times 13= 97.5\, cm^2$

Thus,

$\text{area of triangle ABC = Area of triangle AEC - Area of triangle AEB}$

$=150-97.5$

$= 52.5\,cm^2$

Now, again area of triangle

$ABC=\cfrac12\times BD\times AC$

$\Rightarrow 52.5 =\cfrac {1}{2} \times BD \times AC$

$\Rightarrow 52.5 =\cfrac {1}{2} \times BD\times 25$

$\Rightarrow BD =\cfrac{52.5\times 2}{25}= 4.2\, cm$

Therefore,

$52.5$ square centimeters is the area of triangle

$ABC$ and length is

$DB=4.2\, cm$ .

Find the number of revolution if circumference of a wheel is $22cm$ and travels a distance $110 km$

$Number\space of\space revolutions$

$=\dfrac{total\ distance}{circumference}$

$=\dfrac{110 km}{22}=\dfrac{ 11 0000}{ {22} }=5000$

An archery target has three regions formed by three concentric circles as shown in figure. If the diameters of the concentric circles are in the ratio $1 : 2 : 3,$ then find the ratio of the areas of three regions.

$d_1 : d_2 : d_3 = 1 : 2 : 3$ [Given]

$= 1d : 2d : 3d [ \times \,by \,d]$

$\Rightarrow d_1 : d_2 : d_3 = \dfrac{d}{2} : \dfrac{2d}{2} : \dfrac{3d}{2} [\times \,by \,\dfrac{1}{2}]$

$\Rightarrow r_1 : r_2 : r_3 = r, 2r, 3r$

$\therefore r_1 = r, r_2 = 2r, r_3 = 3r$

Now,

$A_1 = \pi r^2$

$A_2 = \pi (2r)^2 = 4\pi r^2$

$A_3 = \pi(3r)^2 = 9\pi r^2$

Area of region between first and second circles

$= A_2 - A_1 = 4\pi r^2 - \pi r^2 = 3\pi r^2$

Area of region between second and third circles

$= A_3 - A_2 = 9\pi r^2 - 4\pi r^2 = 5\pi r^2$

$\therefore$ Required ratio

$= \pi r^2 : 3\pi r^2 : 5\pi r^2$

On dividing all the three ratios by

$\pi r^2,$ we get required ratio of areas of three regions as

$1 : 3 : 5.$
1764415_1796573_ans_a9af2e3d54984d9cb39f2d7a0820eb91.png

Find the length of the diameter of a circle whose circumference is 44 cm.

1974477_1812409_ans_99b11f5a6963468c9755e087a5133e51.jpg

A circular park, $42 cm$ is a diameter, has a path $3.5$ wide running round it on the outside. Find the cost of travelling the path at $Rs. 4 per\quad m^2$

The diameter of a circular park=

$42 cm$

$\Rightarrow$ the radius of a circular park,

$r= \dfrac{42}{2}=21$
Width of path=

$3.5 m$

$\Rightarrow$ radius of the park with path,

$R= 21+3.5 =24.5m$

$\therefore$ Areaa of path = Area of outer circle-Area of the inner circle
=

$\pi R^2- \pi r^2$
=

$\pi (R^2-r^2)$
=

$\pi (R-r)(R+r)$
=

$\dfrac{22}{7} (24.5-21)(24.5+21)$
=

$\dfrac{22}{7}(3.5)(45.5)$
=

$22 \times 0.5 \times 45.5$
=

$500.5 m^2$
Now, the cost of travelling the path of

$1m^2=Rs4$
The cost of travelling the path of

$500.5m^2=Rs 4 \times 500.5=Rs2002$
1795486_1815493_ans_c92edcc5e8674acf9a369b4523feabf6.PNG

A wire made of silver is looped in the form of circular earring of radius $5.6 cm$ . It is rebent into a square from. Determine the length of the side of the square.

Given:Radius of the circle=

$5.6 cm$
So, the circumference of the circle=

$2 \pi r$
=

$2 \times \frac{22}{7} \times 5.6 =35.2 cm$
Now, The perimeter of square= Circumference of a circle =

$35.2 cm$
Therefore,
Side of a square=

$\dfrac{The\quad perimeter \quad of \quad a \quad square}{4}=\dfrac{35.2}{4}$
=

$8.8 cm$
Hence,the side of a square =

$8.8 cm$

A car has wheels which are $80 cm$ in diameter. How many complete revolution does each wheel make in $10$ minutes when the car is moving at a speed of $80$ km per hour?

Diameter of a car wheel=

$80cm$

$\therefore$ radius =

$r=\dfrac{80}{2}=40cm$
Distance covered in

$1$ revolution= Circumference of wheel

$= 2 \pi r$
=

$2 \times \pi \times 40$
=

$80 \pi \quad cm$
Now, we find total distance covered
We know that,

$speed=\dfrac{Distance}{Time}$
Here,Speed=

$80 km/hr$
Time=

$10$ minutes
=

$\dfrac{10}{60} hour=\dfrac{1}{6} hour$
Putting a value in the formula,We get

$80= \dfrac{Distance}{\dfrac{1}{6}}$

$\Rightarrow$ Distance=

$13.33 km$
=

$13.33 \times 1000 m$
=

$1333.3 m$
=

$1333.3 \times 100 cm$
=

$1333333.3 cm$
Now,
Number of revolutions=

$\dfrac{Total \quad distance}{Distance \quad coverde \quad in \quad 1 \quad revolution}$
=

$\dfrac{1333333.3}{80 \pi}$

=

$\dfrac{1333333.3 \times 7}{80\times 22}$

=

$5303.02$
=

$5303$ (approx).
Hence, number of revolutions =

$5303$

1795532_1815455_ans_b2e9a91acb3f4e37b4d26a1f4cc0d3bf.PNG

The difference between circumference and diameter of circular plot is $105m.$ Find the area of the circular plot.

Given :

Difference between circumference and diameter of circular plot = $105 m$

We know that , the circumference of a circle $2 \pi r$

And diameter of circle = $2r$

According to the question

$2 \pi r - 2r = 105$

$\Rightarrow 2 r( \pi -1) = 105$

$\Rightarrow r = \dfrac {105}{2 (\pi - 1) }$

Now area of circular path $= \pi r^2$

$= \pi ( \dfrac {105}{ 2 ( \pi -1)})^2$

$= \dfrac { 22 \times 105 \times 105}{ 7 \times 4 ( \dfrac {22}{7} -1)^2 }$

$= \dfrac { 11 \times 15 \times 105 \times 49 }{ 2 \times 15 \times 15 }$

$= \dfrac { 11 \times 7 \times 49}{2}$

$= 1886.5 m^2$

The diameter of a circular pond is $17.5 cm$ . It is surrounded by a path of width $3.5m$ . Find the area of the path.

The diameter of circular park =

$17.5 m$

$\Rightarrow$ the radius of circular park

$r = \dfrac { 17.5}{2}$
Width of [path

$= 3.5 m$

$\Rightarrow$ radius of the park with path,

$R = \dfrac {17.5}{2} +3.5 = \dfrac { 17.5 + 7}{2} = 12.25 m$

$\therefore$ Area of path = Area of outer circle - Area of inner circle

$= \pi R^2 - \pi r^2$

$= \pi ( R^2 -r^2 )$

$= \pi ( R-r) (R+ r)$

$= \dfrac {22}{7} ( \dfrac {24.5}{2} - \dfrac {17.5}{2})( \dfrac {24.5}{2} + \dfrac {17.5}{2} )$

$= \dfrac {22}{7} ( \dfrac { 24.5 - 17.5}{2}) ( \dfrac { 24.5 +17.5}{2} )$

$= \dfrac { 22}{7} \times \dfrac {7}{2} \times \dfrac {42}{2}$

$= 231 m^2$

Rajvee walks around a circular park of area $88704 \text{ m}^2$ How long will he take to walk around $10$ rounds at the speed of $4.5$ km per hour?

Area of circular park

$= 88704 m^2$
and Area of circle

$=\pi r^2$

$\implies \pi r^2=88704$

$\Rightarrow \dfrac{22}{7} r^2=88704$

$\Rightarrow r^2=\dfrac{88704 \times 7}{22}$

$\Rightarrow r^2=28224$

$\Rightarrow r= \surd{28224}$

$\Rightarrow r=168~m$
[Taking positive root, because radius can't be negative]
Permeter of circle =

$2 \pi r$

$= 2 \times \dfrac{22}{7} \times 168$

$= 1056~ m$

$=1.056~ km$

$[\because 1000~m=1~km]$
So, total distance after

$10$ rounds

$= 1.056 \times 10=10.56~ km$
Now, we know that
speed

$=\dfrac{Distance}{Time}$
Here, Speed

$= 4.5~ km/hr$
Distance

$= 10.56 ~km$
Putting value in formula, we get

$4.5 = \dfrac{10.56}{\text{Time}}$

$\Rightarrow Time=\dfrac{10.56}{4.5}$

$=2.34$ hour

$=2$ hours

$20$ minutes

$24$ seconds
Hence, Rajvee will take

$2$ hours

$20$ minutes and

$24$ seconds to walk

$10$ rounds.

The area of a circle is $78.5 sq.cm.$ Calculate the circumference of the circle $[Taken \quad \pi=3.14].$

Given:Area of circle =

$78.5 cm^2$

i.e.

$\pi r^2=78.5$

$\Rightarrow 3.14 r^2=78.5$

$\Rightarrow r^2=\dfrac{78.5}{3.14}$

$\Rightarrow r^2 =25$

$\Rightarrow r= 5 \,\,cm$

Now, circumference of circle =

$2 \pi r$

$2 \times 3.14 \times 5$

$31.4 cm$

The length and the breadth of a rectangle are $12 cm$ and $9 cm$ respectively . Find the height of a triangle whose base is $9 cm$ and whose area is one third of rectangle.

Length

$=12cm$

Breadth

$=9cm$

Area of rectangle

$=length\times breadth$

$=12\times9\ cm^2=108cm^2$

$\therefore$ Area of triangle

$=\dfrac{1}{3}$ of Area of rectangle

$=\dfrac{1}{3}\times 108\ cm^2=36cm^2$

Base of the triangle

$=9cm$
Area of triangle

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

$\Rightarrow 36 =\dfrac{1}{2}\times9\times height$

$\Rightarrow height =8 \ cm$

One side of a parallelogram is $18 cm$ and its area is $153 cm^{2}$ . Find the distance of the given side from its opposite side.

It is given that,
Area of parallelogram =

$153 cm^{2}$
One side =

$18 cm$
So the distance between

$AB$ and

$DC$ =

$area/ base$
Substituting the values,

$=\dfrac{ 153}{18}$

$=\dfrac{ 17}{2}$

$= 8.5 cm$ .

1748793_1831368_ans_a1c1c139d2374c76b46476299a3438df.png

The cost of papering four walls of a room at $75$ paisa per square meter is $Rs. 240$ . The height of the room is $5$ meter. Find the length and the breadth of the room. If they are in the ratio $5:3$ .

Let the lenght and breadth be

$5x$ and

$3x$ respectively.
According to the equation.
The cost of papering the four walls of a room paisa per square is

$Rs. 240$

$240=Area \times 0.75$
Area

$=\dfrac {240}{0.75}$
Area

$=\dfrac {240000}{75}$
Area

$=320\ m$
Area

$=2\times$ Height (Length + Breadth)

$320=2\times 5(5x+3x)$

$320=10\times 8x$

$32=8x$

$x=4$
Length

$=5x$

$=5(4)m$

$=20m$
Breadth

$=3x$

$=3(4)m$

$=12m$

How many cubic meters of earth must be dug out to make a well $28$ m deep and $2.8$ m in diameter? Also, find the cost of plastering its inner surface at Rs $4.50$ per sq. meter.

Given,

Radius of the well $= \dfrac{2.8}2 = 1.4\,m$

Depth of the well $= 28\,m$

Hence, the volume of earth dug out $= \pi r^2h$

$= (22/7) \times 1.4 \times 1.4 \times 28$

$= 17248/100 = 172.48\,m^3$

Area of curved surface $= 2\pi rh$

$= 2 \times \dfrac{22}7 \times 1.4 \times 28$

$= 246.40\,m^2$

Now, the cost of plastering at the rate of Rs $4.50$ per sq. m

$= Rs\,246.40 \times 4.50$

$= Rs\,1108.80$

The diameter of a circle is $28 \mathrm{cm}$ . Find its Area.

Let

$r$ be the radius of the circle.

$\begin{aligned}\text { are } a &=\pi r^{2} \\&=\pi\left(\dfrac{28}{2}\right)^{2} \\&=196 \pi \mathrm{cm}^{2} \\&=616 \mathrm{cm}^{2}\end{aligned}$

The diameters of two circles are $32 \mathrm{cm}$ and $24 \mathrm{cm} .$ Find the radius of the circle having its area equal to sum of the areas of the two given circle.

$\textbf{Step 1: Calculating area of two circles}$

$\text{As we know, area of circle}=\pi {r}^2$

$\text{Area of 1st circle}=\pi {r_1}^2=\pi\times16^2=256\pi$

$\text{Area of 2nd circle}=2\pi {r_2}^2=\pi\times12^2=144\pi$

$\textbf{Step 2: Calculating radius}$

$\text{Area of required circle =Area of circle 1 +circle 2}$

$\implies \text{required Area}=256\pi+144\pi$

$\implies \pi R^2=400\pi$

$\implies R=20cm$

$\textbf{Here, the radius of the required circle is 20cm}$

The circumference of a circular field is 308 m. Find its:
(i) Radius
(ii) Area.

Let

$r$ be the radius of the circular field

(i)

$2 \pi r=308$

$\Rightarrow r=\dfrac{308}{2 \pi}\\$

$\Rightarrow r=\dfrac{308}{2} \times \dfrac{7}{22}\\$

$\Rightarrow r=49 \mathrm{m}$

(ii) area

$=\pi r^{2}$

$=\dfrac{22}{7} \times(49)^{2}$

$=7546 \mathrm{m}^{2}$

A storage tank consists of a circular cylinder with a hemisphere stuck on either end. If the external diameter of the cylinder be $1.4\ m$ . and its length be $8\ m$ . Find the cost of painting it on the outside at rate of $Rs\ 20\ per\ m^{2}$ .

$\textbf{Step-1: Find the total surface area of the tank using given information.}$

$\text{Total surface area of a tank=}$

$2\times$

$\text{curved surface area of hemisphere}$

$\text{+ curved surface area of cylinder}$

$\text{Radius of hemisphere}$

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

$\text{Curved surface area of hemisphere}$

$=2\pi r^{2}=2\times \dfrac{22}{7}\times 0.7\times 0.7=3.08\ m^{2}$

$2\times$

$\text{curved surface area of hemisphere}$

$=2\times 3.08\ m^{2}=6.16\ m^{2}$

$\text{Radius of cylinder,}$

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

$\text{Height of cylinder,}$

$h=8\ m$

$\text{Curved surface area of the cylinder}$

$=2\pi rh =2\times \dfrac{22}{7}\times 0.7\times 8=35.2\ m^{2}$

$\text{Total surface area of the storage tank}$

$=35.2+6.16=41.36\ m^{2}$

$\text{Cost of painting its surface area at}$

$Rs\ 20\ per\ m^{2}=41.36\times 20=Rs. 827.2$

$\textbf{Hence, answer is Rs. 827.2}$

2170361_1836526_ans_bd207a1780b44044b9aa5be2a971af64.png

The radii of two circles are $48 \mathrm{cm}$ and $13 \mathrm{cm}$ . Find the area of the circle which has its circumference equal to the difference of the circumferences of the given two circles.

$\textbf{Step 1: Calculating circumference}$

$\text{As we know, circumference of circle}=2\pi r$

$\text{Circumference of 1st circle}=2\pi r_1=2\pi\times48=96\pi$

$\text{Circumference of 2nd circle}=2\pi r_2=2\pi\times13=26\pi$

$\textbf{Step 2: Calculating radius}$

$\text{Circumference of required circle =Circumference of circle 1 -circle 2}$

$\implies \text{required circumference}=96\pi-26\pi$

$\implies 2\pi R=70\pi$

$\implies R=35cm$

$\textbf{Step 3: Calculating area of required circle}$

$\text{Area}=\pi r^2=\pi\times{35}^2=3850 cm^2$

$\textbf{Hence, the area of the required circle is } \mathbf{3850 cm^2}$

The diameter of a circle is $28 \mathrm{cm}$ . Find its circumference.

Let

$r$ be the radius of the circle.

$\begin{array}{l}2 r=28 \mathrm{cm} \\\text { circumference }=2 \pi r \\\quad=28 \pi \mathrm{cm} \\=88 \mathrm{cm}\end{array}$

The diameters of the front and rear wheels of a tractor are $63 \mathrm{cm}$ and $1.54 \mathrm{m}$ respectively. The rear wheel is rotating at $24 \dfrac{6}{11}$ revolutions per minute. Find :
(i) the revolutions per minute made by the front wheel.
(ii) the distance traveled by tractor in 40 minutes.

According to the question, The diameter of the front and real wheels of tractor are

$63 \mathrm{cm}=0.63 \mathrm{m}$ and

$1.54 \mathrm{m}$ respectively.
Radius of the rear wheel

$=\dfrac{1.54}{2}=0.77 \mathrm{m}$ and radius of the front wheel

$=\dfrac{0.63}{2}=0.315 \mathrm{m}$
Distance travelled by tractor in one revolution of rear wheel

$=$ circumference of the rear wheel

$=2 \pi r$

$=2 \times \dfrac{22}{7} \times 0.77=4.84 \mathrm{m}$
The rear wheel rotates at

$24 \dfrac{6}{11}$ revolutions per minute

$=\dfrac{270}{11}$ revolutions per minute since in one revolution the distance travelled by the rear wheel

$=4.84 \mathrm{m}$
So, in

$\dfrac{270}{11}$ revolutions, the tractor travels

$\dfrac{270}{11} \times 4.84=118.8 \mathrm{m}$
Let the number of revolutions made by the front wheel be

$x$ .
(i)
Now, number of revolutions made by the front wheel in one minute

$\times$ Circumference of the wheel

$=$ the distance travalled by the tractor in one minute

$\Rightarrow x \times 2 \times \dfrac{22}{7} \times 0.315=118.8$

$\Rightarrow x=\dfrac{118.8 \times 7}{2 \times 22 \times 0.315}=60$
(ii) Distance travelled by the tractor in 40 minutes
= Number of revolutions madeby the rear wheelin 40 minutes

$\times$ circumference of the rear wheel

$=\dfrac{270}{11} \times 40 \times 4.84=4752 \mathrm{m}$

The radii of two circles are $25 \mathrm{cm}$ and $18 \mathrm{cm}$ . Find the radius of the circle which has a circumference equal to the sum of circumferences of these two circles.

$\textbf{Step 1: Calculating circumference of two circles}$

$\text{As we know, the circumference of circle}=2\pi r$

$\text{Circumference of 1st circle}=2\pi r_1=2\pi\times25=50\pi$

$\text{Circumference of 2nd circle}=2\pi r_2=2\pi\times18=36\pi$

$\textbf{Step 2: Calculating radius}$

$\text{Circumference of required circle =Circumference of circle 1 +circle 2}$

$\implies \text{required circumference}=50\pi+36\pi$

$\implies 2\pi R=86\pi$

$\implies R=43cm$

$\textbf{Hence, the radius of the required circle is 43cm}$

The radii of two circles are in the ratio 3:If the difference between their areas is $2695 \pi \mathrm{cm} ^2$ find the area of the smaller circle.

Let the radius of the two circles be

$3 \mathrm{r}$ and

$8 \mathrm{r}$ . area of the circle having radius

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

$=9 \pi r^{2}$

area of the circle having radius

$8 r=\pi (8 r)^{2} =$

$64 \pi r^{2}$

$\begin{aligned}64 \pi r^{2}-9 \pi r^{2} &=2695 \pi \\55 r^{2} &=2695 \\r^{2} &=49 \\r &=7 \mathrm{cm}\end{aligned}$

Hence radius of the smaller circle is

$3 \times 7=21 \mathrm{cm}$

Area of the smaller circle is given by

$A=\pi r^{2}=\dfrac{22}{7} \times 21^{2}=1386 \mathrm{cm}^{2}$

A wheel has a diameter of 84 cm. Find how many complete revolutions must it make to cover 3.168 km.

Diameter of the wheel

$=84 \mathrm{cm}$ Thus, radius of the wheel

$=42\ \mathrm{cm}$

Circumference of the wheel

$=2 \times \dfrac{22}{7} \times 42=264 \ \mathrm{cm}$

$264\ \mathrm{cm},$ wheel is covering one revolution.

Thus, in

$3.168\ \mathrm{km}=3.168 \times 100000 \ \mathrm{cm},$

number of revolutions covered by the wheel

$=\dfrac{3.168}{264} \times 100000=1200$

A metal wire, when bent in the form of an equilateral triangle of largest area, encloses an area of $484 \sqrt{3} \mathrm{cm}$ If the same wire is bent into the form of a circle of largest area, find the area of this circle.

$\textbf{Step 1: Finding length of the wire}$

$\text{Let the side of the equilateral be }a$

$\text{We know that the area of an equilateral triangle is given by}$

$A=\dfrac{\sqrt{3}}{4}a^2$

$\text{But }A=484\sqrt{3}\text{ (Given)}$

$\therefore \dfrac{\sqrt3}{4}a^2=484\sqrt3$

$a=44$

$\therefore\text{ Perimeter of triangle}=\text{Length of wire}=3a=132\; cm$

$\textbf{Step 2: Finding circle of largest area with this wire}$

$\text{The circle of largest area with this wire will have perimeter equal to the length of the wire}$

$\text{Perimeter of circle}=2\pi r$

$2\pi r=132$

$2\times\dfrac{22}{7}r=132$

$r=21\;cm$

$\text{Area of circle}=\pi r^2=\dfrac{22}{7}\times21\times21=1386\;cm^2$

$\textbf{Hence, the circle of largest area which can be formed with this wire has an area}\mathbf{=1386\;cm^2}$

In the given figure, the area of the shaded portion is $770 \mathrm{cm}$ . If the circumference of the outer circle is $132 \mathrm{cm},$ find the width of the shaded portion.

Given that the circumference of the outer circle is

$132 \mathrm{cm}$

Thus, we have,

$2 \pi R=132 \mathrm{cm}$

$\Rightarrow R=\dfrac{132 \times 7}{2 \times 22}$

$\Rightarrow R=21 \mathrm{cm}$

Area of the bigger circle

$=\pi R^{2}$

$\begin{array}{l}=\dfrac{22}{7} \times 21^{2} \\=1386 \mathrm{cm}^{2}\end{array}$

Also given the area of the shaded portion.

Thus the area of the inner circle = Area of the outer circle - Area of the shaded portion

$\begin{array}{l}=1386-770 \\=616 \mathrm{cm}^{2}\end{array}$

$\Rightarrow \pi r^{2}=616$

$\Rightarrow r^{2}=\dfrac{616 \times 7}{22}$

$\rightarrow r^{2}=196$

$\Rightarrow r=14 \mathrm{cm}$

Thus, the width of the shaded portion

$=21-14=7 \mathrm{cm}$

The shaded portion of the figure given alongside shows two concentric circles. If the circumference of the bigger and smaller circles be $396 \mathrm{cm}$ and $374 \mathrm{cm},$ find the area of the shaded portion.

${\textbf{Step -1: Find the radius of both bigger and smaller circle.}}$

${\text{As we have given,}}$

${\text{the circumference of the bigger circle is 396cm}}{\text{.}}$

${\text{and the circumference of the smaller circle is 374cm}}{\text{.}}$

${\text{Let}}$ $R$ ${\text{be the radius of bigger circle and }}$ $r$ ${\text{be the radius of smaller circle}}{\text{. }}$

${\text{Therefore, from the given data}}$

${\text{Circumference of bigger circle is given by,}}$

$\Rightarrow 396 = 2 \times \dfrac{{22}}{7} \times R$ $\boldsymbol{\left[\mathbf {\because {\textbf{Circumference of a circle = 2}}\pi {\textbf{r}}} \right]}$

$\Rightarrow R = \dfrac{{7 \times 396}}{{2 \times 22}}$

$\Rightarrow R = \dfrac{{7 \times 198}}{{22}}$

$\Rightarrow R = 7 \times 9$

$\Rightarrow R = 63{\text{cm}} \ldots \left( 1 \right)$

${\text{Now, Circumference of smaller circle is given by,}}$ $= 2\pi r$

$\Rightarrow 374 = 2 \times \dfrac{{22}}{7} \times r$ $\boldsymbol{\left[\mathbf {\because {\text{Circumference of a circle = 2}}\pi {\text{r}}} \right]}$

$\Rightarrow r = \dfrac{{7 \times 374}}{{2 \times 22}}$

$\Rightarrow r = \dfrac{{7 \times 187}}{{22}}$

$\Rightarrow r = 59.5{\text{cm}} \ldots \left( 2 \right)$

${\textbf{Step -2: To find area of shaded part take a difference of are of both the circles.}}$

${\text{Are}}{{\text{a}}_{{\text{Bigger circle}}}} - {\text{Are}}{{\text{a}}_{{\text{smaller circle}}}}$ $=$ ${\text{Area of shaded part}}$

$\Rightarrow \pi {R^2} - \pi {r^2}$ $=$ ${\text{Area of shaded part}}$ $\boldsymbol{\left[ {{\textbf{Using the formula: Area of a circle = }}\pi {{\textbf{r}}^2}} \right]}$

$\Rightarrow \pi \left( {{R^2} - {r^2}} \right)$ $=$ ${\text{Area of shaded part}}$

$\Rightarrow \dfrac{{22}}{7}\left( {{{\left( {63} \right)}^2} - {{\left( {59.5} \right)}^2}} \right)$ $=$ ${\text{Area of shaded part}}$

$\Rightarrow {\text{Area of shaded part = }}\dfrac{{22}}{7}\left( {63 + 59.5} \right)\left( {63 - 59.5} \right)$

$\Rightarrow {\text{Area of shaded part = }}\dfrac{{22}}{7}\left( {122.5} \right)\left( {3.5} \right)$

$\Rightarrow {\text{Area of shaded part = 1347}}{\text{.5c}}{{\text{m}}^2}$

${\textbf{Hence, the area of shaded part is 1347}}{\textbf{.5c}}{{\textbf{m}}^\mathbf2}\textbf.$

The circumference of a given circular park is 55 m. It is surrounded by a path of uniform width 3.5 ${m}$ . Find the area of the path.

Let

$r$ be the radius of the circular park.

$2 \pi r=55$

$\begin{aligned}r &=\dfrac{55}{2 \pi} \\&=8.75 \mathrm{m}\end{aligned}$

area of the park

$=\pi \times(8.75)^{2}=240.625 \mathrm{m}^{2}$

Radius of the outer circular region including the path is given by

$\begin{aligned} R &=8.75+3.5 \\ &=12.25 \mathrm{m} \end{aligned}$

Area of that circular region is

$A=\pi \times(12.25)^{2}=471.625 \mathrm{m}^{2}$

Hence area of the path is given by Area of the path

$=471.625-240.625=231 \mathrm{m}^{2}$

There are two circular gardens A and B. The circumference of garden $A$ is 1.760 km and the area of garden B is 25 times the area of garden A. Find the circumference of garden B.

Let

$r$ be the radius of the circular garden

$A$ .

Since the circumference of the garden

$A$ is

$1.760 \mathrm{Km}=1760 \mathrm{m}$ , we have,

$2 \pi r=1760 m$

$\Rightarrow r=\dfrac{1760 \times 7}{2 \times 22}=280 \mathrm{m}$

Area of garden

$A=\pi r^{2}=\dfrac{22}{7} \times 280^{2} \mathrm{m}^{2}$

Let R be the radius of the circular garden B.

since the area of garden

$B$ is 25 times the area of garden

$A$ , we have,

$\pi R^{2}=25 \times \pi r^{2}$

$\rightarrow \pi R^{2}=25 \times \pi \times 280^{2}$

$\rightarrow R^{2}=1960000$

$\Rightarrow R=1400 \mathrm{m}$

Thus circumference of garden

$B=2 \pi R=2 \times \dfrac{22}{7} \times 1400=8800 m=8.8 \mathrm{Km}$

The radius of a circle is $5 \mathrm{m}$ . find the circumference of the circle whose area is 49 times the area of the given circle.

Area of the circle having radius

$85 \mathrm{m}$ is

$\begin{aligned} A &=\pi \times(85)^{2} \\ &=7225 \pi \mathrm{m}^{2} \end{aligned}$

Let the radius of the circle whose area is 49 times of the given circle

$=r$

$\pi r^{2}=49 \times\left(\pi \times 5^{2}\right)$

$\begin{aligned} r^{2} &=(7 \times 5)^{2} \\ r &=35 \end{aligned}$

$\begin{aligned} S &=2 \pi r \\ &=2 \pi \times 35 \\ &=220 \mathrm{m} \end{aligned}$

The diameters of three circles are in the ratio $3: 5: 6 .$ If the sum of the circumferences of these circles be $308 \mathrm{cm} ;$ find the difference between the areas of the largest and the smallest of these circles.

Let the diameter of the three circles be

$3 \mathrm{d}, 5 \mathrm{d}$ and

$6 \mathrm{d}$

$\begin{aligned}\pi \times 3 d+\pi \times 5 d+\pi \times 6 d &=308 \\14 \pi d &=308 \\d &=7\end{aligned}$

radius of the smallest circle

$=\dfrac{21}{2}=10.5$

$\begin{aligned}\text { Area } &=\pi \times(10.5)^{2} \\&=346\end{aligned}$

radius of the largest circle

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

$\begin{aligned}\text { Area } &=\pi \times(21)^{2} \\&=1385.5 \\\text { difference } &=1385.5-346 \\&=1039.5\end{aligned}$

Each wheel of a car is of diameter 80 cm. How many completer revolutions does each wheel make in 10 minutes when the car is traveling at a speed of $66 \mathrm{km}$ per hour?

The car travels in 10 minutes

$=\dfrac{66}{6}$

$=11 \mathrm{km}$

$=1100000 \mathrm{cm}$

Circumference of the wheel = distance covered by the wheel in one revolution Thus, we have,

Circumference

$=2 \times \dfrac{22}{7} \times \dfrac{80}{2}=251.43 \mathrm{cm}$

Thus, the number of revolutions covered by the wheel in

$1100000 \mathrm{cm}=\dfrac{1100000}{251.43} \approx 4375$

The figure given below depits a racing track whose left and right ends are semicircle.
The distance between the two inner parallel line segment is $60\ m$ and they are each $106\ m$ long. If the track is $10\ m$ wide, find:
The distance around the track along its inner edge.

The distance around the track along its inner edge:

$=AB+Arc BEC+CD+$ Area of Arc

$DFA$

$=106+\dfrac {1}{2}\times 2\pi r+106+\dfrac {1}{2}\times 2\pi r$

$=212+\dfrac {1}{2}\times 2\times \dfrac {22}{7}\times 30+\dfrac {1}{2}\times 2\times \dfrac {22}{7}\times 30$

$=212+2\times \dfrac {22}{7}\times 30$

$=212+\dfrac {1320}{7}$

$=\dfrac {2804}{7}m$ .
The Area of Track:
= (Area of

$GHIJ-$ Area of

$ABCD$ ) + (Area of semicircle

$HKI-$ Area of semicircle

$BEC$ ) + (Area of semicircle

$GIJ-$ Area of semicircle

$AFD$ )

$=106\times 80-106\times 60+\dfrac {1}{2}\times \dfrac {22}{7}\times (40)^2$

$-\dfrac {1}{2}\times \dfrac {22}{7}\times (30)^2 +\dfrac {1}{2}\times \dfrac {22}{7}\times (40)^2$

$-\dfrac {1}{2}\times \dfrac {22}{7}\times (30)^2$

$=106(80-60)+\dfrac {22}{7}\times (40)^2-\dfrac {22}{7}\times (30)^2$

$=106\times 20+\dfrac {22}{7}(40^2 -30^2)$

$=2120+\dfrac {22}{7}(40+30)(40-30)$

$=2120+\dfrac {22}{7}\times 10\times 70$

$=2120+2200$

$=4320\ sq.cm$

$\therefore$ Total area of Track

$=4320\ sq.cm$

The figure given below depits a racing track whose left and right ends are semicircle.
The distance between the two inner parallel line segment is $60\ m$ and they are each $106\ m$ long. If the track is $10\ m$ wide, find:
The area of the track.

The distance around the track along its inner edge:

$=AB+Arc BEC+CD+$ Area of Arc

$DFA$

$=106+\dfrac {1}{2}\times 2\pi r+106+\dfrac {1}{2}\times 2\pi r$

$=212+\dfrac {1}{2}\times 2\times \dfrac {22}{7}\times 30+\dfrac {1}{2}\times 2\times \dfrac {22}{7}\times 30$

$=212+2\times \dfrac {22}{7}\times 30$

$=212+\dfrac {1320}{7}$

$=\dfrac {2804}{7}m$ .
The Area of Track:
= (Area of

$GHIJ-$ Area of

$ABCD$ ) + (Area of semicircle

$HKI-$ Area of semicircle

$BEC$ ) + (Area of semicircle

$GIJ-$ Area of semicircle

$AFD$ )

$=106\times 80-106\times 60+\dfrac {1}{2}\times \dfrac {22}{7}\times (40)^2$

$-\dfrac {1}{2}\times \dfrac {22}{7}\times (30)^2 +\dfrac {1}{2}\times \dfrac {22}{7}\times (40)^2$

$-\dfrac {1}{2}\times \dfrac {22}{7}\times (30)^2$

$=106(80-60)+\dfrac {22}{7}\times (40)^2-\dfrac {22}{7}\times (30)^2$

$=106\times 20+\dfrac {22}{7}(40^2 -30^2)$

$=2120+\dfrac {22}{7}(40+30)(40-30)$

$=2120+\dfrac {22}{7}\times 10\times 70$

$=2120+2200$

$=4320\ sq.cm$

$\therefore$ Total area of Track

$=4320\ sq.cm$

Find the area of the shaded region in the given figure, if radii of the two concentric circles with center $O$ are $7\ cm$ and $14\ cm$ respectively and $\angle AOC=40^\circ$ .

Given: Radius of small circle

$r= 7\ cm$

Radius of big circle

$R=14\ cm$

$\angle AOC =\theta =40^\circ$

Area of

$ABCD=$ Area of

$OAC-$ Area of

OBD$$

$=\dfrac {\theta}{360}\times \pi R^2 - \dfrac {\theta}{360}\times \pi r^2$

$=\dfrac {40}{360}\times \dfrac {22}{7}\times (14)^2-\dfrac {40}{360}\times \dfrac {22}{7}\times (7)^2$

$=\dfrac {40}{360}\times \dfrac {22}{7}\times 14\times 14-\dfrac {40}{360}\times \dfrac {22}{7}\times \dfrac {22}{7}\times 7\times 7$

$=\dfrac {40}{360}\times 22\times 28-\dfrac {40}{360}\times 154$

$=\dfrac {40}{360}\times 616-\dfrac {40}{360}\times 154$

$=\dfrac {1}{9}\times 616-\dfrac {1}{9}\times 154$

$=\dfrac {616}{9}-\dfrac {154}{9}$

$=\dfrac {616-154}{9}$

$=\dfrac {462}{9}$

${=\dfrac {154}{3}}$

$=51.3\ sq.cm$

Diagonals $AC$ and $BD$ of a quadrilateral $ABCD$ intersect each other at $P$ show that
$ar. (\triangle APB)\times ar. (\triangle CPD)=ar. (\triangle APD)\times ar. (\triangle BPC) .$

Given: Diagonals

$AC$ and

$BD$ of a quadrilateral

$ABCD$ intersect each other at

$P$ .
To Prove:

$ar.(\triangle APB)\times ar.(\triangle CPD)=ar.(\triangle APD)\times ar.(\triangle BPC)$
Construction: Draw

$AM\bot DB, CN\bot DB$

Proof:

$ar.(\triangle APB)\times ar.(\triangle CPD)=$

$=\left(\dfrac 12\times PB\times AM\right) \times \left(\dfrac 12\times PD\times CN\right)$

$=\dfrac 14 \times PB \times AM\times PD\times CN....(i)$

$ar.(\triangle APD)\times ar.(\triangle BPC)=$

$=\left(\dfrac 12\times PD\times AM\right)\times \left(\dfrac 12\times PB\times CN \right)$

$=\dfrac 14 \times PD\times AM\times PB\times CN....(ii)$

From

$(i)$ and

$(ii),$

$ar.(\triangle APB)\times ar.(\triangle CPD)=ar.(\triangle APD)\times ar.(\triangle BPC)$

Hence,

$Proved.$

There is a slide in a park. One of its side walls has been painted in some colour with a message "KEEP THE PARK GREEN AND CLEAN". If the sides of the wall are $15m,11m$ and $6m$ , find the area painted in colour.

Sides

$a= 15\ m, b = 11\ m, \ c = 6\ m$

$\therefore s = \dfrac {a +b+c}{2}$

$= \dfrac {15+11+6}{2}$

$= \dfrac {32}{2}$

$s = 16\ m$

$\therefore$ Area of triangular field (painted area)

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

$= \sqrt { 16(16-15)(16-11)(16-6) }$

$= \sqrt { 16 \times 1 \times 5 \times 10}$

$= 4 \sqrt {50}$

$= 4 \sqrt {25 \times 2}$

$A = 20 \sqrt {2}\ sq.m$

The circumference of the circle is equal to the perimeter of a square. If area of the square is $484\,sq\,m$ , then find the area of the circle.

Let side of square

$= x\,m$
then perimeter of square

$4 \times x$
and area of square

$= x^2$
According to question
Area of square

$=484\,sq\,m$

$x^2 = 484$

$x= \sqrt{484}$

$= 22\,m$
Perimeter of square

$= 4x = 4 \times 22 = 88\,m$
Let radius of circle is r
circumference of circle = perimeter of square

$\Rightarrow 2\pi r = 88$

$\Rightarrow2 \times \dfrac{22}{7} \times r = 88$

$r =\dfrac{88 \times 7}{2 \times 22}= 14 m$
Thus, area of the circle

$= \pi r^2$

$=\dfrac{22}{7} \times 14 \times 14 = 616\,sq\,m.$
Thus, area of the circle

$= 616\,sq\,m.$

A wooden bookshelf has external dimensions as follows: Height $= 110\ cm,$ Depth $= 25\ cm,$ Breadth $= 5\ cm.$ The thickness of the plank is $5 cm$ everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is $20$ paise per $cm^2$ and the rate of painting is $10$ paise per $cm^2$ , find the total expenses required for polishing and painting the surface of the bookshelf.

To calculate the external surface area of the shelf while leaving out the front face of the shelf $= lh+2(lb+bh)$

$= [85\times110+2(85×25+25×110)]$

$= (9350+9750)$

$= 19100\ cm^2$

Hence, the external surface area of shelf is $19100 \ cm^2$

Area of front face $= [85×110-75×100+2(75×5)]$

$= 1850+750$

$= 2600\ cm^2$

Thus, the area is $2600\ cm^2$

Area to be polished surface $= (19100+2600)\ cm^2$

$= 21700 cm^2$

Given cost of polishing $1\ cm^2$ area $= Rs.\ 0.20$

Therefore, the cost of polishing $21700\ cm^2$ area $Rs.\ (21700×0.20)$

$=Rs.\ 4340$

Dimensions of row of the book shelf

Length $(l) = 75\ cm$

Breadth $(b) = 20\ cm$ and

Height $(h) = 30\ cm$

Area to be painted in one row $= 2(l+h)b+lh$

$= [2(75+30)× 20+75×30]$

$= (4200+2250)$

$= 6450$

So, area is $6450 cm^2$

Painting area

Area to be painted in 3 rows $= (3×6450)\ cm^2$

$= 19350\ cm^2$

Cost of painting $1\ cm^2$ area $= Rs.\ 0.10$

Cost of painting $19350\ cm^2$ area $= Rs.\ (19350 x 0.1)$

$= Rs.\ 1935$

Total expense required for polishing and painting $= Rs.\ (4340+1935)$

$= Rs.\ 6275$

Find the circumference of circle inscribed in a square of sides $14\,cm.$

Given: Side of square

$= 14\,cm$

Diameter of circle inscribed in square

$=$ side of square

$= 14\,cm$

Thus, radius of circle is

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

The circumference of circle

$= 2\pi r$

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

$= 44\,cm$

Thus circumference of the inscribed circle

$= 44\,cm$

The external and internal diameters of hemispheric bowl are $43 \mathrm{cm}$ and $42 \mathrm{cm}$ respectively. If the coloring cost is paise 7 per $\mathrm{cm}^{2}$ , find the total cost of coloring the bowl.

Given,
External diameter of hemispherical bowl

$=43 \mathrm{cm}$

$\therefore$ External radius

$\left(\mathrm{r}_{1}\right)=\dfrac{43}{2} \mathrm{cm}$
Internal diameter

$=42 \mathrm{cm}$

$\therefore$ Internal radius

$\left(\mathrm{r}_{2}\right)=\dfrac{42}{2}=21 \mathrm{cm}$
External surface area of bowl

$=2 \pi r_{1}^{2}$
And internal surface area

$=2 \pi r_{2}{ }^{2}$
Total surface area

$=2 \pi\left(r_1^{2}+r_{2}^{2}\right)$

$\begin{array}{l}=2 \times \dfrac{22}{7}\left[\left(\dfrac{43}{2}\right)^{2}+(21)^{2}\right] \\=2 \times \dfrac{22}{7} \times(462.25+441)\end{array}$

$=2 \times \dfrac{22}{7} \times 903.25 \mathrm{cm}^{2}$

$\therefore$ cost of colouring

$1 \mathrm{cm}^{2}=7$ paise

$=0.07$

$\begin{array}{l}\text { cost for } \dfrac{2 \times 22 \times 903.25}{7} \mathrm{cm}^{2} \\=2 \times \dfrac{22}{7} \times 903.25 \times 0.07 \\=₹ 397.43\end{array}$
Hence, cost of coloring the bowl = ₹ 397.43

A rhombus shaped filed has green grass for $18$ cows to graze. If each side of the rhombus is $30 m$ and its longer diagonal is $48 m$ , how much area of grass field will each cow be getting?

$ABCD$ is a rhombus.

$Area ( \Delta ABD) =Area ( \Delta BDC)$
In

$\Delta ABD , a = 30 m , b =30 m , c = 48 m$

$\therefore$ perimeter

$a +b +c = 30 +30 +48 = 108 m$

$\therefore s = \dfrac { a +b +c}{2} = \dfrac {108}{2} = 54 m$

$\therefore$ Area of

$\Delta ABD , A$

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

$= \sqrt {54(54-30)(54-30)(54-48) }$

$=\sqrt { 54 \times 24 \times 24 \times 6}$

$= \sqrt {(24)^2 \times 324}$

$= 24 \times 18$

$= 432 sq.m.$
But area of whole field

$= 2 \times$ area of

$\Delta ABD$

$=2 \times 432$

$=64 sq.m.$

$\therefore$ Are required for growing grass for

$18$ cows is

$64 sq.m.$

$\therefore$ Area required for growing grass for

$1$ cow....??

$\dfrac {864}{18} = 48 sq.m.$

The front compound wall of a house is decorated by wooden spheres of diameter $21 cm$ , placed on small supports as shown in figure
Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius $1.5 cm$ and height $7 cm$ and is to be painted black. Find the the cost of paint required if silver paint costs $25$ paise per $am^2$ and black paint costs $5$ paise per $cm^2$ .

Diameter of wooden frame

$d = 21 cm$
Radius

$r = \dfrac {21}{2}$
Outer surface area of wooden spheres,

$A = 4 \pi r^2$

$= 4 \times \dfrac {22}{7} \times ( \dfrac {21}{2})^2$

$= 4 \times \dfrac {22}{7}\times \dfrac {21}{1} \times \dfrac {21}{2}$

$\therefore A= 1386 cm^2$
Radius of cylinder

$r_1 = 1.5 cm$
height

$h = 7cm$
Curved surface area of cylinder support

$A = 2 \pi rh$

$= 2 \times \dfrac {22}{7} \times (1.5)^2 \times 7$

$= 66 cm^2$
Surface area of sphere

$A = \pi r^2$

$= \dfrac {22}{7} \times (1.5)^2$

$= \dfrac {22}{7} \times 2.25$

$= 7.07 cm^2$
Area of silver painted

$= [ 8 \times (1386 - 7.07)] cm^2$

$= 8 \times 1378.93$

$= 11031.44 cm^2$
Cost of silver paint

$= 11031.44 \times 0.25$

$= rs 2757.86$
Area of black paint

$=( 8 \times 66) cm^2$

$=528 cm^2$
Cost of black paint

$= 528 \times 0.05$

$=Rs 26.40$
Total cost of silver and black paint

$=Rs 2757.76 + Rs.26.40$

$=Rs.2784.26$

The triangular side walls of a flyover have been used for advertisements. The sides of the walls are $122 m, 22m$ and $120m$ . The advertisements yield an earning of $Rs.5000$ per $m^2$ per year. A company hired one of its walls for $3$ months. How much rent did it pay?

Sides of triangle

$a = 122 m , b= 22 m, c= 120 cm$

$\therefore s = \dfrac {a+b+c}{2}$

$= \dfrac {122 +22 +120}{2}$

$=\dfrac {264}{2}$

$s = 132 m$

$\therefore$ area of triangle A

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

$= \sqrt{ 132 (132 -122)(132-22)(132-120) }$

$= \sqrt { 132 \times 10 \times 110 \times 12 }$

$= \sqrt {1320 \times 1320}$

$= \sqrt { (1320)^2 }$

$\therefore A = 1320 sq.cm$
Rent for each year, each sq m =

$rs 5000$
Rent for each month each sq.m =

$\frac {5000}{12}$
Rent for each quarterly, each sq.m

$= 3 \times \frac {5000}{12} = rs 1250$

$\therefore$ Rent for

$1320 sq. m =1250 \times 1320 =Rs . 1650000$

$\therefore$ Amount company has to pay is

$rs . 1650000$ .

A pathway $4$ metre wide has been constructed centrally in a rectangular garden parallel to its sides of length $60$ metre and width $40$ metre. Find the area of the pathway and find the total expenditure to cover this pathway with sand at the rate of $Rs\,100$ per sq. metre.

Length of the garden

$= 60\,m$
Breadth of the garden

$= 40 \,m$
and width of the path

$= 4\,m$

$\therefore$ Area of the path along length

$= 60 \times 4 = 240\,m^2$
Area of the path along breadth

$= 40 \times 4\,m = 160\,m^2$
Common area

$= 4m \times 4m = 16\,m^2$

$\therefore$ Total area of the path

$= 240 + 160 - 16 = 384\,m^2$

$\because$ Expenditure of lying sand per

$m^2$ Rs

$100$ .

$\therefore$ Total expenditure of sand on

$384\,m^2 = 384 \times 100 = Rs\,38,400$
1882315_1880567_ans_b9005a4314824350bdde9f2883c44096.png

The length, breadth, and height of a room are $15.35\,m$ , $4.65\,m$ , and $6.50\,m$ respectively. If there are $4$ doors each of size $1.5\, m \times 1.3\,m$ and $3$ windows each of size $1.5\, m \times 1.2\,m$ , then find the cost to paint the four walls of room at the rate of Rs $3.40$ per sq. m.

Here

$l = 15.35\,m$ ,

$b = 4.65\,m$ and

$h = 6.50\,m$

$\therefore$ Area of four walls

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

$= 2 \times (15.35 + 4.65) \times 6.50$

$= 2 \times 20 \times 6.50$

$= 260 m^2$

Area of four doors

$= 4 \times 1.5 \times 1.3$

$= 7.8\,m^2$

Area of

$3$ windows

$= 3 \times 1.5 \times 1.2$

$= 5.4\,m^2$

The area of the four walls excluding door and windows

$= 260 - (7.8 + 5.4)$

$= 260 - 13.2$

$= 246.8\,m^2$

$\because$ Cost to paint

$1\,m^2$ is Rs

$3.40$

$\therefore$ Cost to paint

$246.8\,m^2$

$= 246.8 \times 3.40$

$= Rs\,839.12$

A triangular plot whose two sides are $8\,cm$ and $15\,cm$ and its perimeter is $40\,cm$ , then find the area of the triangular plot.

Perimeter of triangular plot = sum of all the three sides

$40 = 8 + 15 + third\,side$

$\Rightarrow$ Third side

$= 40 - 23 = 17\,cm$
Area of triangular plot=

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

$=\sqrt{20(20-8)(20-15)(20-17)}$

$=\sqrt{20 \times 12 \times 5 \times 3}$

$=2 \times 3 \times 3 \times 17$

$=5 \times 4 \times 3=60\,m^2$

The length of two adjacent sides of a plot in the form of parallelogram are $39$ meters and $25$ meters and its diagonal is $56$ metre. Find the cost of leveling the plot at the rate of Rs $100$ per sq. metre.

Area of parallelogram ABCD

$= 2 \times$ (area of

$\Delta ABC$ ) …(i)

First of all, we need to find the area of

$\Delta ABC$ which is the same as the area of

$\Delta ADC$ .

$\therefore s=\dfrac{39+25+56}{2}=\dfrac{120}{2}=60$

$\therefore\,Area\ of \Delta ABC$

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

$=\sqrt{60(60-39)(60-25)(60-56)}$

$=\sqrt{60 \times 21 \times 25 \times 4}$

$=2 \times 2 \times 3 \times 5 \times 7$

$=420\,sq.\,m$

$\therefore$ Using relation (i), we will get the area of parallelogram ABCD
i.e. area of parallelogram ABCD

$= 2 \times 420$

$= 840\,sq.\,m.$
∵ Cost of levelling

$1\,sq.\,m$ is Rs

$100$
∴ Cost of levelling

$840\,sq.\,m$ is

$840 \times 100$
= Rs

$84,000$

1882664_1880198_ans_493859cfbc3c4385aff91fe459105fbc.png

The sides of a ground in the form of a cyclic quadrilateral are $72\,m$ , $154\,m$ , $80\,m$ , and $150\,m$ respectively. Find the cost of paving the tile on the ground at Rs $5$ per sq. metre.

Here

$a = 72\,m$ ,

$b = 154\,m$ ,

$c = 80\,m$ and

$d = 150\, m$

$s=\dfrac{a+b+c+d}{2}$

$s=\dfrac{72+154+80+150}{2}$

$s=\dfrac{456}{2}=228\,m$

$\therefore$ Area of ground which is in the form of cyclic quadrilateral.

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

$=\sqrt{(228-72)(228-154)(228-80)(228-150)}$

$=\sqrt{156 \times 74 \times 148 \times 78}$

$= 78 \times 74 \times 4$

$=11,544\,sq.\,m.$

$\because$ Cost of paving the tiles on the ground

$1\,sq.\,m$ is Rs

$5$ .

$\therefore$ Cost of paving the tiles on the ground

$11,544\,sq.\,m$ is

$11,544 \times 5 = Rs\,57,720$ .

A rectangular water tank is $10$ metre long, $8$ metre wide and $2$ metre deep. Find the total cost to repair its four walls and bottom at the rate of Rs $15$ per square metre.

Floor area of water tank

= length

$\times$ breadth

$= 10 \times 8 = 80\,m^2$

Area of four walls of the water tank

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

$= 2 \times (10 + 8) \times 2$

$= 2 \times 18 \times 2$

$= 72 sq.\,m$

Total area

$= (72 + 80)\,m^2$

$= 152\,m^2$

$\because$ Cost of repairing

$1\,m^2$ is Rs

$15$ .

$\therefore$ Cost of repairing

$152\,m^2 = Rs\,15 \times 152$

$= Rs\,2,280$

A room is $10\,m$ long, and $5\,m$ wide. How many squares of area $50\,sq.\,cm$ are required? Also, find the cost of paving squares @ of Rs $20$ each.

Area of room = length

$\times$ breadth

$= 10\,m \times 5\,m$

$= 50\,sq.\,m$

Area of one square

$= 50\,sq.\,m$

i.e.

$\dfrac{50}{100 \times 100}sq.\,m$

$\therefore$ Number of squares required

$=\dfrac{50\,sq.\,m}{\dfrac{50}{100 \times 100}sq.\,m}=\dfrac{50}{50} \times 100 \times 100$

$=10,000$ squares

$\because$ Cost of paving

$10000$ squares @ Rs

$20$ $.

$= 10000 \times 20$

$= Rs\,2,00,000$

The area of a parallelogram-shaped field is $2000\,sq.\,m$ . If its base is $50\,m$ then find its height.

Area of parallelogram-shaped field
= base

$\times$ corresponding altitude

$2000\,m^2= 50\,m \times h$

$\Rightarrow h=\dfrac{2000\,m^2}{50\,m}=40\,m$
Hence, height of the field is

$40\,m$ .

Find the length of the side of an equilateral triangle whose area is $9\sqrt{3}\,sq.\,cm$ .

Area of equilateral triangle

$\displaystyle =\sqrt{\dfrac34}\displaystyle a^{2}$

$9\sqrt{3}=\sqrt{\dfrac34}\displaystyle a^{2}$

$\displaystyle \Rightarrow a=6cm$

Find the base of an isosceles triangle whose area is $12\,sq.\,cm$ , and length of one of the equal side is $5\,cm$ .

Let equal sides be

$(a) = 5\,cm$ and base

$(b) =?$
Area of an isosceles triangle

$= 12\,sq.\,cm$
Area of an isosceles triangle

$=\dfrac{b}{4}\sqrt{4a^2-b^2}$

$12=\dfrac{b}{4}\sqrt{4 \times (5)^2-b^2}$

$48=b\sqrt{100-b^2}$

Squaring both sides

$2304=b^2(100-b^2)$
or

$b^4-100b^2+2304=0$
or

$b^4-64b^2-36b^2+2304=0$
or

$b^2(b^2-64)-36(b^2-64)=0$
or

$(b^2-64)(b^2-36)=0$
either

$(b^2-64)=0 \Rightarrow b^2=64 \Rightarrow b=\pm 8$
or

$(b^2-36)=0 \Rightarrow b^2=36 \Rightarrow b=\pm 6$
Hence

$base=8\,cm$ , or

$6\,cm$ .

Prove that the area of a triangle having sides $48\,m$ , $10\,m$ and $12\,m$ is $15\sqrt{7}\,m^2$ .

The sides of a triangle are

$8\,m$ ,

$10\,m$ and

$12\,m$ respectively.

Let it be

$a = 8\,m$ ,

$b = 10\,m$ and

$c = 12\,m$

$\therefore$ Semi perimeter(s)

$=\dfrac{a+b+c}{2}$

$=\dfrac{8+10+12}{2}=15\,m$

$\therefore$ Area of the triangle

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

$=\sqrt{15(15-8)(15-10)(15-12)}$

$=\sqrt{15 \times 15 \times 7}$

$=15\sqrt{7}$

Hence, area of the traingle

$=15\sqrt{7}\,m^2$

If the area and base of a parallelogram are $174.60\,sq.\,m$ and $18\,m$ respectively, write its height.

Area of a parallelogram = base

$\times$ height

$height=\dfrac{area}{base}=\dfrac{174.60}{18}=9.7\,m$

Find the area of a triangular field, the length of whose sides are $275\,m$ , $660\,m$ , and $715\,m$ . What is the cost of cultivating the field at the rate of $\,200$ per hectare?

Suppose sides of a triangular field are

$a = 275\,m,\,b = 660\,m$ and

$c = 715 m$

$\therefore$ Semi perimeter(s)

$=\dfrac{a+b+c}{2}$

$=\dfrac{275+660+715}{2}$

$\Rightarrow$

$s=825\,m$

Area of triangular field

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

Area of triangular field

$=\sqrt{825(825-275)(825-660)(825-715)}$

Area of triangular field

$=\sqrt{825 \times 550 \times 165 \times 110}\,m^2=90759\,m^2$

$\because$ Cost of cultivating

$1$ hectare

i.e.

$10000\, m^2=Rs.200$

$\therefore$ Cost of cultivating

$90750\,m^2\,area=\dfrac{200}{10000} \times 90750=Rs.1815$

Find the altitude of a triangular region whose base is $40\,m$ , and area is $560\,m^2$ .

Area of triangular region

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

$560\,m^2 =\dfrac{1}{2} \times 40\,m \times altitude$

$\Rightarrow 560\,m^2 = 20\,m \times altitude$

$\Rightarrow altitude = 28\,m$

A room is $8\,m\,\,50\,cm$ long, and $6\,m\,\,50\,cm$ wide. What should be the length of the mat of width $25$ cm to cover the whole floor? Find the total cost of the mat at the rate of $\,20\,per\,m$ .

Length of room

$= 8.50\,m$

Breadth of room

$= 6.50\,m$

Area of room

$= l \times b = (8.50 \times 6.50)\,m^2$

Let length of the mat be x.

Area of mat

$=\bigg(\dfrac{25}{100} \times x\bigg)$

$\therefore 8.50 \times 6.50 \,m^2=\dfrac{25}{100} \times x$

$\Rightarrow \dfrac{8.50 \times 6.50}{25}=x$

$\Rightarrow x=221\,m$

Cost of

$1$ m of mat is

$₹\,20$

Cost of

$221\,m$ of mat

$= 221 \times 20 = ₹\,4,420$ .

The diameters of three circles are in the ratio $3:5:6.$ If the sum of the circumferences of these circles be 308 cm; find the difference between the areas of the largest and the smallest of these circles.

${\textbf{Step -1: Assume a variable and find its value}}{\textbf{.}}$

${\text{Given the ratio of diameter of circle as 3 : 5 : 6 and circumference = 308 cm}}{\text{.}}$

${\text{Let us assume a variable = x}}{\text{.}}$

${\text{So, the diameter of the circle = 3x, 5x, 6x}}{\text{.}}$

$\Rightarrow \pi \left( {3x} \right) + \pi \left( {5x} \right) + \pi \left( {6x} \right) = 308$

$\Rightarrow 14\pi x = 308$

$\mathbf{\left[ {\pi = \dfrac{{22}}{7}} \right]}$

$\therefore x = 7$

$\text{ cm}$

${\textbf{Step -2: Find the diameter and then radius to get the area of the circle}}{\textbf{.}}$

${\text{Diameter of largest circle = 6}} \times {\text{7}}$

${\text{ = 42 cm}}$

${\text{Radius of circle = }}\dfrac{{42}}{2}$

$= 21$

$\text{ cm}$

${\text{Diameter of smallest circle = 3}} \times {\text{7}}$

$= 21$

$\text{ cm}$

${\text{Radius of circle = }}\dfrac{{21}}{2}$

$= 10.5$

$\text{ cm}$

${\text{Difference = Area of largest circle }} - {\text{ Area of smallest circle}}$

$= \pi {\left( {21} \right)^2} - \pi {\left( {10.5} \right)^2}$

$\mathbf{\left[ {\pi = \dfrac{{22}}{7}} \right]}$

$= 1039.5{\text{ c}}{{\text{m}}^2}$

${\textbf{Hence, the difference of area of largest and smallest circle is 1039}}{\textbf{.5 c}}{{\textbf{m}}^2}.$

In figure, $OABC$ is a square of side $7 \ cm.$ If $OAPC$ is a quadrant of a circle with centre $O,$ then find the area of the shaded region. $\left [ Use \ \pi = \dfrac{22}{7} \right ].$

Given that OABC is a square of side $7 \ cm.$
Therefore, Area of square $OABC = (7)^2\space\ cm^2= 49 \space\ cm^2$
Also, it is given that OAPC is a quadrant of circle with centre O.
Radius of the quadrant of the circle $= OA = 7 \ cm$
Therefore, Area of the quadrant of circle $=\dfrac{1}{4}(\pi r^2)$
$=\dfrac{1}{4}(\pi \times 7^2)$
$=\dfrac{49\pi}{4}$
$=\dfrac{49\times 22}{4\times 7}$
$=\dfrac{77}{2}$ $cm^2$
Therefore, Area of the shaded region $=$ Area of Square − Area of Quadrant of circle.
$=49-\dfrac{77}{2}$
$=\dfrac{98-77}{2}$
$=\dfrac{21}{2}$

$\Rightarrow 10.5$ $cm^2$
So, the area of the shaded region is $10.5 \ cm^2$

A path of 4 m width runs round a circular grassy plot whose circumference a $163 \dfrac{3} {7}$ m Find
(i) the area of the path
(ii) the cost of gravelling the path at the rate of $Rs. 1.50$ per square meter
(iii) the cost of turfing the plot at the rate of 45 paise per $m^{2}$ .

Circumference of plot

$=163\cfrac { 3 }{ 7 } m$

$=\cfrac { 1144 }{ 7 } m$

$\Rightarrow 2\pi r=\cfrac { 1144 }{ 7 }$

$\Rightarrow r=\cfrac { 1144\times 7 }{ 7\times 44 }$

$\Rightarrow r =26m$

i) Area of path

$=\pi \left( { R }^{ 2 }-{ r }^{ 2 } \right)$

$=\cfrac { 22 }{ 7 } \left( { \left( 26+4 \right) }^{ 2 }-{ \left( 26 \right) }^{ 2 } \right)$

$=\cfrac { 22 }{ 7 } \times 224$

$=704 \,\,m^2$

ii) Cost of graveling

$=$

$\left( 704\times 1.50 \right)$

$=$

$Rs. 1056$

iii) Area of plot

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

$=\cfrac { 22 }{ 7 } \times 26\times 26$

$=2124.57 \,\,m^2$

$\therefore$ Cost of turfing at the rate of 45 paise per

$m^2$

$=Rs. \left( 2124.57\times 0.45 \right)$

$=Rs. 956.06$

Concentric circles of radii $1,2,3,......,100cm$ are drawn. The interior of the smallest circle is coloured red and the angular regions are coloured alternately green and red, so that no two adjecent regions are of the same colour. Then find the total area of green regions in sq.cm.

Smallest circle is colured Red

Then Area of Green region for 1st two Circles

$=\pi\times{2}^{2}-\pi\times{1}^{2}$

$=\pi\times 1\times 3$

Then Area of next green region

$=\pi\times{4}^{2}-\pi\times{3}^{2}=\pi\left(4-3\right)\left(4+3\right)=7\pi$

and so on

Last green region area

$=\pi\times{100}^{2}-\pi\times{99}^{2}=\pi\left(100-99\right)\left(100+99\right)=199\pi$

Total Green area

$=\pi\times 1 \times 3 + \pi \times 1 \times 7 +\pi \times 1 \times 11+...+\pi\times 1\times 199$

$=\pi\left(3 + 7 + 11 +...+ 199\right)$

$a = 3,\, n = 50,\, d = 4$

$S =\left(\dfrac{50}{2}\right)\left(3 + 199\right) =50\times 101$

Area of green region

$=50\times 101 \pi {cm}^{2} = 5050\pi{cm}^{2}=15871.4{cm}^{2}$

A tent is in the shape of a cylinder surmounted by a conical cap.If the height and diameter of the cylinder part are 2.0 m and 6 m, respectively and the slant height of the cap is $2.4 m$ .Find the area of the canvas used for making the tent.Also,find the cost of the canvas of the tent at the rate of $rs \quad 300$ per ${m}^{2}$ .

2039323_1260230_ans_67bac6e83cd443eab0885bfeadf8dbf8.JPG

The radius of a circular park is $50\ m$ .A circular concrete footpath of width $5\ m$ is construction around the park. Find the cost of construction at the rate of $Rs 50\ per\ m^{2}$ ?

Area of the circle

$=πr^{ 2 },r=5m$

$A=\dfrac { 22 }{ 7 } \ast 5\ast 5=78.57m^{ 2 }$

$\therefore$ the area of the path=

$π(R+r)^{ 2 }$

$=\dfrac { 22 }{ 7 } (50+5)^{ 2 }$

$=\dfrac { 22 }{ 7 } \ast { 55 }^{ 2 }$

$=9507.14m^{ 2 }$

Cost of construction

$=50\ast 9507.14=Rs.475357$

The moon is about $384000\ km$ from the earth and its path around the earth is nearly circular. Find circumference of the path travelled by the moon every month.

So radius(r)

$=384000$ km

So length of path is circumference

$=2\pi r =2\times 3.14\times 384000$

$=2\times 314\times 3840$

$=2411520$ km.

A gardener wants to fence a circular garden of diameter $21\ m$ . Find the length of the rope he needs to purchase if he makes $2$ rounds of the fence. Also find the cost of the rope, if it costs $Rs.\ 4$ per meter. $(Take\ \pi=3.14)$

The diameter of the garden is $21\;{\rm{m}}$ , then the radius will be $\dfrac{{21}}{2}\;{\rm{m}}$ .

The length of the rope used will be the circumference of the circular garden,

${\rm{length}} = 2\pi r$

$= 2 \times \frac{{22}}{7} \times \frac{{21}}{2}$

$= 66\;{\rm{m}}$

As he makes 2 rounds of fence, then the length of the rope used is $2 \times 66 = 132\;{\rm{m}}$ .

The cost of $1\;{\rm{m}}$ rope is ${\rm{Rs}}\;4$ .

The cost of $132\;{\rm{m}}$ is ${\rm{Rs}}\;4 \times 132 = {\rm{Rs}}\;528$ .

An elastic belt is placed around the rim of a pulley of radius 5 cm. from one point C on the belt, the elastic belt is pulled directly away from the centre O of the pulley until it is at P, 10 cm from the point O. Find the length of the belt that is still in contact with the pulley. Also find the shaded area.

${\rm{Reflex}}\,\,\angle AOB = 360^\circ - \angle AOB = 360^\circ - 120^\circ = 240^\circ = \alpha$

Length of belt in contact with pulley = Length of major are

$\begin{array}{l} =\frac { \alpha }{ { 360 } } \times 2\pi r \\ =\frac { { 240 } }{ { 360 } } \times 2\times 3.14\times 5 \\ =20.933\, \, cm \end{array}$

In Right angle Triangle,

$\begin{array}{l} \Delta OAP,\, \, { { Since } }=\frac { { AP } }{ { OP } } \Rightarrow \frac { { \sqrt { 3 } } }{ 2 } =\frac { { AP } }{ { 10 } } \\ \Rightarrow AP=5\sqrt { 3 } \\ Area\, \, of\, \, \Delta OAP\Rightarrow \frac { 1 }{ 2 } \times b\times h\Rightarrow \frac { 1 }{ 2 } \times 5\sqrt { 3 } \times 5=\frac { { 25\sqrt { 3 } } }{ 2 } \\ Area\, \, of\, \, \left( { \Delta OAP } \right) =ar\left( { \Delta OBP } \right) =\frac { { 25\sqrt { 3 } } }{ 2 } \\ Area\, \, of\, \, { { minor } }\, \, { { sector } }\left( { \Delta ACB } \right) =\frac { \theta }{ { 360 } } \times \pi { r^{ 2 } } \\ =\frac { { 129 } }{ { 360 } } \times 3.14\times 5\times 5=26.17\, \, c{ m^{ 2 } } \\ Shaded\, \, area=ar\left( { \Delta OAP } \right) +ar\left( { \Delta OBP } \right) -area\, \, of\, \, { { minor } }\, \, { { sector } }\Delta OAC \\ =\frac { { 25\sqrt { 3 } } }{ 2 } +\frac { { 25\sqrt { 3 } } }{ 2 } -26.17 \\ =25\sqrt { 3 } -26.17 \\ =25\times 1.73-26.17 \\ =43.25-26.17\Rightarrow 17.08\, \, c{ m^{ 2 } } \end{array}$

1221716_1205294_ans_6253dad739d542d3a02ace2b220b3e6f.JPG

A wire when bent in the form of a square enclose an area of 484 $\mathrm{cm^{2}}$ . If the same wire is bent in the form of a circle. Find the area of the circle.

Given that, a wire, when bent, is able to enclose a square having area

$484\ cm^2$

We know that, area of square

$= sid{e^2}$

$484 = {a^2}$

$a = 22\,\,cm$

We know that perimeter of the square

$=4a$

$\begin{array}{l} =4\times 22 \\ =88\, \, cm \end{array}$

Therefore the length of the wire

$= 88\ cm$

Now,

given that the same wire is bent into a form of circle

circumference of circle = length of the wire

$\begin{array}{l} 2\pi r=88 \\ 2\times \dfrac { { 22 } }{ 7 } \times r=88 \\ r=\dfrac { { 88\times 7 } }{ { 22\times 2 } } \\ r=14\, \, cm \end{array}$

Area of circle

$\begin{array}{l} =\pi { r^{ 2 } } \\ =\dfrac { { 22 } }{ 7 } \times { \left( { 14 } \right) ^{ 2 } } \\ =\dfrac { { 22 } }{ 7 } \times 14\times 14 \\ =\dfrac { { 22 } }{ 7 } \times 196 \\ =616\ c{ m^{ 2 } } \end{array}$

Hence, the area of the circle is

$616\ cm^2$ .

The diameter of the wheel of a car is $77$ cm. How many revolution will it make to travel $121$ km?

2039546_1297447_ans_f5f7d3b4414b4fe593d99e0d55910e8f.jpg

A round table cover has six equal design as shown in fig 12.14 .If the radius of the cover is 28 cm,find the cost of making the designs at the rate of $0.35 per \quad {cm}^{2}$ (use $\sqrt {3}=1.17$ )

2039357_1268903_ans_478d7a94689f4538b0969d4c9a30a505.jpg

In the adjoining figure, $ABCD$ is a parallelogram whose two adjacent sides are $6\ cm$ and $4\ cm$ . If the height corresponding to the base $AB$ is $3\ cm$ find:
(i) the area of parallelogram $ABCD$
(ii) the height corresponding to the base $AD$

According to given data:

Given data:

$AB=6\,cm$

$AD=4\,cm$

height

$=3\,cm$

Now,

(i) Area of the parallelogram

$ABCD=base\times height$

$=AB\times 3$

$=6\times 3$

$=18\,cm^2$

(II) The height corresponding to base

$AD$ .

We know that,

$Area=AD\times height$

$\therefore 18=4\times x\,cm$

$\therefore x=\dfrac{18}{4}=4.5\,cm$

Therefore Area parallelogram

$ABCD$ is

$18\,.cm^2$ and height the corresponding to base

$AD$ is

$4.5cm$ .

2118742_1454589_ans_f7bae6d4b36644898e3c14bca3e43fa9.png

Find the area of the following circles if the diameter is $49 m$ .

Diam

$=49m$

$\Rightarrow$ Radius

$=\dfrac{49}{2}m$

$\Rightarrow$ Area

$=\dfrac{22}{7}\times \left( \dfrac{49}{2}\right)^2=1886.5.$

Hence, the answer is

$1886.5.$

Ritu went to a park 130m long 90 m wide. She took on complete of it. What distance did she travel?

2040882_1453151_ans_c9eef593a08f48c282f16871564661b1.jpg

A conical tent is 9 m high and the radius of its base is 12 m.
(1) What is the cost of the canvas required to make it ,if a square metre canvas costs Rs. 10?
(2) How many persons can be accomodated in the tent , if each person requires 2 square metre on the ground and ${ 15 }m^{ 3 }$ of space to breathe in ?

2039552_1303126_ans_3b05e4d5795649aeb4344127d3f8a941.jpg

The wheel of a motor cycle is of radius 35 cm. How many revolutions per minute must the wheel make so as to keep a speed of 66 km / hr ?

$\textbf{Calculate total number of revolutions per minute :}$

$\text{Given data,}$

$\text{r = 35 cm}$

$\text{We know that, circumference = 2}\pi r$

$= 2 \times \dfrac{22}{7} \times 35$

$= 44 \times 5$

$= 220\,cm$

$= 2.2\,m$

$\therefore \text{Distance covered in one complete revolution is 2.2m}$

$\text{We have given that speed of the wheel = 66 km/hr}$

$= \dfrac{66 \times 1000}{60}$

$= 1100 \text { m/min}$

$\therefore\dfrac{\text{Revolution}}{\text{min}} = \dfrac{1100}{2.2}$

$\Rightarrow\dfrac{\text{Revolution}}{\text{min}} = 500$

$\textbf{Hence, Revolution per min = 500}$

In figure $ABC$ is a right triangle right angled at $A,\ BCED,\ ACFG$ and $ABMN$ are square on the sides, $BC, \ CA$ and $AB$ respectively. Line segment $AX\bot DE$ meets $BC$ at $Y$ . Show that
$ar (BYXD)=2\ ar(\triangle MBC)$

1664528_1670539_ans_e5d5156114c941f49caf5e18cd8eaac8.jpeg

(3,-1) (2,6) and (-5,7) are the mid points of the sides of the triangle ABC . The area of triangle ABC is .....s.q. units.

1323043_1585206_ans_9e24088489364067aa9786040354437c.jpg

In figure $ABC$ is a right triangle right angled at $A,\ BCED,\ ACFG$ and $ABMN$ are square on the sides, $BC, \ CA$ and $AB$ respectively. Line segment $AX\bot DE$ meets $BC$ at $Y$ . Show that
$\triangle MBC\cong \triangle ABD$

1664504_1670536_ans_43b228baf3d5461e9123c235467fa01b.jpeg

Find circumfernce of circles with
(a) radius = 14 cm
(b) Diameter =28 mm.

1311843_1588466_ans_67698d1ea4484a72b0f4e5e650d1f229.jpg

In figure $ABC$ and $BDE$ are two equilateral triangles such thta $D$ is the mid-point of $BC.AE$ intersects $BC$ in $F$ . Prove that
$ar(\triangle FED)=\dfrac {1}{8} ar(\triangle AFC)$

1664090_1670526_ans_f83aa1db69504a6298414b5af9be8eda.jpeg

Find the radius of a semicircular if its area is
(i) 1232 $cm^{2}$
(ii) 7700 $cm^{2}$

1314611_1508827_ans_720ddabe67a84a0d87cfe1767debed8b.jpg

Find the area of a circle whose circumference is $74$ feet. [Assume that $\pi = 3.14159....,\dfrac{1}{\pi}=.31831$ and $\log \pi =.49715$ ]

1596901_1734431_ans_071834897898486e890926568a84ca32.jpg

A floor is 4.5 m long and 3.6 m wide. A 6 cm square tile costs 23.What will be the cost to cover the floor with these tiles?

2001974_1790264_ans_25db43e3c61b48618e5aa49b091e22b8.png

.

$A\left( \triangle ABC \right) =\frac { 1 }{ 2 } \times AC\times BD$

$\therefore A\left( \triangle ABC \right) =\frac { 1 }{ 2 } \times 40\times 16$

$\therefore A\left( \triangle ABC \right) =320$ sq.m.

$A\left( \Box PQRS \right) =PQ\times PS$

$\therefore A\left( \Box PQRS \right) =10\times 8$

$\therefore A\left( \Box PQRS \right) =80$ sq.m.

Thus, total area of the shaded region is given by,

$A=A\left( \triangle ABC \right) -A\left( \Box PQRS \right)$

$\therefore A=320-80$

$\therefore A=240$ sq.m.

In figure $ABC$ is a right triangle right angled at $A,\ BCED,\ ACFG$ and $ABMN$ are square on the sides, $BC, \ CA$ and $AB$ respectively. Line segment $AX\bot DE$ meets $BC$ at $Y$ . Show that
$\triangle FCB\cong \triangle ACE$

1664522_1670543_ans_ccc5df2446ce4348b04612000e16b35f.jpeg

In figure $ABC$ is a right triangle right angled at $A,\ BCED,\ ACFG$ and $ABMN$ are square on the sides, $BC, \ CA$ and $AB$ respectively. Line segment $AX\bot DE$ meets $BC$ at $Y$ . Show that
$ar (BYXD)=ar(ABMN)$

1664517_1670542_ans_4e49975044a644b3be3eec621c5d4ceb.jpeg

A steel wire, when bent in the form of a square, encloses an area of $121 \mathrm{cm}^{2}.$ The same wire is bent in the form of a circle. Find area the circle.

Area of a square = side x side

Side

$\times$ side

$=121 \mathrm{cm}^{2}$

(Side)

$^{2}=121$

Side

$=\sqrt{121}$

Side

$=11 \mathrm{cm}$

We know that.

Perimeter of a square

$=4 \times$ side

$4 \times$ side

$=4 \times 11$

$=44 \mathrm{cm}$

According to the question,

Perimeter of a circle = perimeter of a square

Perimeter of a circle

$=44$

$\Rightarrow 2 \pi r=44 \quad(r$ is radius of a circle

$)$

$\Rightarrow r=\dfrac{44}{2 \pi}=\dfrac{44}{2 \times \dfrac{22}{7}}=7 \mathrm{cm}$

Area of a circle

$=\pi r^{2}=\dfrac{22}{7} \times 7 \times 7=154 \mathrm{cm}^{2}$

No.	Radius $(r)$	Diameter $(d)$	Area of circle
$(1)$	$42\ cm$	________	______
$(2)$	________	$14\ metre$	______
$(3)$	$1.4\ cm$	________	______
$(4)$	$28\ metre$	________	______
$(5)$	$35\ cm$	________	______
$(6)$	_______	$5.6\ metre$	______

Perimeter And Area - Class 7 Maths - Extra Questions

A piece of wire is 34m\dfrac { 3 } { 4 } m long. What will be the total length of 10 such pieces of wire?

Find the circumference of the circle whose radius is 2828 mm. Take π=227\pi=\dfrac{22}{7}.

If the length of longest chord of a circle is 8 cm8\ cm then find its radius.

Find the area of the circle having radius 77 cm.

Find area of circle with circumference 88cm88cm

Find the circumference of the circle having radius 77 cm.

The school ground is Circular in shape if its radius is 21 m Find its area

Find the diameter of aa wheel whose circumference is 176 cm176\ cm.

In a triangle, the measures of the angles are x, x + 20 and 2x what is the value of x?

Find the area of parallelogram whose base is 48 cm and height is 25 cm

The circle of radius 5cm has the area as..

Find the area of a circle of radius 77 cm.

The minute hand of a circular clock is 15 cm15\ cm long. How far does the tip of the minute hand move in 11 hour? (Take π=3.14\pi =3.14)

The circumference (correct upto three decimal places) of a circle is 3.14 m3.14\ m, find its area.

Find the circumference of a circle whose diametre is 10.5 cm10.5\ cm

Ratio of the circumference of a circle to its diameter is denoted by symbol _____ .

Area of a parallelogram with base 24 cm24\ cm and height 4 cm4\ cm is:

If area of a triangular piece of cardboard is 90cm290\, cm^2, then the length of altitude corresponding to 20cm20\, cm long base is _____ cm.

1 hectare == _____ m2 m^2

Circumference 'C' of a circle ean be found by multiplying diameter 'd' with _____ .

1km2=1\, km^2 = _____ m2m^2.

Area of a triangle =12×base×=\dfrac {1}{2} \times base \times ______ .

Find the circumference of a circle whose diameter is 4.9m4.9 \,m

Find the area of a circle whose diameter is 28 cm28 \text{ cm}

The carpet for a room 6.6 m6.6\ m by 5.6 m5.6\ m costs Rs 3960Rs\ 3960 and it was made from a roll 70 cm70\ cm wide. Find the cost of the carpet per metre.

Find the circumference of a circle whose radius is 3.5 m3.5\ m

The area of a circle is (14+6√5)π(14 + 6\sqrt{5})\pi units, Find its radius.

The radius of a circle is 28cm28cm find the circumference and area.

The area of the circular ring included between two concentric circles of radii 14 cm14\ \text{cm} and 10.5 cm10.5\ \text{cm} to the closest integer is ...........

Find the circumference (in cm) of a circle whose diameter is 5.6 m.

A horse is tethered by a rope 1010 m long at a point. Find the area of the region where it can graze. (π=3.14)(\pi = 3.14)

The area of a circle whose diameter is 1712cm\displaystyle 17\dfrac{1}{2} \,\,cm is 240.625cm2240.625\,\, cm^2If true then enter 11 else 0

Take π=3.14\pi = 3.14, then the area of a circle radius = 17.6 cm is 972.6464cm2972.6464\,\,cm^2If true then enter 11 and if false then enter 00

Take π=3.14\pi = 3.14, the area of a circle with diameter 21.6 m21.6\ \text{m} is 366.2496m2366.2496 \,\text{m}^2.If true then enter 11 and if false then enter 00

The radius of the circle is 77 cm , Find the area of the circle in cm\text{cm} 2^2.

11 sq m =t\displaystyle =t sq cm.Find tt

11 sq cm =k\displaystyle = k sq mm.Find kk

Area of square is 10,00010,000 sq m. m. Then its side measures(in m)

The area of a parallelogram is 9.6cm29.6 {cm}^{2}, and the base is 3.2cm3.2 cm. Find the height.

Find the area of parallelogram in which base is 2.5cm2.5 cm, and altitude is 1.4cm1.4 cm.

How many times will a wheel of radius 35 cm35\ cm be rotated to travel 660 m660\ m?(Take π=227)\pi = \dfrac {22}{7}).

The two adjacent sides of a parallelogram are 4.8cm4.8 cm and 3.6cm3.6 cm. The distance between the longer sides is 2.4cm2.4 cm. Find the distance between the other pair.

In the following figure, the radius of the circle is 77 cm and m(arcRYS)=30∘m(arc RYS) = 30^{\circ}, find find:Area of the circle

Sarlaben stiches golden plate on edges of circular bed of 77 metre diameter. Labour to stich golden plate is Rs.3Rs. 3 per 11 metre. Find the labour charge of Sarlaben in rupees.

The radius of a circle is 7 cm7\ cm. Find the circumference of the circle in cm.

On the wall of school's prayer hall, in a circle of 70 cm70\ cm radius, a picture of nature is to be drawn, if the cost of drawing is Rs.5Rs. 5 per 100 sq cm100\ sq\ cm, then find the total cost in rupees.

A heritage stepwell is secured by a circular wire fence. Radius of this circle is 17.5 metre17.5\ metre. Find the cost of fencing(in Rs) at the rate of Rs.50Rs. 50 per metre.

Radius of the circular swimming pool is 70 cm70\ cm. Then find the area of the bottom of the pool in sq cmsq\ cm

Diameter of a bangle is 7 cm7\ cm. Thin stripe of gold is to be fixed on this bangle, find the cost(in Rs) for it at the rate of Rs.100Rs. 100 per 1 cm1\ cm.

No.Radius (r)(r)Diameter (d)(d)CircumferenceArea of circle(1)(1)0.7 m0.7\ m_______________________(2)(2)__________84 cm84\ cm_______________(3)(3)140 metre140\ metre_______________________(4)(4)4.9 cm4.9\ cm_______________________(5)(5)_________70 cm70\ cm_______________

Diameter of the wheel of horse-cart is 1.401.40 metre. How much distance(in metres) is covered (correct upto one decimal place) by this wheel in one rotation?

Radius of the circular ground is 10.5 m10.5\ m. What is the cost(in Rs) to fill up this ground with soil at the rate of Rs.150Rs. 150 per sq m?

The radius of a circle is increasing at the rate of 0.7 cm/sec0.7 \ cm/sec. What is the rate of increase of its circumference ?

What is the area of the circle that can be inscribed in a square of side 6cm?

On a square handkerchief, nine circular designs each of radius 7 cm7\ cm are made (see the given figure). Find the area of remaining portion of the handkerchief. [Use π=227]\left [Use\ \pi = \dfrac {22}{7}\right ].

The vertices A , B , C of the triangles are (2 , 1) , (5 , 2) and (3 , 4) respectively . Find the coordinates of the circumcentre and also the radius of the circumcircle

Find the area of a circle inscribed in a square of side 66 cm.

Find the area of the following circle, given that: Radius=5cmRadius=5cm

The diameters of the two circles are 32 cm and 24 cm. Find the radius of the circle having its area equal to the sum of the areas of the two given circle.

Find the cicumference of the circle whose diameter is 44 units.

If the radius of a circle is 3 cm3\ cm, find its circumference.

Find the area of a circular park whose radius is 4.54.5m.

The diameter of the whell of a can is 77cm77cm.How many revolutions will it make to travel 242242km

Two adjacent sides of parallelogram are 24 cm and 18 cm. If the distance between the longer sides is 12 cm, find the distance between the shorter sides.

Two sides of the parallelogram PQRS are 66 cm and 44 cm. The height corresponding to the base RS is 33 cm. Find area of parallelogram.

Write the formula to find the Area of a sector of angle θ\theta of a circle with radius rr.

There are four circular windows in Nisha's house, each of radius 28cm28 cm are to be fitted. Find the total area required for these windows.

Two sides of the parallelogram PQRS are 66 cm and 44 cm. The height corresponding to the base RS is 33 cm. Find The height corresponding to the base PS.

Find the circumference of Circle if Diameter = 4 cm.

Find the area of the following circle, given that:Radius=14mm(π=227)Radius=14mm ( \pi=\dfrac{22}{7})

Find the area of a circular park of janakpuri whose circumference is 77m77m.

Find the area of the following circle, given that:Diameter=49mDiameter=49m

A car has wheels each of which is 84 cm in diameter. Find how many complete revolutions must each wheel take to cover a distance of 792 m.

Find the ratio of the circumferences of two concentric circles of radii 2m2 m and 3m3 m.

Use π=227\pi =\frac{22}{7}, unless stated otherwise.The circumference of a circle is 39.6 cm. Find its area.

The circumference of a circle is 57.2cm57.2cm. Find its radius.

A circular park is surrounded by a rod 21m21 m wide. If the radius of the park is 105m105 m, find the area of the road.

Find the area of the following circles , given that: (a) radius = 14 mm (take =227)\frac{22}{7}).(c) radius =5 cm.

Find the circumference of a circle whose area is 12474cm212474 cm^2

A piece of wire is $\dfrac { 3 } { 4 } m$ long. What will be the total length of 10 such pieces of wire?

Find the circumference of the circle whose radius is $28$ mm. Take $\pi=\dfrac{22}{7}$ .

If the length of longest chord of a circle is $8\ cm$ then find its radius.

Find the area of the circle having radius $7$ cm.

Find area of circle with circumference $88cm$

Find the circumference of the circle having radius $7$ cm.

Find the diameter of $a$ wheel whose is $176\ cm$ .

Find the area of a circle of radius $7$ cm.

The minute hand of a circular clock is $15\ cm$ long. How far does the tip of the minute hand move in $1$ hour? (Take $\pi =3.14$ )

The circumference (correct upto three decimal places) of a circle is $3.14\ m$ , find its area.

Find the circumference of a circle whose diametre is $10.5\ cm$

Area of a parallelogram with base $24\ cm$ and height $4\ cm$ is:

If area of a triangular piece of cardboard is $90\, cm^2$ , then the length of altitude corresponding to $20\, cm$ long base is _____ cm.

1 hectare $=$ _____ $m^2$

$1\, km^2 =$ _____ $m^2$ .

Area of a triangle $=\dfrac {1}{2} \times base \times$ ______ .

Find the circumference of a circle whose diameter is $4.9 \,m$

Find the area of a circle whose diameter is $28 \text{ cm}$

The carpet for a room $6.6\ m$ by $5.6\ m$ costs $Rs\ 3960$ and it was made from a roll $70\ cm$ wide. Find the cost of the carpet per metre.

Find the circumference of a circle whose radius is $3.5\ m$

The area of a circle is $(14 + 6\sqrt{5})\pi$ units, Find its radius.

The radius of a circle is $28cm$ find the circumference and area.

The area of the circular ring included between two concentric circles of radii $14\ \text{cm}$ and $10.5\ \text{cm}$ to the closest integer is ...........

A horse is tethered by a rope $10$ m long at a point. Find the area of the region where it can graze. $(\pi = 3.14)$

diameter is $\displaystyle 17\dfrac{1}{2} \,\,cm$ is $240.625\,\, cm^2$

If true then enter $1$ else 0

radius = 17.6 cm is $972.6464\,\,cm^2$

If true then enter $1$ and if false then enter $0$

diameter $21.6\ \text{m}$ is $366.2496 \,\text{m}^2$ .

If true then enter $1$ and if false then enter $0$

The radius of the circle is $7$ cm , Find the area of the circle in $\text{cm}$ $^2$ .

$1$ sq m $\displaystyle =t$ sq cm.Find $t$

$1$ sq cm $\displaystyle = k$ sq mm.Find $k$

Area of square is $10,000$ sq $m.$ Then its side measures(in m)

The area of a parallelogram is $9.6 {cm}^{2}$ , and the base is $3.2 cm$ . Find the height.

Find the area of parallelogram in which base is $2.5 cm$ , and altitude is $1.4 cm$ .

How many times will a wheel of radius $35\ cm$ be rotated to travel $660\ m$ ?
(Take $\pi = \dfrac {22}{7})$ .

The two adjacent sides of a parallelogram are $4.8 cm$ and $3.6 cm$ . The distance between the longer sides is $2.4 cm$ . Find the distance between the other pair.

In the following figure, the radius of the circle is $7$ cm and $m(arc RYS) = 30^{\circ}$ , find find:
Area of the circle

Sarlaben stiches golden plate on edges of circular bed of $7$ metre diameter. Labour to stich golden plate is $Rs. 3$ per $1$ metre. Find the labour charge of Sarlaben in rupees.

The radius of a circle is $7\ cm$ . Find the circumference of the circle in cm.

On the wall of school's prayer hall, in a circle of $70\ cm$ radius, a picture of nature is to be drawn, if the cost of drawing is $Rs. 5$ per $100\ sq\ cm$ , then find the total cost in rupees.

A heritage stepwell is secured by a circular wire fence. Radius of this circle is $17.5\ metre$ . Find the cost of fencing(in Rs) at the rate of $Rs. 50$ per metre.

Radius of the circular swimming pool is $70\ cm$ . Then find the area of the bottom of the pool in $sq\ cm$

Diameter of a bangle is $7\ cm$ . Thin stripe of gold is to be fixed on this bangle, find the cost(in Rs) for it at the rate of $Rs. 100$ per $1\ cm$ .

No. Radius $(r)$ Diameter $(d)$ Circumference Area of circle
$(1)$ $0.7\ m$ ____ _
$(2)$ ____ $84\ cm$ _
$(3)$ $140\ metre$ __ _
$(4)$ $4.9\ cm$ __ _
$(5)$ _ $70\ cm$ _

Diameter of the wheel of horse-cart is $1.40$ metre. How much distance(in metres) is covered (correct upto one decimal place) by this wheel in one rotation?

Radius of the circular ground is $10.5\ m$ . What is the cost(in Rs) to fill up this ground with soil at the rate of $Rs. 150$ per sq m?

The radius of a circle is increasing at the rate of $0.7 \ cm/sec$ . What is the rate of increase of its circumference ?

On a square handkerchief, nine circular designs each of radius $7\ cm$ are made (see the given figure). Find the area of remaining portion of the handkerchief. $\left [Use\ \pi = \dfrac {22}{7}\right ]$ .

Find the area of a circle inscribed in a square of side $6$ cm.

Find the area of the following circle, given that: $Radius=5cm$

Find the cicumference of the circle whose diameter is $4$ units.

If the radius of a circle is $3\ cm$ , find its circumference.

Find the area of a circular park whose radius is $4.5$ m.

The diameter of the whell of a can is $77cm$ .How many revolutions will it make to travel $242$ km

Two sides of the parallelogram PQRS are $6$ cm and $4$ cm. The height corresponding to the base RS is $3$ cm. Find area of parallelogram.

Write the formula to find the Area of a sector of angle $\theta$ of a circle with radius $r$ .

There are four circular windows in Nisha's house, each of radius $28 cm$ are to be fitted. Find the total area required for these windows.

Two sides of the parallelogram PQRS are $6$ cm and $4$ cm. The height corresponding to the base RS is $3$ cm. Find The height corresponding to the base PS.

Find the area of the following circle, given that:
$Radius=14mm ( \pi=\dfrac{22}{7})$

Find the area of a circular park of janakpuri whose circumference is $77m$ .

Find the area of the following circle, given that:
$Diameter=49m$

Find the ratio of the circumferences of two concentric circles of radii $2 m$ and $3 m$ .

Use $\pi =\frac{22}{7}$ , unless stated otherwise.
The circumference of a circle is 39.6 cm. Find its area.

The circumference of a circle is $57.2cm$ . Find its radius.

A circular park is surrounded by a rod $21 m$ wide. If the radius of the park is $105 m$ , find the area of the road.

Find the area of the following circles , given that:
(a) radius = 14 mm (take = $\frac{22}{7})$ .
(c) radius =5 cm.

Find the circumference of a circle whose area is $12474 cm^2$

In the figure given below, $AD = 8 cm$ and the area of the parallelogram $ABCD = 44$ $cm^2$ . Find the measure of the corresponding altitude.

Area of a parallelogram is $400\ cm^{2}$ and its height is $16\ cm$ . Find its base.

Find the circumference of a circle of radius $175\ cm.\ \left(Take\ \pi=\dfrac{22}{7}\right)$

The area of a circle is $616\ cm^2$ . Find its circumference.

Find the area enclosed by the circle of radius $(a)$ .

If the area of circle is $176cm^2$ .find the radius of the circle.

Find the height of a parallelogram whose area is $54\ cm^{2}$ and the base is $15\ cm$ .

In $\triangle ABC$ , $seg MN\parallel seg BC$ , if $BM=6$ , $AN=5$ , $AM=2$ , then find $NC$ .

The radius of circle is $7cm$ find its circumference and Area.

Area of circle $=?\ r=0.129\ cm$ .

Find the perimeter of a circle with radius $6.3 \ \mathrm { m }.$ Given $\left( \pi = \frac { 22 } { 7 } \right)$

Find the area of circle of diameter $7\ cm$ .

The area of a circular field is $13.86\ m^{2}$ . Find the circumference of the field .

A circle is inscribed in a square whose each side is $36$ cm. find the area of the circular region