SChand CLASS 9 Chapter 17 Circle, Circumference and Area Exercise 17(B)

   Exercise 17 B

Question 1

Ans: (i) (i) Diameter $=7 \mathrm{~cm}$

So Radius(r) $=\frac{7}{2} \mathrm{~cm}$

So Area of the circle $=1 r^{2}=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \mathrm{~cm}^{2}$

$=\frac{77}{2}=38.5 \mathrm{~cm}^{2}$

 

(ii)  $(r)=14 \mathrm{~cm}$

So Area $=\pi r^{2}=\frac{22}{7} \times 14 \times 14 \mathrm{~cm}^{2}=616 \mathrm{~cm}^{2}$

(iii) Diameter $=2.8 \mathrm{~cm}$

So $\begin{aligned} \text { Radius }(r) &=\frac{2.8}{2}=1.4 \mathrm{~cm} \\ \text { So } \text { Area } &=\pi r^{2}=\frac{22}{7} \times 1.4 \times 1.4 \mathrm{~cm}^{2} \\ &=6.16 \mathrm{~cm}^{2} \end{aligned}$

Question 2

Ans:  length of string $=28 \mathrm{~m}$

So Radius $(r)=28 \mathrm{~m}$

Area which the norse can graze $=\pi r^{2}$

$=\frac{22}{7} \times 28 \times 28 \mathrm{~m}^{2}=2469 \mathrm{~m}^{2}$

Question 3

Ans:(i) Area of a circular field $=154 \mathrm{~cm}^{2}$

Let $r$ be the radius then

$\begin{aligned}& \pi r^{2}=154 \\\Rightarrow & \frac{22}{7} r^{2}=154 \Rightarrow r^{2}=\frac{154 \times 7}{22}=49 \\\Rightarrow & r^{2}=(7)^{2} \Rightarrow r=7\end{aligned}$

$\text { So } \text { Radius }=7 \mathrm{~cm}$

(ii) Area of the circular field $=1386 \mathrm{~cm}^{2}$ 

Let be the radius on the field then

$\begin{aligned}&\Rightarrow \pi r^{2}=1386 \Rightarrow \frac{22}{7} r^{2}=1386 \\&\Rightarrow r^{2}=\frac{1386 \times 7}{22}=63 \times 7=441=(21)^{2} \\&\Rightarrow r=21 \\&\text { So Radius }=21 \mathrm{~cm}\end{aligned}$

Question 4

Sol: Area of a circle $=24.64 \mathrm{~cm}^{2}$

Let r be the radius then

$\begin{aligned}&\pi r^{2}=24.64 \Rightarrow \frac{22}{7} r^{2}=24.64 \\&\Rightarrow r^{2}=\frac{24.64 \times 7}{22}=1.12 \times 7=7.84 \\&\Rightarrow r^{2}=(2.8)^{2} \Rightarrow r=2.8 \\&\text { Now circumference }=2 \pi r \\&=2 \times \frac{22}{7} \times 2.8 \mathrm{~cm}=17.6 \mathrm{~cm}\end{aligned}$

Question 5

Sol: Area of square $=121 \mathrm{~cm}^{2}$

So side as the square (a) $=\sqrt{\text { Area }}=\sqrt{121}=11 \mathrm{~cm}$

So perimeter of the wire $=49=4 \times 11=44 \mathrm{~cm}$ Then perimeter of circular wire which is bent down $=44 \mathrm{~cm}$

So Radius $=\frac{\text { circumference }}{2 \pi}$

$=\frac{44 \times 7}{2 \times 22}=7 \mathrm{~cm}$

Then area of the circle $=\pi r^{2}$

$=\frac{22}{7} \times 7 \times 7=154 \mathrm{~cm}^{2}$

Question 6

Sol: The area of square $=484 \mathrm{sq} . \mathrm{m}$

So side of the square (i) $=\sqrt{\text { arca }}$

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

and perimeter of square $=49=4 \times 22 \mathrm{~m}=88 \mathrm{~m}$

so circumference of the circle $=$ perimeter of the square $=88 \mathrm{~m}$

So Radius $(r)=\frac{\text { circumference }}{2 \pi}=\frac{88 \times 7}{2 \times 22} \mathrm{~m}$

and area of the circle $=\pi r^{2}=\frac{22}{7} \times 14 \times 1459 . \mathrm{m}$ $=61659 \cdot m$

Question 8

Sol: Side of square=12.5cm

Diameter of disc = 7cm

(IMAGE TO BE ADDED)

So Radius $(r)=\frac{7}{2} \mathrm{~cm}$

So Area of square $=a^{2}=\left(12.5 \mathrm{l}^{2} \mathrm{~cm}^{2}\right.$

$=156.25 \mathrm{~cm}^{2}$

Aera of circular disc $=\pi r^{2}$

$\begin{aligned}&=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \mathrm{~cm}^{2} \\&=\frac{77}{2}=38.5 \mathrm{~cm}^{2}\end{aligned}$

So Area of remaining  part $=156.25-38.50$

$=117.75 \mathrm{~cm}^{2}$

weight of one sq. $\mathrm{cm}=0.8 \mathrm{gm}$

So total weight al the remaining portion

$=117.75 \times 0.8 \mathrm{gm}=94.2 \mathrm{gm}$

Question 9

Sol: Ratio in the circumferences of two circles $=2: 3$

Wet circumference of the first circle $=2 x$ and of the second circle $=3 x$

So Radius $\left(r_{1}\right)$ of first circle

$=\frac{\text { circumference }}{2 \pi}=\frac{2 x}{2 \pi}=\frac{x}{\pi}$

and radius $\left(\gamma_{2}\right)$ aj second circle $=\frac{3 x}{2 \pi}$

Now area of the first circle $=\pi r_{1}{ }^{2}$

$=\pi \times\left(\frac{x}{\pi}\right)^{2}=\frac{\pi \times x^{2}}{\pi^{2}}=\frac{x^{2}}{\pi}$

and area of the second circle $=\pi r_{2}{ }^{2}$

$\begin{aligned}&=\pi \times\left(\frac{3 x}{2 \pi}\right)^{2}=\frac{\pi \times 9 x^{2}}{4 \pi^{2}}=\frac{9 x^{2}}{4\pi} \\&\text { So Ratio } \frac{x^{2}}{\pi}: \frac{9 x^{2}}{4 x} \\&=1: \frac{9}{4}=4: 9\end{aligned}$

Question 10

Sol: side of the square park $=100 \mathrm{~m}$

So Area $=(\sin e)^{2}=(100)^{2}=10000 \mathrm{~m}^{2}$

Radius of each quadrant at the corner of the park $=14 \mathrm{~m}$

So area of one quadrant $=\frac{1}{4} \pi r^{2}$

$\frac{1}{4} \times \frac{22}{7} \times(14) \times(14) \mathrm{m}^{2}=154 \mathrm{~m}^{2}$

and area of 4 qud rants $=154 \times 4=616 \mathrm{~m}^{2}$ so Area of the remaining portion of the park $=10000-616=9384 \mathrm{~m}^{2}$

Question 11

Sol: Radius of circular field $(k)=20 \mathrm{~m}$

 width of inside path $=5 \mathrm{~m}$

(IMAGE TO BE ADDED)

So inner radius $(r)=20-5=15 \mathrm{~m}$

So area of path - outer area - inner area $=\pi R^{2}-\pi r^{2}$

$=\pi\left[R^{2}-r^{2}\right]=\frac{22}{7}\left[(20)^{2}-(15)^{2}\right] \mathrm{m}^{2}$

$=\pi(20+15)\left(20-151 \mathrm{~m}^{2}\right.$

$=\frac{22}{7} \times 35 \times 5=550 \mathrm{~m}^{2}$

Question 12

Sol: width of road $=3.5 \mathrm{~m}$ 

circumference of a circular plot $=44 \mathrm{~m}$

(IMAGE TO BE ADDED)

So

 $\begin{aligned} \text { Radius of the plot } &=\frac{\text { circumference }}{2 \pi} \\ &=\frac{44 \times 7}{2 \times 22}=7 \mathrm{~m} \end{aligned}$

So Outer radius $(R)=7+3.5=10.5 \mathrm{~m}$

So Area of the outer Road = outer Area-inner area

$\begin{aligned}&=\pi R^{2}-\pi r^{2} \\&=\pi\left(R^{2}-r^{2}\right)=\pi\left[\left(10.51^{2}-(7)^{2}\right]\right. \\&=\frac{22}{7}(10.5+7)(10.5-7) \mathrm{m}^{2} \\&=\frac{22}{7} \times 17.5 \times 3.5 \mathrm{~m}^{2}=192.5 \mathrm{~m}^{2}\end{aligned}$

cost of paving the road $=Rs10$ pen $m^{2}$

So Total cost $=192.5 \times 10=\mathrm{RS} 1925$

Question 13

Sol: inner diameter of semicircular lawn $=35 \mathrm{dm}$

So Radius $(r)=\frac{35}{2}=17.5 \mathrm{dm}$

width of the flower bed $=3.5 \mathrm{dm}$

(IMAGE TO BE ADDED)

So outer radius $(R)=17.5+.3 .5=21 \mathrm{dm}$ Now arca af the flower bed $=$ outer area $-$ inner area

$\begin{aligned}&=\frac{1}{2} \pi R^{2}-\frac{1}{2} \pi r^{2}=\frac{1}{2} \pi\left[R^{2}-r^{2}\right] \\&=\frac{1}{2} \times \frac{22}{7}\left[(21)^{2}-\left(17.51^{2}\right] \mathrm{dm}^{2}\right. \\&=\frac{11}{7}(21+17.5)(21-17.5) \mathrm{dm}^{2}\end{aligned}$

$\begin{aligned}&=\frac{11}{7} \times 38.5 \times 3.5 \mathrm{dm}^{2} \times 11 \times 5.5 \times 3.5\mathrm{dm}^{2} \\&=211.75 \mathrm{dm}^{2}\end{aligned}$

Question 15

Sol: Area of shaded portion $=346.5 \mathrm{~cm}^{2}$ circumference of the inner circle =88cm

$\text { So } \begin{aligned}\text { inner radius }(r) &=\frac{\text { circumferences }}{2 \pi} \\&=\frac{88 \times 7}{2 \times 22}=14 \mathrm{~cm}\end{aligned}$

Let outer radius $=R$. then

Area of shaded portion $=\pi R^{2}-\pi r^{2}$

$\text { So } \begin{aligned}&346.5=\pi\left[R^{2}-(14)^{2}\right] \\&\Rightarrow 346.5=\frac{22}{7}\left[R^{2}-196\right] \\&\Rightarrow R^{2}-196=\frac{346.5 \times 7}{22}=110.25 \\&\Rightarrow R^{2}=110.25+196=306.25 \\&\text { So } R=\sqrt{306.25}=17.5\end{aligned}$

So Radius as the outer circle $=17.5 \mathrm{~cm}$

Question 16

Sol: Radius of first inner circle $\left(r_{1}\right)=3.5 \mathrm{~cm}$

 and radius of second circle $\left(r_{2}\right)=7 \mathrm{~cm}$

(IMAGE TO BE ADDED)

So Area between these two circles 

$\left.=\pi[(r2))^{2}-(r1)\right]=\frac{22}{1}\left[(1)^{2}-(3.5)^{2}\right]$

$\frac{22}{7}$ (7+3.5)(7-3.5)$(cm)^{2}$

$=\frac{22}{7} \times 10.5 \times 3.5 \mathrm{~cm}^{2}=115.5 \mathrm{~cm}^{2}$

Let the radius of third circle = R 

Area between the last two circle = 115.5cm

${So}\pi\left(R^{2}-(7)^{2}\right)=115.5$

$\Rightarrow \frac{22}{7}\left(R^{2}-119\right)=115.5$

$\Rightarrow R^{2}-49=\frac{115.5 \times 7}{22}=36.75$

$\Rightarrow R^{2}=36.75+49=85.75 \mathrm{~cm}^{2}$

So $R=\sqrt{85.75}=9.26$

So radius of third circle = 9.26= 9.3

Question 17

Sol: perimeter of  a circular field $=650 \mathrm{~m}$

So diameter of the field = $\frac{\text { circumference }}{\pi}$

$=\frac{650}{75}=\frac{650 \times 7}{22} \mathrm{~m}$

If The square plot has its vertices on the circumference of the circle 

(IMAGE TO BE ADDED)

So diagonal of square = diameter of the circle 

$=\frac{650 \times 7}{22} \mathrm{~m}$

So Area of square plot $=\frac{\text { (diagonal) }^{2}}{2}$

$=\frac{1}{2}\left[\frac{650 \times 7}{22}\right]^{2}=\frac{1}{2}\left[\frac{4550}{221}\right]^{2}=\frac{1}{2}(206.818)^{2}$

$=\frac{1}{2}\left(42773.76) / \mathrm{m}^{2}=21386.88 \mathrm{~m}^{2}=21387 \mathrm{~m}^{2}\right.$

Question 18

Sol: Inside perimeter of the running hack with hemispheres ends and straight parallel   sides $=312 \mathrm{~m}$

Length of straight portion $=90 \mathrm{~m}$

$=\frac{312-180}{2}=\frac{132}{2}=66 \mathrm{~m}$

So radius = $\frac{\text { Perimeter } \times 2}{2 \pi}=\frac{66 \times 2 \times 7}{2 \times 22}=21 \mathrm{~m}$

Width of track =2m

So outer radius $=21+2=23 \mathrm{~m}$

So area of the track = Area of semicircular ends + Area of rectangular portion 

$=2 \pi\left[\left(231^{2}-(21)^{2}\right]+90 \times 2 \times 2\right.$

$=2 \pi(23+21)(23-21)+360$

$=2 \pi \times 44 \times 2+360=2 \times 88 \pi+360$

$=(176 \pi+360) \mathrm{m}^{2} .$

Question 19

Sol: Diameter of large semicircle $=10 \mathrm{~cm}$

So Radius $(P)=\frac{10}{2}=5 \mathrm{~cm}$

Diameter of each of the smaller semicircles $=5 \mathrm{~cm}$

So Radius of each semicircle $(r)=\frac{5}{2} \mathrm{~cm}$

(IMAGE TO BE ADDED)

(i) perimeter of the shaded portion

$\begin{gathered}=\frac{1}{2} \times 2 \pi R+2 \times \frac{1}{2} \times 2 \pi r=\pi R+2 \pi r \\=\pi[R+2 \mathrm{r}]=3.14\left(5+2 \times \frac{\mathrm{s}}{2}\right) \\=3.14 \times 10 \mathrm{~cm}=31.4 \mathrm{~cm}\end{gathered}$

(ii) Area of shaded portion

$=\frac{1}{2} \pi R^{2}-\frac{1}{2} \pi r^{2}+\frac{1}{2} \pi r^{2}$

$=\frac{1}{2} \pi R^{2}=\frac{1}{2} \times 3.14 \times 5 \times 5 \mathrm{~cm}^{2}$

$=39.25 \mathrm{~cm}^{2}=39.3 \mathrm{~cm}^{2}$

Question 21

Sol: Length of chord $A B=11 \mathrm{~cm}$

is $O P \perp A B$

So $A P=P B=\frac{14}{2}=7 \mathrm{~cm}$

Let $O A=R$ and $O P=r$

in right $\triangle O A P$

$O A^{2}=O P^{2}+A P^{2}$

$\Rightarrow R^{2}=r^{2}+(7)^{2}$

$\Rightarrow R^{2}-r^{2}=49$..........(i)

Now area of shaded portion 

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

$=\pi\left(R^{2}-r^{2}\right)=\frac{22}{7} \times 49$

$=154 \mathrm{~cm}^{2}$

Question 22

Sol: Radius of the circle (r)= 40cm

$\angle A O B=90^{\circ}$

Area of Quadrant $O A B=\frac{1}{4} \pi r^{2}$ $=\frac{1}{4} \times 3.14 \times 40 \times 40 \mathrm{~cm}^{2}=1256 \mathrm{~cm}^{2}$ Area of $\triangle A O B=\frac{1}{2} O A \times O D$ $=\frac{1}{2} \times$ to $\times 40=800 \mathrm{~cm}^{2}$

So Area of shaded portion $=1256 \mathrm{~cm}^{2}-800 \mathrm{~cm}^{2}=456 \mathrm{~cm}^{2}$

Question 23

Sol: In the figure $O A B$ is a quadrant whose radius OH $=21 \mathrm{~cm}$

and radius of quadrant

$O C D=O C=14 \mathrm{~cm}$

So Area of bed (shaded portion)

$\begin{aligned}&=\frac{1}{4} \pi r^{2}-\frac{1}{4} \pi r^{2} \\&=\frac{1}{4} \pi\left[R^{2}-r^{2}\right]=\frac{1}{4} \times \frac{22}{7}\left(21^{2}-14^{2} 1\right.\\&=\frac{11}{14}[441-196]=\frac{11}{14} \times 245 \mathrm{~m}^{2} \\&=\frac{11 \times 35}{2}=\frac{385}{2} \\&=192.5 \mathrm{~m}^{2}\end{aligned}$

Question 24

Sol: side of a square $A D C D=14 \mathrm{~cm}$

$A B=B C=C D=D A=14 \mathrm{~cm}$

So Radius of each circle $=\frac{14}{2} \mathrm{~cm}=7 \mathrm{~cm}$

So Area of shaded portion =area y 4 circlest area of square -area of 4 quadrant

$=$ Area as 4 circles - area of one circle + area of square

$=$ Area of 3 circles $+$ area of square $=3 \times \pi r^{2}+a^{2}$

$=3 \times \frac{22}{7} \times 7 \times 7+14 \times 14 \mathrm{~cm}^{2}$

$=462+196=658 \mathrm{~cm}^{2}$

$=658 \mathrm{~cm}^{2}$