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

Exercise 17 A

Question 1

Ans:1  (i) Diameter of the circle $(d)=4 \mathrm{~cm}$ so circumference $=\pi d=\frac{22}{7} \times 49=154 \mathrm{~cm}$

(ii) Diameter of the circle $(d)=14 \mathrm{~cm}$ So Circumference $=\pi d=\frac{22}{7} \times 14=44 \mathrm{~cm}$

(iii) Diameter of the circle (d) $=9.8 \mathrm{~cm}$ So circumference $=\pi d=\frac{22}{7} \times 9.8=30.8 \mathrm{~cm}$

(iv) $D=7 \mathrm{~cm}$

So Circumference $=\pi d=\frac{22}{7} \times 7=22 \mathrm{~cm}$

Question 2

Ans: (i) Radius of the circle $(r)=7 \mathrm{~cm}$

So circumference $=2 \pi r=2 \times \frac{22}{7} \times 7=44 \mathrm{~cm}$

(ii) $(r)=28 \mathrm{~cm}$

So circimference $=2 \pi r=2 \times \frac{22}{7} \times 28=176 \mathrm{~cm}$

(iii) $\quad(R)=3.5 \mathrm{~cm}$

so circumference $=2 \pi r=2 \times \frac{22}{7} \times 3.5=22 \mathrm{~cm}$

(iv) $(r)=98 \mathrm{~m}$

so circumference $=2 \pi r=2 \times \frac{22}{7} \times 98 \mathrm{~m}=616 \mathrm{~m}$

Question 3

Ans:(i) Radius $(r)=4.5 \mathrm{~cm}$ circumference $=2 \pi r=2 \times 3.14 \times 4.5$ $=28.26 \mathrm{~cm}=28.3 \mathrm{~cm}$

(ii)

 $\begin{aligned} \text { (d) } &=15 \mathrm{~cm} \\ \text { so } \text { circumference } &=\pi \mathrm{d}=3.14 \times 15 \mathrm{~cm} \\ &=47.10=47.1 \mathrm{~cm} \end{aligned}$

Question 4

Ans: (i) circumference $=22 \mathrm{~cm}$

Let of be the diameter of the circle

So $d=\frac{\text { circumference }}{\pi}=\frac{22}{\frac{22}{7}}=\frac{22 \times 7}{22}=7 \mathrm{~cm}$

so Diameter $=7 \mathrm{~cm}$

(ii)

$\begin{aligned} \text { circumference } &=8.8 \mathrm{~cm} \\ \text { so } \text { Diameter } &=\frac{\text { circumference }}{\pi}=\frac{8.8}{\frac{22}{7}} \\=\frac{8.8 \times 7}{22} &=0.4 \times 7=2.8 \mathrm{~cm} \end{aligned}$

(iii) circumference $=44 \mathrm{~cm}$

$\text { so } D=\frac{\text { circum furence }}{\pi}=\frac{\frac{14}{22}}{7}=\frac{44 \times 7}{22}=14 \mathrm{~cm}$

Question 5

Ans : (i)circumference $=100 \mathrm{~m}$

let the radius of the circle $=r$

$\begin{aligned}&\text { So } 2 \pi r=100 \\&\Rightarrow 2 \times 3.14 \times r=100 \\&\Rightarrow r=\frac{100}{2 \times 3.14}=\frac{100 \times 100}{2 \times 314} \\&=\frac{10000}{628}=15.92=15.9 \mathrm{~m} \\&\text { So } \text { Radius }=15.9 \mathrm{~m}\end{aligned}$

(ii) circumference $=6.4 \mathrm{dm}$

Let radius of the circle $=r$

$\begin{aligned}&\text { Then. } 2 \pi r=6.4 \\&\Rightarrow 2 \times 3.14 \times r=6.4 \\&\Rightarrow 6.28 r=6.4 \Rightarrow r=\frac{6.4}{6.28} \\&\Rightarrow r=1.019=1.02 \mathrm{dm} \\&\text { So } r=1.02 \mathrm{dm}\end{aligned}$

Question 6

Ans: 

$\begin{aligned} \text { Diameter of Venus planet } &=12.278 \mathrm{~km} \\ \text { so circumference } &=\pi d=\frac{22}{7} \times 12278 \\=22 \times 1754 &=38588 \mathrm{~km} \end{aligned}$

Question 7

(i) Radius of the semicircle $=3.85 \mathrm{~cm}$

(IMAGE TO BE ADDED)

So perimeter $=\pi r+2 r$

$=\frac{22}{7} \times 3.85+2 \times 3.85$

$=22 \times 0.55+7.70$

$=12.10+7.70=19.80 \mathrm{~cm}$

(ii)Let $r$ be the radius of the circle

$\text { so } \begin{aligned}& 2 \pi r-2 r=16.8 \\\Rightarrow & 2 r(\pi-1)=16.8 \\\Rightarrow & 2 r\left(\frac{22}{7}-1\right)=16.8 \Rightarrow 2 r\left(\frac{15}{7}\right)=16.8 \\=& r=\frac{16.8 \times 7}{2 \times 15}=\frac{168 \times 7}{10 \times 30}=\frac{56 \times 7}{100} \\=& \frac{392}{100}=3.92\end{aligned}$

So radius of the circle = 3.92cm 

Question 8

Ans: Radius of the circler wire $=42 \mathrm{~cm}$

so circumference $=2 \pi r=2 \times \frac{22}{7} \times 42 \mathrm{~cm}=264 \mathrm{~cm}$

on bending it into a square,

The perimeter of square $=264 \mathrm{~cm}$

so side of the square $=\frac{\text { perimeter }}{4}$

$=\frac{264}{4}=66 \mathrm{~cm}$

Question 9

Ans : Radius of the 

circle $=3.5 \mathrm{~cm}$

(IMAGE TO BE ADDED)

So perimeter $=2 \pi r \times \frac{1}{4}+2 r$

$\begin{aligned}=\frac{1}{2} \pi r &+2 r=\frac{1}{2} \times \frac{22}{7} \times 3.5+2 \times 3.5 \\&=5.5+7.0=12.5 \mathrm{~cm}\end{aligned}$

Question 10

(i)  The inner circumference of the circular track $=440 \mathrm{~m}$

(IMAGE TO BE ADDED)

So $(R)=\frac{\text { circumference }}{2 \pi}$

$=\frac{440 \times 7}{2 \times 22}=70 \mathrm{~m}$

width of track $=14 \mathrm{~m}$

So Radius of the Outer circle (R)

$=70+14=84 m$

and diameter $=2 R=2 \times 84=168 \mathrm{~m}$

(ii)The inner edge of the circular track $=440 \mathrm{~m}$ 

So inner radius $(r)=\frac{\text { circumference }}{2 \pi}$

$=\frac{440 \times 7}{22 \times 2}=70 \mathrm{~m}$

with of track $=10 \mathrm{~m}$

So outer radius $(R)=70+10=80 \mathrm{~m}$

so length of outer edge

$\begin{aligned}=2 \pi R &=2 \times \frac{22}{7} \times 80=\frac{3520}{7} \mathrm{~m} \\&=502.86 \mathrm{~m}\end{aligned}$

Question 11

Ans; Circumference of roller $=3 \mathrm{~m}$

Distance travelled $=21 \mathrm{~m}$

So Number of revolution will be

$=\frac{\text { Distance }}{\text { circumference }}=\frac{21}{3}=7$

Question 12

Ans: Diameter of the cycle wheel (d) = 70 cm

So circumference $=\pi d=\frac{22}{7} \times 70=220 \mathrm{~cm}$

Distance travelled in 25 revolutions 

$=220 \times 25=5500 \mathrm{~cm}=55 \mathrm{~m}$

So distance of 55m travelled in 10 sec 

So The distance travelled in 1 hour

$=\frac{55}{10} \times 60 \times 60 \mathrm{~m}=19800 \mathrm{~m}=19.8 \mathrm{~km}$

so speed $=19.8 \mathrm{~km} / \mathrm{h}$

Question 13

Sol: Distance of wheel (d)= 91cm

So circumference = $\pi d=\frac{22}{7} \times 91=286 \mathrm{~cm}$

Speed $=10 \mathrm{~km} / \mathrm{hr}$.

Distance travelled in 1 hour $=10 \mathrm{~km}$ So Distance travelled in $1 \mathrm{~min}=\frac{10}{60} \mathrm{~km}$

$=\frac{10 \times 1000}{60} m=\frac{500}{3} m$

So Number of revolution in one min

$\begin{aligned}&=\frac{500}{3} \times \frac{100}{286} \\&=\frac{500 \times 50}{3 \times 143}=\frac{25000}{429}=58 \frac{118}{429}\end{aligned}$

Question 14

Ans: Radius of circular field $(\gamma)=35 \mathrm{~m}$

so circumference $=2 \pi r=2 \times \frac{22}{7} \times 35$

$=220 \mathrm{~m}$

Distance travelled in 4 round $=220 \times 4 \mathrm{~m}=880 \mathrm{~m}$ speed of walking $=5 \mathrm{~km} / \mathrm{h}$

So Time taken to travelled $880 \mathrm{~m}=\frac{880}{5000} \mathrm{hr}$

$\begin{aligned}&=\frac{880 \times 60}{5000}=\frac{264}{25}=10.76 \mathrm{~min} \\&=10 \mathrm{~min} 34 \mathrm{sec} \text { (Approx) }\end{aligned}$

Question 15

Sol : (d) =84m

So circumference $=\pi_{0} 1=\frac{22}{7} \times 84=267 \mathrm{~cm}$ Distance covened in $1 \mathrm{sec}=4$ revolutions $=4 \times 264=1056 \mathrm{~cm}$

So Distance covered in 1 hour

$\begin{aligned}&=1056 \times 60 \times 60 \mathrm{~cm}=\frac{1056 \times 60 \times 60}{100 \times 1000}\mathrm{~km} \\&=\frac{38016}{1000}=38.016 \mathrm{~km}\end{aligned}$

So speed of the cart= $38.016 \mathrm{~km} / \mathrm{hr}$

Question 16

Diameter of semicircular part $=70 \mathrm{~m}$

So circumference of two semicircular part

$=2 \times \frac{1}{2} \pi d=\pi d=\frac{22}{7} \times 70=220 \mathrm{~m}$

So length of one complete round 

$=A B C+D E F+A F+C D$

$=220+100+100 m=420 m$

Question 17

Ans : Perimeter of the circular field = 660m

(IMAGE TO BE ADDED)

So $\begin{aligned} &(r)=\frac{\text { circumfenence }}{2 \pi} \\=& \frac{660 \times 7}{2 \times 22}=15 \times 7=105 \mathrm{~m} \end{aligned}$

Join AC

So Length of diagonal of square $A D C D$

$=$ diameter of the circle $=2 r$

$=2 \times 105=210 \mathrm{~m}$

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

$=\frac{12101^{2}}{2}=\frac{210 \times 210}{2} \mathrm{~m}^{2}=22,050 \mathrm{~m}^{2}$