SChand CLASS 9 Chapter 17 Circle, Circumference and Area TEST

TEST 

Question 1

Ans: Number of revolution $=1000$ 

Distance travelled $=440 \mathrm{~m}$

So perimeter of the what $=\frac{440 \times 100}{1000}=14 . \mathrm{cm}$

and diameter $=\frac{\text { perimeter }}{\pi}$

$=\frac{44 \times 7}{22}=14 \mathrm{~cm}$

Question 2

Ans : Area of a semi-circular field $=30859 \mathrm{~m}$

$\begin{aligned}&r=\sqrt{\frac{\text { area \times 2 }}{\pi}}=\sqrt{\frac{308 \times 2 \times 7}{22}} \mathrm{~cm} \\&=\sqrt{14 \times 14}=\sqrt{196 \mathrm{~cm}}=14 \mathrm{~cm} \\&\text { perimeter }=\frac{1}{2} \times 2 \pi r+2 r=\pi r+2 r=\frac{22}{7} \times 14 \times 2 \\&=44+28=72 \mathrm{~cm}\end{aligned}$

Option (d) is correct 

Question 3

Ans : Length of wire $=36 \mathrm{~cm}$

So circumference of $\mathrm{semi}$ - circle $=36 \mathrm{~cm}$

$\begin{aligned}&\text { radius }=r \\&\text { so } \pi r+2 r=36 \\&\Rightarrow \frac{22}{7} r+2 r=36 \Rightarrow \frac{36}{7} r=36 \\&\Rightarrow r=\frac{36 \times 7}{36}=7 \mathrm{~cm}\end{aligned}$

Option (c) is correct 

Question 4

Ans: Radius of a circular wire = 42cm

So its perimeter =2 \pi r

$=2 \times \frac{22}{7} \times 42 \mathrm{~cm}=264 \mathrm{~cm}$

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

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

option (a) is correct

Question 5

Ans: Area of square metal wire $=484 \mathrm{~cm}^{2}$

$\begin{aligned}&\text { So } \text { side }=\sqrt{\text { Area }}=\sqrt{484} \mathrm{~cm}=22 \mathrm{~cm} \\&\text { So perimeter }=4 \times \text { side }=4 \times 22=88 \mathrm{~cm} \\&\text { So perimeter of circle }=88 \mathrm{~cm} \\&\text { So } \text { Radius of circle }=\frac{\text { perimeter } \mathrm{kr}}{2 \pi} \\&=\frac{88 \times 7}{2 \times 22}=14 \mathrm{~cm} \\&\text { and area }=\pi r^{2}=\frac{22}{7} \times 14 \times 14 \mathrm{~cm}^{2} \\&=616 \mathrm{~cm}^{2}\end{aligned}$

Option (a) is correct 

Question 7

Sol: A circle is circumscribed by a rectangle with

 sides $16 \mathrm{~cm}$ and $12 \mathrm{~cm}$

(IMAGE TO BE ADDED)

So Diagonal $A C=\sqrt{A B^{2}+B C^{2}}$

$\begin{aligned}&=\sqrt{16^{2}+12^{2}}=\sqrt{256+114 \mathrm{~cm}} \\&=\sqrt{400}=20 \mathrm{~cm}\end{aligned}$

So diameter of circle = 20m

And radius = $\frac{20}{2}=10 \mathrm{~cm}$
Area of circle = $\pi r^{2}=\pi \times 10 \times 10$ = $100 \pi$ $c m^{2}$

Question 8

Sol: (R)=50m

and radius of wheel of a bicycle $=50 \mathrm{~cm}$ 

Distance travelled by wheel of o cycle $=1$ hr.

Now circumference of $p a t h=2 \pi R$

$=\frac{2 \times 22}{7} \times 50 \mathrm{~m}=\frac{2200}{7} \mathrm{~m}$

and circumference of wheel

$=\frac{2 \times 22}{7} \times 50 \mathrm{~cm}=\frac{2200}{7} \mathrm{~cm}$

So Nam ben of revolution $=\frac{2200 \times 100 \times 7}{7 \times 2200}$

$=100$

Number of revolution in is min

$=\frac{100 \times 15}{60}=25(1 \mathrm{hr}=60 \mathrm{~min})$

Question 9

Sol: The given figure consists is of 4 -small semi-circles 

Radii of each small semi-circle $=12 \mathrm{~cm}$ and rodius of big semi-circles

$=42 \times 2=84 \mathrm{~cm}$

(IMAGE TO BE ADDED)

perineter of shaded portion

$\begin{aligned}&=4 \times \pi r+2 \times \pi R \\&=4 \times \frac{22}{7} \times 42+2 \times \frac{22}{7} \times 84 \mathrm{~cm} \\&=528+528=1056 \mathrm{~cm}\end{aligned}$

Question 10

Ans: Radius of each quadrant $=4 \mathrm{~cm}$

 Length of cardboard $=18 \mathrm{~cm}$ 

and width $=10 \mathrm{~cm}$

(IMAGE TO BE ADDED)

Now circumference of four quadrants

$\begin{aligned}&=4 \times \frac{1}{2} \times \pi r \\&=2 \times \frac{22}{7} \times 4=\frac{176}{7}=25-1 \mathrm{~cm}\end{aligned}$

perimeter of remaining portion

$=2(18+10)-4 \times 8=56-32=24 \mathrm{~cm}$

Total perimeter $=25.1+24=49.1 \mathrm{~cm}$