# Circle Questions for RRB NTPC PDF

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## Circle Questions for RRB NTPC PDF

Download RRB NTPC Top-15 Circle Questions PDF. Questions based on asked questions in previous exam papers very important for the Railway NTPC exam.

Question 1: A sector of a circle subtending angle 36° has an area 3.85 $cm^{2}$. The length of the arc is

a) 2.2 cm

b) 2.5 cm

c) 2.7 cm

d) None of these

Question 2: A wire when bent in the form of a square encloses an area of 484 sq. cm. If the same wire is bent in the form of a circle, what is the area enclosed by it?

a) 264 sq. cm

b) 616 sq. cm

c) 488 sq. cm

d) 492 sq. cm

Question 3: If the circumference of a circle is 22 cm, find the area of the semicircle.

a) 38.5 sq.cm

b) 19.25 sq.cm

c) 44 sq.cm

d) 77 sq.cm

Question 4: What is the ratio of radius of incircle and circumcircle for an equilateral triangle with an area of $22\surd3$ square units?

a) 1/2

b) 1/4

c) 1/3

d) 2/3

Question 5: If the radius (r) of a circle is increased by ‘x’ units, what is the number of units by which the circumference of the circle is increased?

a) $\pi$

b) $2\pi$

c) $2\pi r$

d) $2\pi x$

Instructions

Consider the following information and questions based on it.
P, W, Q, X, R, Y, Z and S sit randomly in a circle facing each other.
1. Y sits exactly between Q and R.
2. P does not sit next to either X or Z.
3. Z is to the immediate right of Q.
4. S sits third from the left of X and the right of Y.

Question 6: If they all were facing outside the circle, W would be to the left of

a) X

b) S

c) P

d) Cannot be determined

Question 7: If the area of a circle is $9\pi$ sq. cm then its circumference is

a) 9 cm

b) $6 \pi$ cm

c) $3 \pi$ cm

d) 6 cm

Question 8: In the circle below, chord $\overline{AB}$ is extended to meet the tangent $\overline{DE}$ at D. If $\overline{AB}$ = 9 cm and $\overline{BD}$ = 3 cm, find the length of $\overline{DE}$.

a) 5 cm

b) 4 cm

c) $\sqrt{27}$ cm

d) 6 cm

Question 9: The area of a circle is 616 sq.m. Find its diameter. $(\pi=22/7)$

a) 7m

b) 14m

c) 28m

d) 56m

Question 10: A circle touches all the sides of a quadrilateral PQRS. whose sides, PQ = 2cm, QR = 3cm and RS = 4cm, What is the length of PS?

a) 3 cm

b) 2 cm

c) 1 cm

d) 4 cm

Question 11: The area of the circumcircle of a right angle triangle whose sides are given as 2 cm, $2\sqrt{3}$ cm and 4 cm, is given by

a) $6 \pi$ $cm^2$

b) $4 \pi$ $cm^2$

c) $16 \pi$ $cm^2$

d) $12 \pi$ $cm^2$

Question 12: 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:$\pi=\frac{22}{7}$)

a) $154\ cm^2$

b) $153\ cm^2$

c) $155\ cm^2$

d) $150\ cm^2$

Question 13: The area of a circle got increased by 22 cm when the radius was increased by 1 cm. What was the original radius?

a) 5 cm

b) 9 cm

c) 3 cm

d) 7 cm

Question 14: If a chord of length 24 cm is at a distance of 5 cm from centre, then find the radius of the circle.

a) 17 cm

b) 15 cm

c) 25 cm

d) 13 cm

Question 15: What is the area of a circular track that runs around a circle with circumference 440 m, having a width of 7 m?

a) 3856 $m^2$

b)

3234 .

c) 3900 $m^2$

d) 3204 $m^2$

The area of the arc = $\frac{36}{360} * \pi * r^2$ = 3.85

Solving, we get r = 3.5

Length of the arc = $\frac{36}{360} * 2 * \pi * r$ = 2.2

Let say, radius of the circle is r.

Area of the square =484 sq. cm.

So,Side of the square$\sqrt{484\ }=22$ cm.

same wire is used to form the circle.

So, perimeter of the circle will be same as perimeter of the square.

So, $2\times\pi\ \times r=4\times\ 22.$

or,$r=\frac{88\ }{2\pi\ }=14.$

So,Area of the circle$\pi r^2=\pi\times14^2=\ \frac{\ 22}{7}\times14^2=616\ cm^2.\$

B is correct choice.

According to question :

2×(22/7)×r=22.

or,r=(7/2)=3.5.

So,Area of circle=$πr^2$

=$(22/7)(3.5)^2$.

=38.5 $cm^2$

So,Area of semicircle = (38.5/2)=19.25$cm^2$

B is correct choice.

We know that

radius of incircle of equilateral triangle = $\frac{side}{2\sqrt{3}}$

radius of circumcircle of equilateral triangle = $\frac{side}{\sqrt{3}}$

hence there ratio = 1:2

Circumference of circle = $2\pi\ r$

Circumference of circle when radius increased by x = $2\pi\ \left(r+x\right)=2\pi\ r+2\pi\ x$

Hence, circumference of the circle increased by $2\pi x$ units.

W seats left of P.

$\pi\ r^2=9\pi\$

$r^2=9$

r = 3

Circumference = $2\pi\ r$= $2\pi\ \times\ 3=6\pi\$

Area of the circle = $\pi\ r^2$

616 = $\frac{22}{7}\times\ r^2$

$r^2=196$

r = 14 m

Hence diameter = 2r = 28 m

C which is the length of the hypotenuse of a right angled triangle of a circumcircle passing through the centre of the circumcircle.

Sides are given as 2 cm, $2\sqrt{3}$ cm and 4 cm,

Since hypotenuse of a right angled triangle is the longest side i.e c = 4 cm which makes it the diameter of the circumcircle.

Therefore, area of the circumcircle = $\pi \times R^2 = \pi \times 2^2$

$=\pi \times 4 cm^2$

Option B is correct.

Given

Encloses area by forming the square = 121$cm^2$

we know that

Area of square =$a^2$

so,

$a^2$=121

a=11 cm

now perimeter of square = perimeter of circle according to question 

from here we get r=7 cm

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

= $\frac{22}{7}\cdot7^2=154cm^2$ Ans

area of circle = $\pi\times r^2$

ATQ

$\pi(r + 1)^2- \pi\times r^2 = 22$

$\pi((r + 1)^2- r^2) = 22$

$\frac{1+ 2r}{7}=1$

2r = 6

r = 3cm

it becomes a right angle triangle

chord = 24 cm

so

base of triangle = half of 24 = 12 cm

height = 5cm

apply pythagoras’s thearom

hypotenuse = $\sqrt(5^2 + 12^2)$

= 13cm

Circumference=$2\times \frac{22}{7} \times R=440$

=$\frac {22}{7} \times 77 \times 77 – \frac {22}{7} \times 70 \times 70$