Class 9 Maths: Circles – Extra Questions with Answers

Ans: A

Solution:
Let the original radius be r. The original area is πr².
The new radius is 1.25r. The new area is π(1.25r)² = 1.5625πr².
The increase in area is 1.5625πr² – πr² = 0.5625πr².
The percentage increase is (0.5625πr² / πr²) * 100% = 56.25%.
Answer: 56.25%

Ans: D

Solution:
Arc length $s = r\theta$, where $\theta$ is in radians.
$45^\circ = \frac{45}{180}\pi = \frac{\pi}{4}$ radians.
$11 = r \cdot \frac{\pi}{4}$
$r = \frac{44}{\pi}$

Answer: $\frac{44}{\pi}$ cm

Ans: B

Solution:
Draw a radius to each end of the chord, forming an isosceles triangle. Draw a perpendicular from the center of the circle to the chord, bisecting the chord. This forms two right triangles. The hypotenuse of each right triangle is the radius (10 cm), and one leg is half the chord length (6 cm). Use the Pythagorean theorem: distance² + 6² = 10². Distance² = 100 – 36 = 64. Distance = 8 cm.
Answer: 8 cm

Ans: D

Solution:
Let r1 be the radius of the inner lane, and r2 be the radius of the outer lane.
The circumference of the inner lane is 2πr1.
The circumference of the outer lane is 2πr2.
The outer lane is 44 meters longer than the inner lane: 2πr2 = 2πr1 + 44
2π(r2 – r1) = 44
r2 – r1 = 44 / (2π)
r2 – r1 = 22 / π
The width of the track is r2 – r1. If we assume pi to be 22/7 then the width is 22/(22/7) = 7
Answer: 7 meters

Ans: C

Solution:
Let the side length of the square be s. The diagonal of the square is the diameter of the circle, which is 2*2=4. Using the Pythagorean theorem, s^2 + s^2 = 4^2, so 2s^2 = 16, and s^2 = 8. The area of the square is s^2.

Answer: 8

Ans: D

Solution: Since AB is perpendicular to BC, triangle ABC is a right-angled triangle with hypotenuse AC. The circle passing through A, B, and C has AC as its diameter. Using the Pythagorean theorem, AC = √(AB² + BC²) = √(9² + 12²) = √(81 + 144) = √225 = 15 cm.

Answer: 15

Ans: A

Solution:
Let the radius of the circle be $r$.
For the longer chord (24 cm), the distance from the center is 5 cm.
Half the length of the chord is 12 cm.
Using the Pythagorean theorem, $r^2 = 12^2 + 5^2 = 144 + 25 = 169$.
So, $r = 13$ cm.
For the shorter chord (10 cm), half the length is 5 cm.
Let the distance from the center be $d$.
Then $r^2 = 5^2 + d^2$.
$13^2 = 5^2 + d^2$
$169 = 25 + d^2$
$d^2 = 169 – 25 = 144$
$d = 12$ cm.

Answer: 12 cm

Ans: C

Solution:
Circumference = Distance / Revolutions = 4.4 km / 500 = 4400 m / 500 = 8.8 m
2 * pi * radius = 8.8 m
radius = 8.8 m / (2 * pi) = 8.8 m / (2 * 3.14) ≈ 1.4 m

Answer: 1.4 m

Ans: A

Solution: The four circles form a square with side length 16 cm (two radii). The area of the square is 16*16 = 256 cm². The four quarter circles make a full circle with radius 8 cm, and its area is π*8² = 64π cm². The area enclosed is the area of the square minus the area of the full circle, 256-64π.
Answer: 256-64π

Ans: D

Solution:
Consider two congruent circles with centers O and O’, and equal chords AB and A’B’ respectively. Since the circles are congruent, their radii are equal (OA = OB = O’A’ = O’B’). Triangles OAB and O’A’B’ are congruent by SSS (OA=O’A’, OB=O’B’, AB=A’B’). Therefore, the corresponding angles at the centers are equal (∠AOB = ∠A’O’B’) because corresponding angles of congruent triangles are equal.

Answer: ∠AOB = ∠A’O’B’