CBSE Class 10 Areas related to Circles Sure Shot Questions Set 01

Points to Remember

• 1. Circumference of a circle = \( 2\pi r \)

• 2. Area of circle = \( \pi r^2 \)
• 3. Length of an arc of a sector of a circle with angle of degree measure \( \theta = \frac{\theta}{360} \times 2\pi r \)
• 4. Area of a sector of a circle with angle of degree measure \( \theta = \frac{\theta}{360} \times \pi r^2 \)
• 5. Area of segment of a circle = Area of corresponding sector – Area of the corresponding triangle.

Multiple Choice Questions

Question. If the difference between the circumference and the radius of a circle is 37 cm, then using \( \pi = \frac{22}{7} \), the radius of the circle (in cm) is :
(a) 154
(b) 44
(c) 14
(d) 7
Answer: (d)
Sol. Let the radius be \( r \text{ cm} \).
Thus, circumference = \( 2\pi r \)
Now, \( 2\pi r - r = 37 \) (Given)
\( \Rightarrow r \left( \frac{44}{7} - 1 \right) = 37 \)
\( \Rightarrow r \left( \frac{44 - 7}{7} \right) = 37 \)
\( \Rightarrow r = \frac{7 \times 37}{37} = 7 \text{ cm} \)
So, the correct option is (d).

Question. If \( \pi \) is taken as \( \frac{22}{7} \), the distance (in metres) covered by a wheel of diameter 35 cm, in one revolution is :
(a) 2.2
(b) 1.1
(c) 9.625
(d) 96.25
Answer: (b)
Sol. Given, diameter = 35 cm
Now, One revolution = Circumference
\( = \pi d \)
\( = \frac{22}{7} \times 35 \)
\( = 110 \text{ cm} = 1.1 \text{ m} \)
So, the correct option is (b).

Question. If the sum of circumferences of two circles with radii \( R_1 \) and \( R_2 \) respectively, is equal to the circumference of a circle of radius \( R \), then
(a) \( R_1 + R_2 > R \)
(b) \( R_1 + R_2 < R \)
(c) \( R_1 + R_2 = R \)
(d) \( R_1 - R_2 = R \)
Answer: (c)
Sol. According to question,
\( 2\pi R_1 + 2\pi R_2 = 2\pi R \)
\( \Rightarrow 2\pi (R_1 + R_2) = 2\pi R \)
\( \Rightarrow R_1 + R_2 = R \)
So, the correct option is (c).

Question. The circumference of a circle exceeds its diameter by 180 cm. Then, its radius will be :
(a) 45 cm
(b) 42 cm
(c) 30 cm
(d) 32 cm
Answer: (b)
Sol. Let the radius of circle be \( r \).
Then,
\( 2\pi r - 2r = 180 \) (Given)
\( \Rightarrow 2r (\pi - 1) = 180 \)
\( \Rightarrow 2r \left( \frac{22}{7} - 1 \right) = 180 \)
\( \Rightarrow 2r \left( \frac{15}{7} \right) = 180 \)
\( \Rightarrow r = \frac{180 \times 7}{2 \times 15} = 42 \)
So, the correct option is (b).

Question. An umbrella has 8 ribs that are equally placed. If the umbrella is placed as 2D-structure, then the angle formed between two consecutive ribs is :
(a) 45°
(b) 36°
(c) 40°
(d) 50°
Answer: (a)
Sol. Since, the ribs of umbrella divides it into 8 equal parts,
\(\therefore\) Angle between two consecutive ribs = \( \frac{360^\circ}{8} \)
\( = 45^\circ \)
So, the correct option is (a).

Fill in the Blanks

Question. If the perimeter of a semi-circular protractor is 18 cm, then its radius is ...........
Answer: 3.5 cm

Question. A sector of a circle becomes segment also, if the degree measure of the sector angle is ..............
Answer: 180°

Question. If the area of a circle is \( 49\pi \), then its perimeter is ............
Answer: 44 units

Question. The area of the square that can be inscribed in a circle of radius 4 cm is ..............
Answer: 32 cm²

True/False

Question. The distance around a circle or the length of a circle is called its circumference.
Answer: True

Question. Perimeter of a protractor is represented by the formula \( \pi r + 2r \).
Answer: True

Question. Area of quadrant is one-half of area of circle.
Answer: False
Area of quadrant is one-fourth of area of circle.

Question. The perimeter of quadrant of a circle of radius \( r \) is \( \frac{r}{2} (\pi + 4) \).
Answer: True

Very Short Answer Type Questions

Question. Find the diameter of semi-circular protractor if its perimeter is 36 cm. 
Answer: Sol. Given : Perimeter of semi-circular protractor = 36 cm
We know, Perimeter of semi-circular protractor = \( \pi r + 2r \)
\(\therefore \pi r + 2r = 36 \)
\( \Rightarrow r(\pi + 2) = 36 \)
\( \Rightarrow r \left( \frac{22}{7} + 2 \right) = 36 \)
\( \Rightarrow r \left( \frac{36}{7} \right) = 36 \)
\( \Rightarrow r = 7 \)
So, Diameter = \( 2r = 2 \times 7 = 14 \text{ cm} \).

Question. Find the area of annulus whose inner and outer radii are 6 cm and 8 cm.
Answer: Sol. Given,
Inner radius of annulus = 6 cm
Outer radius of annulus = 8 cm
Area of annulus = \( \pi(R^2 - r^2) \)
= \( \pi(8^2 - 6^2) \)
= \( \pi(64 - 36) \)
= \( 28\pi \)
= \( 28 \times \frac{22}{7} \)
= \( 88 \text{ cm}^2 \) Ans.

Question. If the area of a circle is numerically equal to twice its circumference then find the diameter of the circle.
Answer: Sol. Let the radius of the circle be \( r \).
Given, the area of a circle is numerically equal to twice its circumference.
Thus \( \pi r^2 = 2(2\pi r) \)
or \( r = 4 \) units
Thus, the diameter of the circle = 8 units. Ans.

Question. Find the area of a sector of a circle with radius 6 cm, if its angle of sector is 60°. 
Answer: Sol. Given : \( r = 6 \text{ cm} \) and \( \theta = 60^\circ \)
Now, Area of sector = \( \frac{\theta}{360} \times \pi r^2 \)
= \( \frac{60}{360} \times \frac{22}{7} \times (6)^2 \)
= \( \frac{1}{6} \times \frac{22}{7} \times 6 \times 6 \)
= \( 18.86 \text{ cm}^2 \) Ans.

Short Answer Type Questions-I

Question. The radii of two circles are 4 cm and 3 cm respectively. Then, find the diameter of the circle having an area equal to the sum of areas of the two circles (in cm).
Answer: Sol. Let radii of two circles be \( r_1 \) and \( r_2 \) respectively.
Then, \( r_1 = 4 \text{ cm} \) and \( r_2 = 3 \text{ cm} \) [Given]
Let the radius of the larger circle = \( R \)
Thus, \( \pi R^2 = \pi[(r_1)^2 + (r_2)^2] \)
\( \Rightarrow R^2 = [(4)^2 + (3)^2] \)
\( \Rightarrow R^2 = [16 + 9] \)
\( \Rightarrow R^2 = 25 \)
\( \Rightarrow R = 5 \text{ cm} \)
[as radius cannot be negative]
Thus, the diameter of the larger circle = 10 cm. Ans.

Question. The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of circumferences of the two circles. 
Answer: Sol. Given, the radii of two circles are 19 cm and 9 cm.
Let the radius of required circle be \( R \).
Then,
Circumference of required circle = Circumferences of two circles
\( \Rightarrow 2\pi R = 2\pi (19) + 2\pi (9) \)
\( \Rightarrow 2\pi R = 2\pi (19 + 9) \)
\( \Rightarrow R = 28 \text{ cm} \) Ans.

Question. Find the area of a quadrant of a circle whose circumference is 88 cm.
Answer: Sol. Given,
Circumference of circle = 88 cm
\( \Rightarrow 2\pi r = 88 \text{ cm} \)
\( r = \frac{88}{2 \times \frac{22}{7}} \)
= \( \frac{88 \times 7}{2 \times 22} \)
\( r = 14 \text{ cm} \)
Area of quadrant = \( \frac{1}{4} \pi r^2 \)
= \( \frac{1}{4} \times \frac{22}{7} \times 14 \times 14 \)
= \( 154 \text{ cm}^2 \). Ans.

Multiple Choice Questions

Question. The area of the circle is 154 cm². The radius of the circle is
(a) 7 cm
(b) 14 cm
(c) 3.5 cm
(d) 17.5 cm
Answer: (a)
Explanation: Area of circle = 154 cm²
\( \Rightarrow \pi r^2 = 154 \) cm²
\( \Rightarrow \frac{22}{7} \times r^2 = 154 \Rightarrow r^2 = 154 \times \frac{7}{22} \)
\( \Rightarrow r^2 = 7 \times 7 = 49 \therefore r = \sqrt{49} = 7 \) cm

Question. If angle of sector is 60°, radius is 3.5 cm then length of the arc is
(a) 3 cm
(b) 3.5 cm
(c) 3.66 cm
(d) 3.8 cm
Answer: (c)
Explanation: Here \( r = 3.5 \) cm \( = \frac{35}{10} = \frac{7}{2} \) cm, \( \theta = 60^\circ \)
Length of arc = \( \frac{\theta}{360^\circ} \times 2\pi r \)
\( = \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times \frac{7}{2} = \frac{1}{6} \times 22 = \frac{11}{3} = 3.66 \) cm

Question. The area of a quadrant of a circle whose circumference is 22 cm, is
(a) \( \frac{11}{8} \) cm²
(b) \( \frac{77}{2} \) cm²
(c) \( \frac{77}{4} \) cm²
(d) \( \frac{77}{8} \) cm²
Answer: (d)
Explanation: Here circumference, \( 2\pi r = 22 \) cm
\( 2 \times \frac{22}{7} \times r = 22 \Rightarrow r = 22 \times \frac{7}{22} \times \frac{1}{2} = \frac{7}{2} \) cm
\( \therefore \) Area of quadrant of circle = \( \frac{1}{4} \pi r^2 \)
\( = \frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{77}{8} \) cm²

Question. If \( \theta \) is the angle in degrees of a sector of a circle of radius ‘r’, then area of the sector is
(a) \( \frac{\pi \theta r^2}{180^\circ} \)
(b) \( \frac{\pi \theta r^2}{360^\circ} \)
(c) \( \frac{2\pi \theta r}{180^\circ} \)
(d) \( \frac{2\pi \theta r}{360^\circ} \)
Answer: (b)

Question. A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 7 m long rope. The area of that part of the field in which the horse can graze, is
(a) 77 cm²
(b) \( \frac{77}{2} \) cm²
(c) 154 cm²
(d) \( \frac{77}{4} \) cm²
Answer: (b)
Explanation: Here \( r = 7 \) m, \( \theta = 90^\circ \)
\( \therefore \) Area of sector = \( \frac{\theta}{360^\circ} \times \pi r^2 = \frac{90^\circ}{360^\circ} \times \frac{22}{7} \times 7 \times 7 \)
\( = \frac{1}{4} \times 22 \times 7 = \frac{77}{2} \) m²

Question. The area of the circle whose diameter is 21 cm is
(a) 346.5 cm²
(b) 37.68 cm²
(c) 18.84 cm²
(d) 19.84 cm²
Answer: (a)
Explanation: Here diameter = 21 cm
\( \therefore \) Radius, \( r = \frac{21}{2} \) cm
Area of the circle, \( A = \pi r^2 \)
\( \therefore A = \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} = 11 \times 3 \times \frac{21}{2} = \frac{693}{2} = 346.5 \) cm²

Question. The area of the sector of a circle with radius 6 cm and of angle 60° is
(a) 9.42 cm²
(b) 37.68 cm²
(c) 18.84 cm²
(d) 19.84 cm²
Answer: (c)
Explanation: Here \( r = 6 \) cm, \( \theta = 60^\circ \)
Area of the sector = \( \frac{\theta}{360^\circ} \times \pi r^2 \)
\( \therefore \) Area = \( \frac{60^\circ}{360^\circ} \times 3.14 \times 6 \times 6 \)
\( = \frac{1}{6} \times 3.14 \times 6 \times 6 = 3.14 \times 6 = 18.84 \) cm²

Question. The area of a circle whose circumference is 22 cm, is
(a) 11 cm²
(b) 38.5 cm²
(c) 22 cm²
(d) 77 cm²
Answer: (b)
Explanation: Circumference of circle = 22 cm,
\( 2\pi r = 22 \) cm
\( \Rightarrow 2 \left( \frac{22}{7} \right) r = 22 \Rightarrow r = \frac{22 \times 7}{2 \times 22} = \frac{7}{2} \) cm
\( \therefore \) Area of circle = \( \pi r^2 = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = \frac{77}{2} = 38.5 \) cm²

Question. The area of a circle is 154 cm². Its diameter is
(a) 7 cm
(b) 14 cm
(c) 21 cm
(d) 28 cm
Answer: (b)
Explanation: Here area of the circle, \( A = 154 \) cm², Radius, \( r = ? \)
Area of the circle = 154 cm² ...(Given)
\( \therefore \pi r^2 = 154 \Rightarrow \frac{22}{7} \times r^2 = 154 \)
\( \Rightarrow r^2 = 154 \times \frac{7}{22} = 7 \times 7 \Rightarrow r = 7 \) cm
\( \therefore \) Diameter of the circle = \( 2 \times r = 2 \times 7 = 14 \) cm

Question. The length of the minute hand of a clock is 14 cm. The area swept by the minute hand in 5 minutes is
(a) 153.9 cm²
(b) 102.6 cm²
(c) 51.3 cm²
(d) 205.2 cm²
Answer: (c)
Explanation: Angle swept by the minute hand in 1 minute = \( (360^\circ \div 60) = 6^\circ \)
\( \therefore \) Angle swept by the minute hand in 5 min. (\( \theta \)) = \( 6^\circ \times 5 = 30^\circ \)
Length of minute hand (\( r \)) = 14 cm
\( \therefore \) Area swept = \( \frac{\theta}{360^\circ} \pi r^2 \)
\( = \frac{30^\circ}{360^\circ} \times \frac{22}{7} \times 14 \times 14 = \frac{154}{3} = 51.3 \) cm²

Question. The radii of two circles are 19 cm and 9 cm respectively. The radius of the circle which has circumference equal to the sum of the circumference of two circles is
(a) 35 cm
(b) 10 cm
(c) 21 cm
(d) 28 cm
Answer: (d)
Explanation: Let the radii of two circles be \( r_1 \) and \( r_2 \) and the radius of large circle be \( r \).
\( \therefore r_1 = 19 \) cm, \( r_2 = 9 \) cm
Circumference of two circles = \( C_1 + C_2 = C \) (where C = large circle)
\( = 2\pi r_1 + 2\pi r_2 = 2\pi \times 19 + 2\pi \times 9 = 38\pi + 18\pi = 56\pi \)
\( \therefore \) Circumference of large circle = \( 56\pi \)
\( \Rightarrow 2\pi r = 56\pi \Rightarrow r = 28 \)
\( \therefore \) Radius of large circle = 28 cm

Question. The area of the circle that can be inscribed in a square of side 6 cm, is
(a) \( 18\pi \) cm²
(b) \( 12\pi \) cm²
(c) \( 9\pi \) cm²
(d) \( 14\pi \) cm²
Answer: (c)
Explanation: Size of square = 6 cm, radius = \( \frac{6}{2} = 3 \) cm;
Area of the circle = \( \pi r^2 = \pi \times 3 \times 3 = 9\pi \) cm²

Assertion-Reason Questions

Question. Assertion: The length of the minute hand of a clock is 7 cm. Then the area swept by the minute hand in 5 minutes is \( 12\frac{5}{6} \) cm².
Reason: The length of an arc of a sector of angle \( \theta \) and radius \( r \) is given by \( l = \frac{\theta}{360^\circ} \times 2\pi r \).
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Answer: (b)
Explanation: Angle made by minute hand in 5 minutes = \( \frac{360^\circ}{60} \times 5 = 30^\circ \)
Area swept by minute hand in 5 minutes
\( = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{30^\circ}{360^\circ} \times \frac{22}{7} \times 7 \times 7 = \frac{77}{6} = 12\frac{5}{6} \) cm² ...[Here \( r = 7 \) cm]

Question. Assertion: A wire is looped in the form of a circle of radius 28 cm. It is bent into a square. Then the area of the square is 1936 cm².
Reason: Angle described by a minute hand in 60 minutes = 360°.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Answer: (d)
Explanation: We have, length of wire = \( 2\pi r \)
\( 2 \times \frac{22}{7} \times 28 = \) length of wire ...[Here, \( r = 28 \) cm]
length of wire = 176 cm
Now, perimeter of square = length of wire = 176
\( \Rightarrow 4a = 176 \Rightarrow a = 44 \)
\( \therefore \) Area of Square = \( (44)^2 = 1936 \) cm²

Question. Assertion: If the outer and inner diameter of a circular path is 10 m and 6 m then area of the path is \( 16\pi \) m².
Reason: If R and r be the radius of outer and inner circular path = \( \pi(R^2 - r^2) \).
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Answer: (a)
Explanation: Area of the path = \( \pi \left[ \left(\frac{10}{2}\right)^2 - \left(\frac{6}{2}\right)^2 \right] = \pi(25 - 9) = 16\pi \)

Question. Assertion: If the circumference of a circle is 176 cm, then its radius is 28 cm.
Reason: Circumference = \( 2\pi \times \) radius.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Answer: (a)
Explanation: Circumference = \( 2 \times \frac{22}{7} \times \) radius = 176
\( \therefore \) radius = \( \frac{176 \times 7}{2 \times 22} = 28 \) cm

Question. Assertion: If a wire of length 22 cm is bent is the shape of a circle, then area of the circle so formed is 40 cm².
Reason: Circumference of the circle = length of the wire.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Answer: (d)
Explanation: We have, \( 2\pi r = 22 \)
\( \Rightarrow \) radius, \( r = 22 \times \frac{7}{22} \times \frac{1}{2} = 3.5 \) cm
\( \therefore \) Area of the circle = \( \frac{22}{7} \times 3.5 \times 3.5 = 38.5 \) cm²