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

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.