Area of Segment PYQs
Class 10 Previous Year Questions (2014 – 2026)
Topic Overview
A segment is the region between a chord and its corresponding arc. Area of Minor Segment = Area of Sector - Area of Triangle. For $\theta = 60^{\circ}$, use equilateral triangle area $\frac{\sqrt{3}}{4} a^2$. For $\theta = 90^{\circ}$, use $\frac{1}{2} r^2$.
Q1
2026
00:00
Find the area of the minor segment of a circle of radius 7 cm and central angle 90°. (Use $\pi = 22/7$)
Area of Sector $= (90/360) \times (22/7) \times 7 \times 7 = 38.5$ cm².
Area of $\triangle OAB = 1/2 \times 7 \times 7 = 24.5$ cm².
Area of Segment $= 38.5 - 24.5 = 14$ cm².
Ans: 14 cm²
Q2
2025
00:00
A chord of length 14 cm is drawn in a circle of radius 14 cm. Find the area of the minor segment.
Radius $= 14$, Chord $= 14 \Rightarrow \triangle OAB$ is equilateral ($\theta = 60^{\circ}$).
Area of Sector $= (60/360) \times (22/7) \times 14 \times 14 = 102.67$ cm².
Area of $\triangle OAB = (\sqrt{3}/4) \times 14^2 = 49\sqrt{3} \approx 84.87$ cm².
Area of Segment $= 102.67 - 84.87 = 17.8$ cm².
Ans: 17.8 cm²
Q3
2024
00:00
A chord of a circle of radius 10 cm subtends a right angle at the center. Find the area of the corresponding minor segment. (Use $\pi = 3.14$)
Area of Sector $= (90/360) \times 3.14 \times 10 \times 10 = 78.5$ cm².
Area of $\triangle OAB = 1/2 \times 10 \times 10 = 50$ cm².
Area of Segment $= 78.5 - 50 = 28.5$ cm².
Ans: 28.5 cm²
Q4
2023
00:00
A chord of a circle of radius 15 cm subtends an angle of 60° at the center. Find the area of the corresponding minor segment. (Use $\pi = 3.14, \sqrt{3} = 1.73$)
Area of Sector $= (60/360) \times 3.14 \times 15 \times 15 = 117.75$ cm².
Area of Equilateral $\triangle OAB = (\sqrt{3}/4) \times 15^2 = (1.73/4) \times 225 = 97.31$ cm².
Area of Segment $= 117.75 - 97.31 = 20.44$ cm².
Ans: 20.44 cm²
Q5
2022
00:00
Find the area of the minor segment of a circle of radius 14 cm, when the angle of the corresponding sector is 60°.
Area of Sector $= (60/360) \times (22/7) \times 14 \times 14 = 102.67$ cm².
Area of $\triangle OAB = (\sqrt{3}/4) \times 14 \times 14 = 49\sqrt{3} \approx 84.87$ cm².
Area $= 102.67 - 84.87 = 17.8$ cm².
Ans: 17.8 cm²
Q6
2021
00:00
In a circle of radius 21 cm, an arc subtends an angle of 60° at the center. Find the area of the segment formed by the corresponding chord.
Area of Sector $= (60/360) \times (22/7) \times 21 \times 21 = 231$ cm².
Area of $\triangle OAB = (\sqrt{3}/4) \times 21^2 = 441\sqrt{3}/4$ cm².
Area $= 231 - 441\sqrt{3}/4$ cm².
Ans: $231 - 441\sqrt{3}/4$ cm²
Q7
2020
00:00
Find the area of the minor segment of a circle of radius 12 cm and central angle 120°. (Use $\pi = 3.14, \sqrt{3} = 1.73$)
Area of Sector $= (120/360) \times 3.14 \times 12 \times 12 = 150.72$ cm².
Area of $\triangle OAB = 1/2 \times r^2 \times \sin 120^{\circ} = 1/2 \times 144 \times \sqrt{3}/2 = 36\sqrt{3} = 62.28$ cm².
Area $= 150.72 - 62.28 = 88.44$ cm².
Ans: 88.44 cm²
Q8
2024
00:00
A round table cover has six equal designs as shown in figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹0.35 per cm². (Use $\sqrt{3} = 1.7$)
Six equal designs form six segments with $\theta = 360/6 = 60^{\circ}$.
Area of one segment $= \text{Sector} - \text{Equilateral } \triangle$.
Area $= (60/360) \times (22/7) \times 28^2 - (1.7/4) \times 28^2 = 410.67 - 333.2 = 77.47$ cm².
Total Area for 6 designs $= 6 \times 77.47 = 464.82$ cm².
Cost $= 464.82 \times 0.35 = ₹162.68$.
Ans: ₹162.68
Q9
2022
00:00
A chord of a circle of radius 15 cm subtends an angle of 60° at the center. Find the areas of the corresponding minor and major segments. (Use $\pi = 3.14, \sqrt{3} = 1.73$)
Area of minor segment $= 20.44$ cm² (as in Q2).
Area of circle $= 3.14 \times 15 \times 15 = 706.5$ cm².
Area of major segment $= 706.5 - 20.44 = 686.06$ cm².
Ans: Minor = 20.44 cm², Major = 686.06 cm²
Q10
2018
00:00
In a circle of radius 21 cm, an arc subtends an angle of 60° at the center. Find (i) length of arc (ii) area of sector (iii) area of segment.
(i) Arc length $= (60/360) \times 2 \times (22/7) \times 21 = 22$ cm.
(ii) Area of sector $= (60/360) \times (22/7) \times 21 \times 21 = 231$ cm².
(iii) Area of segment $= 231 - (441\sqrt{3}/4)$ cm².
Ans: (i) 22 cm, (ii) 231 cm², (iii) $231 - 441\sqrt{3}/4$ cm²
Find the area of the minor segment of a circle of radius 7 cm and central angle 90°. (Use $\pi = 22/7$)
Area of Sector $= (90/360) \times (22/7) \times 7 \times 7 = 38.5$ cm².
Area of $\triangle OAB = 1/2 \times 7 \times 7 = 24.5$ cm².
Area of Segment $= 38.5 - 24.5 = 14$ cm².
Ans: 14 cm²
Q2
2025
00:00
A chord of length 14 cm is drawn in a circle of radius 14 cm. Find the area of the minor segment.
Radius $= 14$, Chord $= 14 \Rightarrow \triangle OAB$ is equilateral ($\theta = 60^{\circ}$).
Area of Sector $= (60/360) \times (22/7) \times 14 \times 14 = 102.67$ cm².
Area of $\triangle OAB = (\sqrt{3}/4) \times 14^2 = 49\sqrt{3} \approx 84.87$ cm².
Area of Segment $= 102.67 - 84.87 = 17.8$ cm².
Ans: 17.8 cm²
Q3
2024
00:00
A chord of a circle of radius 10 cm subtends a right angle at the center. Find the area of the corresponding minor segment. (Use $\pi = 3.14$)
Area of Sector $= (90/360) \times 3.14 \times 10 \times 10 = 78.5$ cm².
Area of $\triangle OAB = 1/2 \times 10 \times 10 = 50$ cm².
Area of Segment $= 78.5 - 50 = 28.5$ cm².
Ans: 28.5 cm²
Q4
2023
00:00
A chord of a circle of radius 15 cm subtends an angle of 60° at the center. Find the area of the corresponding minor segment. (Use $\pi = 3.14, \sqrt{3} = 1.73$)
Area of Sector $= (60/360) \times 3.14 \times 15 \times 15 = 117.75$ cm².
Area of Equilateral $\triangle OAB = (\sqrt{3}/4) \times 15^2 = (1.73/4) \times 225 = 97.31$ cm².
Area of Segment $= 117.75 - 97.31 = 20.44$ cm².
Ans: 20.44 cm²
Q5
2022
00:00
Find the area of the minor segment of a circle of radius 14 cm, when the angle of the corresponding sector is 60°.
Area of Sector $= (60/360) \times (22/7) \times 14 \times 14 = 102.67$ cm².
Area of $\triangle OAB = (\sqrt{3}/4) \times 14 \times 14 = 49\sqrt{3} \approx 84.87$ cm².
Area $= 102.67 - 84.87 = 17.8$ cm².
Ans: 17.8 cm²
Q6
2021
00:00
In a circle of radius 21 cm, an arc subtends an angle of 60° at the center. Find the area of the segment formed by the corresponding chord.
Area of Sector $= (60/360) \times (22/7) \times 21 \times 21 = 231$ cm².
Area of $\triangle OAB = (\sqrt{3}/4) \times 21^2 = 441\sqrt{3}/4$ cm².
Area $= 231 - 441\sqrt{3}/4$ cm².
Ans: $231 - 441\sqrt{3}/4$ cm²
Q7
2020
00:00
Find the area of the minor segment of a circle of radius 12 cm and central angle 120°. (Use $\pi = 3.14, \sqrt{3} = 1.73$)
Area of Sector $= (120/360) \times 3.14 \times 12 \times 12 = 150.72$ cm².
Area of $\triangle OAB = 1/2 \times r^2 \times \sin 120^{\circ} = 1/2 \times 144 \times \sqrt{3}/2 = 36\sqrt{3} = 62.28$ cm².
Area $= 150.72 - 62.28 = 88.44$ cm².
Ans: 88.44 cm²
Q8
2024
00:00
A round table cover has six equal designs as shown in figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹0.35 per cm². (Use $\sqrt{3} = 1.7$)
Six equal designs form six segments with $\theta = 360/6 = 60^{\circ}$.
Area of one segment $= \text{Sector} - \text{Equilateral } \triangle$.
Area $= (60/360) \times (22/7) \times 28^2 - (1.7/4) \times 28^2 = 410.67 - 333.2 = 77.47$ cm².
Total Area for 6 designs $= 6 \times 77.47 = 464.82$ cm².
Cost $= 464.82 \times 0.35 = ₹162.68$.
Ans: ₹162.68
Q9
2022
00:00
A chord of a circle of radius 15 cm subtends an angle of 60° at the center. Find the areas of the corresponding minor and major segments. (Use $\pi = 3.14, \sqrt{3} = 1.73$)
Area of minor segment $= 20.44$ cm² (as in Q2).
Area of circle $= 3.14 \times 15 \times 15 = 706.5$ cm².
Area of major segment $= 706.5 - 20.44 = 686.06$ cm².
Ans: Minor = 20.44 cm², Major = 686.06 cm²
Q10
2018
00:00
In a circle of radius 21 cm, an arc subtends an angle of 60° at the center. Find (i) length of arc (ii) area of sector (iii) area of segment.
(i) Arc length $= (60/360) \times 2 \times (22/7) \times 21 = 22$ cm.
(ii) Area of sector $= (60/360) \times (22/7) \times 21 \times 21 = 231$ cm².
(iii) Area of segment $= 231 - (441\sqrt{3}/4)$ cm².
Ans: (i) 22 cm, (ii) 231 cm², (iii) $231 - 441\sqrt{3}/4$ cm²