Question 1
Find the area of a 60° sector of a circle of radius 6 cm.
Steps only
Area = 60/360 × π × 6² = 6π = 132/7
Answer
132/7 cm².
CLASS 10 · MATHEMATICS · CHAPTER 11
Areas Related to Circles
Find arc lengths and the areas of sectors, segments and circular designs. Short calculations come first, with the reason for each formula available in the explanation.
20 worked answersAll 1 exercise sets
What you will learn
- Find sector area from its central angle.
- Separate a segment into a sector and triangle.
- Apply circle-area formulas to practical designs.
20 worked answers · short steps first, explanations when you need them
Exercise 11.1
Textbook pages 158–159
Question 2
Find the area of a quadrant of a circle whose circumference is 22 cm.
Steps only
2πr = 22, so r = 7/2 cm
Quadrant area = 1/4 × 22/7 × (7/2)² = 77/8
Answer
77/8 cm².
Question 3
A minute hand is 14 cm long. Find the area it sweeps in 5 minutes.
Steps only
Angle in 5 minutes = 5/60 × 360° = 30°
Area = 30/360 × 22/7 × 14² = 154/3
Answer
154/3 cm².
Question 4(i)
A chord of a circle of radius 10 cm subtends 90° at the centre. Find the minor-segment area using π = 3.14.
Steps only
Sector area = 90/360 × 3.14 × 10² = 78.5
Triangle area = 1/2 × 10 × 10 = 50
Segment area = 78.5 − 50 = 28.5
Answer
28.5 cm².
Question 4(ii)
For the same circle, find the area of the major sector using π = 3.14.
Steps only
Major angle = 360° − 90° = 270°
Area = 270/360 × 3.14 × 10² = 235.5
Answer
235.5 cm².
Question 5(i)
In a circle of radius 21 cm, a 60° arc is formed. Find its length.
Steps only
Arc length = 60/360 × 2 × 22/7 × 21 = 22
Answer
22 cm.
Question 5(ii)
For the same 21 cm circle and 60° arc, find the sector area.
Steps only
Area = 60/360 × 22/7 × 21² = 231
Answer
231 cm².
Question 5(iii)
For the same circle, find the area of the segment formed by the 60° arc and its chord.
Steps only
Triangle area = √3/4 × 21² = 441√3/4
Segment area = 231 − 441√3/4
Answer
231 − 441√3/4 cm² (about 40.04 cm²).
Question 6(i)
A chord of a 15 cm radius circle subtends 60°. Find the minor-segment area using π = 3.14 and √3 = 1.73.
Steps only
Sector = 60/360 × 3.14 × 15² = 117.75
Triangle = 1.73/4 × 15² = 97.3125
Minor segment = 20.4375
Answer
20.44 cm² approximately.
Question 6(ii)
For the same circle, find the major-segment area.
Steps only
Circle area = 3.14 × 15² = 706.5
Major segment = 706.5 − 20.4375 = 686.0625
Answer
686.06 cm² approximately.
Question 7
A chord of a circle of radius 12 cm subtends 120° at the centre. Find the corresponding minor-segment area using π = 3.14 and √3 = 1.73.
Steps only
Sector = 120/360 × 3.14 × 12² = 150.72
Triangle = 1/2 × 12² × sin 120° = 36√3 = 62.28
Segment = 150.72 − 62.28 = 88.44
Answer
88.44 cm².
Question 8(i)
A horse is tied at a corner of a 15 m square field with a 5 m rope. Find the grazing area using π = 3.14.
Steps only
Grazing area = 1/4 × 3.14 × 5² = 19.625
Answer
19.625 m².
Question 8(ii)
How much does the grazing area increase if the rope is 10 m instead of 5 m?
Steps only
Increase = 1/4 × 3.14 × (10² − 5²)
= 1/4 × 3.14 × 75 = 58.875
Answer
58.875 m².
Question 9(i)
A circular brooch has diameter 35 mm and five wire diameters. Find the total wire length.
Steps only
Circle wire = πd = 22/7 × 35 = 110 mm
Five diameters = 5 × 35 = 175 mm
Total = 285 mm
Answer
285 mm.
Question 9(ii)
The five diameters divide the brooch into ten equal sectors. Find the area of each sector.
Steps only
Radius = 17.5 mm
Each sector = 1/10 × 22/7 × 17.5² = 96.25
Answer
96.25 mm².
Question 10
An umbrella of radius 45 cm has eight equally spaced ribs. Find the area between two consecutive ribs.
Steps only
Angle = 360°/8 = 45°
Area = 1/8 × 22/7 × 45² = 22275/28
Answer
22275/28 cm², about 795.54 cm².
Question 11
Two non-overlapping wipers, each 25 cm long, sweep through 115°. Find the total cleaned area.
Steps only
Total = 2 × 115/360 × 22/7 × 25²
= 158125/126 ≈ 1254.96
Answer
Approximately 1254.96 cm².
Question 12
A lighthouse warns ships over an 80° sector to a distance of 16.5 km. Find the warned sea area using π = 3.14.
Steps only
Area = 80/360 × 3.14 × 16.5²
= 189.97
Answer
189.97 km².
Question 13
A round cover of radius 28 cm has six equal segment-shaped designs. Find their total cost at ₹0.35 per cm², using √3 = 1.7.
Steps only
Six sectors together = πr² = 22/7 × 28² = 2464 cm²
Six triangles = 6 × √3/4 × 28² = 1999.2 cm²
Design area = 464.8 cm²; cost = 464.8 × 0.35 = ₹162.68
Answer
₹162.68.
Question 14
Choose the formula for the area of a sector of angle p° in a circle of radius R.
Steps only
Sector area = p/360 × πR²
Answer
Option (C).