Mathematics strengthens logical reasoning and analytical thinking, and Chapter 11: Areas Related to Circles plays a crucial role in building strong geometrical foundations in Class 10 Maths. This chapter focuses on the areas of sectors, segments, and the relationship between circles and other geometric shapes, including finding areas and arc lengths.
These NCERT Solutions for Class 10 Maths Chapter 11 include important CBSE board questions asked between 2020 and 2025. All solutions are explained step by step in clear and simple language, helping students understand concepts thoroughly, avoid common mistakes, and score well in board examinations.
NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles
Q.
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 the circumferences of the two circles.
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Find the area of the shaded region in the following figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.
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To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)
Q.
A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
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An umbrella has 8 ribs which are equally spaced (see the following figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
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A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the following figure. Find:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.
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The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
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In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
- the length of the arc
- area of the sector formed by the arc
- area of the segment formed by the corresponding chord
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A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment (ii) major sector. (Use π = 3.14)
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Find the area of a quadrant of a circle whose circumference is 22 cm.
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Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
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Tick the correct answer in the following and justify your choice: If the perimeter and the area of a circle are numerically equal, then the radius of the circle is
(A) 2 units (B) π units (C) 4 units (D) 7 units
Q.
The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?
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The following figure depicts an archery target marked with its five scoring areas from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing Gold score is 21 cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.
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1. Q. 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 the circumferences of the two circles.
Solution:
Sum of circumferences = 2π·19 + 2π·9 = 2π(28)
So 2πR = 2π(28) ⇒ R = 28 cm
✅ Answer: 28 cm
2) Q. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
Solution:
πR² = π(8²) + π(6²) = π(64 + 36) = π(100)
⇒ R² = 100 ⇒ R = 10 cm
✅ Answer: 10 cm
3) (Archery target)
Figure:
Q. Gold diameter = 21 cm, other 4 bands each 10.5 cm wide. Find area of each scoring region.
Solution (π = 22/7):
Gold radius r₁ = 10.5 cm
Next radii: r₂=21, r₃=31.5, r₄=42, r₅=52.5
Area of each band = π(R² − r²)
Gold = π(10.5²) = 346.5 cm²
Red = π(21² − 10.5²) = 1039.5 cm²
Blue = π(31.5² − 21²) = 1732.5 cm²
Black = π(42² − 31.5²) = 2425.5 cm²
White = π(52.5² − 42²) = 3118.5 cm²
✅ Answers: Gold 346.5, Red 1039.5, Blue 1732.5, Black 2425.5, White 3118.5 (cm²)
4) Q. Wheel diameter = 80 cm. Revolutions in 10 min at 66 km/h?
Solution:
d = 80 cm = 0.8 m ⇒ circumference = πd = 0.8π m
Distance in 10 min = 66 × (10/60) = 11 km = 11000 m
Revolutions = 11000 / (0.8π) = 13750/π
Using π = 22/7 ⇒ 13750 × 7 / 22 = 4375
✅ Answer: 4375 revolutions
5) Q. If perimeter and area of a circle are numerically equal, radius is: (A)2 (B)π (C)4 (D)7
Solution:
2πr = πr² ⇒ r = 2
✅ Answer: (A) 2 units
6) Q. Area of sector (r=6 cm, angle=60°).
Solution:
Area = (60/360)πr² = (1/6)π(36) = 6π cm²
✅ Answer: 6π cm² (≈ 18.84 cm²)
7) Q. Area of a quadrant if circumference = 22 cm.
Solution (π=22/7):
2πr = 22 ⇒ r = 22/(2π) = 22/(2×22/7)=3.5 cm
Quadrant area = (1/4)πr² = (1/4)×(22/7)×(3.5)² = 9.625 cm²
✅ Answer: 9.625 cm²
8) (π=3.14)
Q. r=10 cm chord subtends 90° at centre. Find (i) minor segment (ii) major sector.
Solution:
Sector(90°) = (90/360)πr² = (1/4)×3.14×100 = 78.5
Triangle (with sides 10,10, angle 90°) = 1/2×10×10 = 50
(i) Minor segment = 78.5 − 50 = 28.5 cm²
(ii) Major sector (270°) = (270/360)×3.14×100 = 235.5 cm²
✅ Answers: 28.5 cm², 235.5 cm²
9) Q. r=21 cm, arc subtends 60° at centre. Find (i) arc length (ii) sector area (iii) segment area.
Solution (π=22/7, √3≈1.732):
(i) Arc length = (60/360)×2πr = (1/6)×2π×21 = 7π = 22 cm
(ii) Sector area = (60/360)×πr² = (1/6)×π×441 = 231 cm²
(iii) Triangle area = (1/2)r² sin60 = (1/2)×441×(√3/2) ≈ 220.5×0.866 ≈ 190.7
Segment = 231 − 190.7 ≈ 40.3 cm²
✅ Answers: 22 cm, 231 cm², ≈40.3 cm²
10) (π=3.14, √3=1.73)
Q. r=15 cm, chord subtends 60°. Find areas of corresponding minor & major segments.
Solution:
Sector(60°) = (60/360)πr² = (1/6)×3.14×225 = 117.75
Triangle area = (1/2)r² sin60 = (1/2)×225×(1.73/2) = 112.5×0.865 = 97.3125
Minor segment = 117.75 − 97.3125 = 20.44 cm² (approx)
Circle area = 3.14×225 = 706.5
Major segment = 706.5 − 20.44 = 686.06 cm² (approx)
✅ Answers: Minor ≈20.44 cm², Major ≈686.06 cm²
11) (π=3.14, √3=1.73)
Q. r=12 cm, chord subtends 120°. Find area of corresponding segment.
Solution:
Sector(120°) = (120/360)πr² = (1/3)×3.14×144 = 150.72
Triangle area = (1/2)r² sin120 = (1/2)×144×sin60 = 72×0.865 = 62.28
Segment = 150.72 − 62.28 = 88.44 cm²
✅ Answer: 88.44 cm²
12) (π=3.14)
Figure:
Q. Square field side 15 m, horse tied at corner, rope 5 m. (i) grazing area (ii) increase if rope 10 m.
Solution: (quarter circle)
(i) Area = (1/4)π(5²)= (1/4)×3.14×25 = 19.625 m²
(ii) New area = (1/4)π(10²)= (1/4)×3.14×100 = 78.5
Increase = 78.5 − 19.625 = 58.875 m²
✅ Answers: 19.625 m² and 58.875 m²
13) Figure:
Q. Brooch: circle diameter 35 mm + 5 diameters. Find (i) total wire length (ii) area of each of 10 sectors.
Solution (π=22/7):
r = 17.5 mm, d=35 mm
(i) Wire length = circumference + 5 diameters
= πd + 5d = 35(π+5) = 35(22/7 + 5) = 35(57/7) = 285 mm
(ii) Circle area = πr² = (22/7)×(17.5²)= (22/7)×306.25 = 962.5 mm²
Each sector (10 equal) = 962.5/10 = 96.25 mm²
✅ Answers: 285 mm; 96.25 mm²
14) Figure:
Q. Umbrella radius 45 cm, 8 ribs equally spaced. Find area between two consecutive ribs.
Solution:
Angle per sector = 360/8 = 45°
Area = (45/360)πr² = (1/8)π(45²)= (1/8)π(2025)=253.125π
✅ Answer: 253.125π cm² ≈ 795.3 cm²
15) (π=3.14)
Q. Two wipers, each blade 25 cm, angle 115°, no overlap. Total area cleaned per sweep?
Solution:
Area per wiper = (115/360)π(25²) = (115/360)×3.14×625 ≈ 626.91
Total = 2×626.91 ≈ 1253.82 cm²
✅ Answer: 1253.82 cm² (approx)
16) (π=3.14)
Q. Lighthouse spreads light over 80° sector to distance 16.5 km. Area warned?
Solution:
Area = (80/360)π(16.5²) = (2/9)×3.14×272.25 ≈ 189.97 km²
✅ Answer: ≈189.97 km²
17) (√3=1.7, π=22/7)
Figure:
Q. Round table cover radius 28 cm. Six equal designs shown. Find cost at ₹0.35 per cm².
Solution: (designs = area(circle) − area(regular hexagon))
Area circle = πr² = (22/7)×28² = (22/7)×784 = 2464 cm²
Hexagon inscribed ⇒ side a = r = 28
Area hexagon = (3√3/2)a² = (3×1.7/2)×784 = 2.55×784 = 1999.2 cm²
Design area total = 2464 − 1999.2 = 464.8 cm²
Cost = 464.8 × 0.35 = ₹162.68
✅ Answer: ₹162.68 (approx)
18) Q. Tick correct: Area of sector of angle p° and radius R is:
(A) (p/180)×2πR (B) (p/180)×πR² (C) (p/360)×2πR (D) (p/720)×2πR²
Solution:
Correct formula = (p/360)πR²
Option D: (p/720)×2πR² = (p/360)πR² ✅
✅ Answer: (D)
19) Figure:
Q. Find area of shaded region if PQ = 24 cm, PR = 7 cm and O is centre.
Solution (as per figure: RQ is diameter ⇒ ∠RPQ = 90°):
In right ΔPQR: RQ = √(PQ² + PR²) = √(24² + 7²) = √(576+49)=√625=25
So radius r = 25/2 = 12.5 cm
Chord PQ = 24
Chord formula: PQ = 2r sin(θ/2) ⇒ 24 = 25 sin(θ/2) ⇒ sin(θ/2)=0.96
⇒ θ ≈ 147.48° (angle subtended at centre by chord PQ)
Shaded region = minor segment for chord PQ
Segment area = sector − triangle
Sector = (θ/360)πr² = (147.48/360)×3.14×(12.5²) ≈ 200.99
Triangle OQP = (1/2)r² sinθ = 0.5×156.25×sin147.48° ≈ 42.00
Segment ≈ 200.99 − 42.00 = 158.99 cm²
✅ Answer: ≈ 159.0 cm²
20) Figure:
Q. Two concentric circles radii 7 cm and 14 cm, ∠AOC=40°. Find area of shaded region.
Solution: (annular sector)
Area = (40/360)π(14² − 7²)
= (1/9)π(196 − 49) = (1/9)π(147) = (49/3)π
Using π = 22/7 ⇒ (49/3)×(22/7)= (7×22)/3 = 154/3 = 51.33 cm²
✅ Answer: ≈ 51.33 cm²