Chapter 11 of 16

Updated 6 August 2026

NCERT Solutions

Areas Related to Circles — NCERT Solutions

CBSE · Class 10 · Mathematics

NCERT Solutions for Areas Related to Circles, CBSE Class 10 Mathematics: 14 textbook questions solved step by step. Covers Exercise 11.1.

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A labeled diagram illustrating the definitions of minor sector, major sector, minor segment, and major segment of a circle, with the center, radius, chord, and arc clearly marked. — Learn better with visuals Super Tutor pairs illustrations like this with notes and quizzes for Areas Related to Circles.

14 Questions Solved · 1 Section

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Exercise 11.1

1Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.Show solution

Given: Radius $r = 6$ cm, angle of sector $\theta = 60°$

Formula: Area of sector $= \dfrac{\theta}{360} \times \pi r^2$

Solution:
$\text{Area of sector} = \frac{60}{360} \times \frac{22}{7} \times 6 \times 6$
$= \frac{1}{6} \times \frac{22}{7} \times 36$
$= \frac{22 \times 36}{6 \times 7} = \frac{792}{42} = \frac{132}{7}$
$= 18\frac{6}{7} \text{ cm}^2$

Answer: Area of the sector $= \dfrac{132}{7}$ cm² $= 18\dfrac{6}{7}$ cm²

2Find the area of a quadrant of a circle whose circumference is 22 cm.Show solution

Given: Circumference $= 22$ cm

Step 1: Find the radius.
$2\pi r = 22 \implies r = \frac{22}{2\pi} = \frac{22 \times 7}{2 \times 22} = \frac{7}{2} \text{ cm}$

Step 2: Find the area of the quadrant.
A quadrant corresponds to $\theta = 90°$ .
$\text{Area of quadrant} = \frac{90}{360} \times \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$
$= \frac{1}{4} \times \frac{22}{7} \times \frac{49}{4} = \frac{1}{4} \times \frac{22 \times 49}{28} = \frac{1}{4} \times \frac{1078}{28} = \frac{1078}{112} = \frac{77}{8}$

Answer: Area of the quadrant $= \dfrac{77}{8}$ cm² $= 9\dfrac{5}{8}$ cm²

3The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.Show solution

Given: Length of minute hand (radius) $r = 14$ cm

Step 1: Find the angle swept in 5 minutes.
The minute hand completes $360°$ in 60 minutes.
$\text{Angle in 5 minutes} = \frac{360°}{60} \times 5 = 30°$

Step 2: Find the area swept.
$\text{Area} = \frac{\theta}{360} \times \pi r^2 = \frac{30}{360} \times \frac{22}{7} \times 14 \times 14$
$= \frac{1}{12} \times \frac{22}{7} \times 196$
$= \frac{1}{12} \times \frac{22 \times 196}{7} = \frac{1}{12} \times 22 \times 28 = \frac{616}{12} = \frac{154}{3}$

Answer: Area swept by the minute hand in 5 minutes $= \dfrac{154}{3}$ cm² $= 51\dfrac{1}{3}$ cm²

4A 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)Show solution

Given: Radius $r = 10$ cm, $\theta = 90°$ , $\pi = 3.14$

(i) Area of minor segment:

Area of minor sector (with $\theta = 90°$ ):
$= \frac{90}{360} \times \pi r^2 = \frac{1}{4} \times 3.14 \times 10 \times 10 = \frac{314}{4} = 78.5 \text{ cm}^2$

Area of triangle OAB (right-angled at O, with OA = OB = 10 cm):
$= \frac{1}{2} \times OA \times OB = \frac{1}{2} \times 10 \times 10 = 50 \text{ cm}^2$

Area of minor segment:
$= \text{Area of minor sector} - \text{Area of } \triangle OAB$
$= 78.5 - 50 = 28.5 \text{ cm}^2$

(ii) Area of major sector:
Angle of major sector $= 360° - 90° = 270°$
$= \frac{270}{360} \times \pi r^2 = \frac{3}{4} \times 3.14 \times 100 = \frac{942}{4} = 235.5 \text{ cm}^2$

Answers:

Area of minor segment $= 28.5$ cm²
Area of major sector $= 235.5$ cm²

5In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (i) the length of the arc (ii) area of the sector formed by the arc (iii) area of the segment formed by the corresponding chordShow solution

Given: Radius $r = 21$ cm, $\theta = 60°$

(i) Length of the arc:
$l = \frac{\theta}{360} \times 2\pi r = \frac{60}{360} \times 2 \times \frac{22}{7} \times 21$
$= \frac{1}{6} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{6} \times 132 = 22 \text{ cm}$

(ii) Area of the sector:
$= \frac{\theta}{360} \times \pi r^2 = \frac{60}{360} \times \frac{22}{7} \times 21 \times 21$
$= \frac{1}{6} \times \frac{22}{7} \times 441 = \frac{1}{6} \times 1386 = 231 \text{ cm}^2$

(iii) Area of the segment:

Since $\theta = 60°$ and OA = OB = 21 cm, triangle OAB is equilateral (all sides = 21 cm).

$\text{Area of } \triangle OAB = \frac{\sqrt{3}}{4} \times (21)^2 = \frac{\sqrt{3}}{4} \times 441 = \frac{441\sqrt{3}}{4} \text{ cm}^2$

$\text{Area of segment} = 231 - \frac{441\sqrt{3}}{4} = \left(231 - \frac{441\sqrt{3}}{4}\right) \text{ cm}^2$

Answer: Area of segment $= \left(231 - \dfrac{441\sqrt{3}}{4}\right)$ cm²

6A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and √3 = 1.73)Show solution

Given: Radius $r = 15$ cm, $\theta = 60°$ , $\pi = 3.14$ , $\sqrt{3} = 1.73$

Area of minor sector:
$= \frac{60}{360} \times 3.14 \times 15^2 = \frac{1}{6} \times 3.14 \times 225 = \frac{706.5}{6} = 117.75 \text{ cm}^2$

Area of triangle OAB:
Since $\theta = 60°$ and OA = OB = 15 cm, $\triangle OAB$ is equilateral.
$= \frac{\sqrt{3}}{4} \times 15^2 = \frac{1.73}{4} \times 225 = \frac{389.25}{4} = 97.3125 \approx 97.31 \text{ cm}^2$

Area of minor segment:
$= 117.75 - 97.31 = 20.44 \text{ cm}^2$

Area of circle:
$= \pi r^2 = 3.14 \times 225 = 706.5 \text{ cm}^2$

Area of major segment:
$= 706.5 - 20.44 = 686.06 \text{ cm}^2$

Answers:

Area of minor segment $\approx 20.44$ cm²
Area of major segment $\approx 686.06$ cm²

7A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π = 3.14 and √3 = 1.73)Show solution

Given: Radius $r = 12$ cm, $\theta = 120°$ , $\pi = 3.14$ , $\sqrt{3} = 1.73$

Area of sector:
$= \frac{120}{360} \times 3.14 \times 12^2 = \frac{1}{3} \times 3.14 \times 144 = \frac{452.16}{3} = 150.72 \text{ cm}^2$

Area of triangle OAB:
Draw OM $\perp$ AB. Since $\theta = 120°$ , $\angle AOM = 60°$ .
$OM = r\cos 60° = 12 \times \frac{1}{2} = 6 \text{ cm}$
$AM = r\sin 60° = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3} \text{ cm}$
$AB = 2 \times AM = 12\sqrt{3} \text{ cm}$
$\text{Area of } \triangle OAB = \frac{1}{2} \times AB \times OM = \frac{1}{2} \times 12\sqrt{3} \times 6 = 36\sqrt{3}$
$= 36 \times 1.73 = 62.28 \text{ cm}^2$

Area of segment:
$= 150.72 - 62.28 = 88.44 \text{ cm}^2$

Answer: Area of the corresponding segment $= 88.44$ cm²

8A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope. Find (i) the area of that part of the field in which the horse can graze. (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π = 3.14)

Given: Side of square field $= 15$ m, length of rope $= 5$ m (radius), $\pi = 3.14$

The horse is tied at a corner of the square. The angle at each corner of a square is $90°$ .

(i) Grazing area with rope = 5 m:
The horse can graze in a sector of radius 5 m and angle 90°.
$\text{Area} = \frac{90}{360} \times \pi r^2 = \frac{1}{4} \times 3.14 \times 5^2 = \frac{1}{4} \times 3.14 \times 25 = \frac{78.5}{4} = 19.625 \text{ m}^2$

(ii) Grazing area with rope = 10 m:
$\text{Area} = \frac{1}{4} \times 3.14 \times 10^2 = \frac{1}{4} \times 3.14 \times 100 = \frac{314}{4} = 78.5 \text{ m}^2$

Increase in grazing area:
$= 78.5 - 19.625 = 58.875 \text{ m}^2$

Answers:

(i) Area of grazing with 5 m rope $= 19.625$ m²
(ii) Increase in grazing area $= 58.875$ m²

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9A 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. Find: (i) the total length of the silver wire required. (ii) the area of each sector of the brooch.

Given: Diameter $= 35$ mm, so radius $r = 17.5$ mm; 5 diameters divide the circle into 10 equal sectors.

(i) Total length of silver wire:

Circumference of circle $= 2\pi r = 2 \times \dfrac{22}{7} \times 17.5 = 2 \times \dfrac{22}{7} \times \dfrac{35}{2} = 2 \times 55 = 110$ mm
Length of 5 diameters $= 5 \times 35 = 175$ mm
Total length $= 110 + 175 = 285$ mm

(ii) Area of each sector:
The circle is divided into 10 equal sectors, so each sector has angle $= \dfrac{360°}{10} = 36°$ .
$\text{Area of each sector} = \frac{36}{360} \times \pi r^2 = \frac{1}{10} \times \frac{22}{7} \times (17.5)^2$
$= \frac{1}{10} \times \frac{22}{7} \times 306.25 = \frac{1}{10} \times \frac{6737.5}{7} = \frac{6737.5}{70} = \frac{962.5}{10} = 96.25 \text{ mm}^2$

Answers:

(i) Total length of silver wire $= 285$ mm
(ii) Area of each sector $= 96.25$ mm²

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10An umbrella has 8 ribs which are equally spaced. Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Given: Radius $r = 45$ cm, number of ribs $= 8$

The 8 equally spaced ribs divide the circle into 8 equal sectors.
$\text{Angle of each sector} = \frac{360°}{8} = 45°$

Area between two consecutive ribs (one sector):
$= \frac{45}{360} \times \pi r^2 = \frac{1}{8} \times \frac{22}{7} \times 45 \times 45$
$= \frac{1}{8} \times \frac{22}{7} \times 2025 = \frac{22 \times 2025}{56} = \frac{44550}{56} = \frac{22275}{28} = 795.535... \approx 795.54 \text{ cm}^2$

Answer: Area between two consecutive ribs $= \dfrac{22275}{28}$ cm² $\approx 795.54$ cm²

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11A 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.

Given: Length of blade (radius) $r = 25$ cm, angle $\theta = 115°$ , number of wipers $= 2$

Area cleaned by one wiper in one sweep:
$= \frac{\theta}{360} \times \pi r^2 = \frac{115}{360} \times \frac{22}{7} \times 25 \times 25$
$= \frac{115}{360} \times \frac{22}{7} \times 625$
$= \frac{115 \times 22 \times 625}{360 \times 7} = \frac{1581250}{2520} = \frac{158125}{252} \approx 627.48 \text{ cm}^2$

Total area cleaned by two wipers:
$= 2 \times \frac{158125}{252} = \frac{158125}{126} = \frac{158125}{126} \approx 1254.96 \text{ cm}^2$

More precisely:
$= \frac{115}{360} \times \frac{22}{7} \times 625 \times 2 = \frac{23 \times 22 \times 625}{72 \times 7} = \frac{316250}{504} = \frac{158125}{252}$

Answer: Total area cleaned $= \dfrac{158125}{126}$ cm² $\approx 1254.96$ cm²

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12To 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)

Given: Radius $r = 16.5$ km, angle $\theta = 80°$ , $\pi = 3.14$

Area of the sector:
$= \frac{\theta}{360} \times \pi r^2 = \frac{80}{360} \times 3.14 \times (16.5)^2$
$= \frac{2}{9} \times 3.14 \times 272.25$
$= \frac{2}{9} \times 854.865 = \frac{1709.73}{9} = 189.97 \text{ km}^2$

Answer: Area of the sea over which ships are warned $\approx 189.97$ km²

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13A round table cover has six equal designs as shown in Fig. 11.11. 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 √3 = 1.7)

Given: Radius $r = 28$ cm, 6 equal designs, rate $= ₹0.35$ per cm², $\sqrt{3} = 1.7$

The 6 equal designs are 6 segments formed by 6 equal chords (each chord subtends $60°$ at the centre, since $360°/6 = 60°$ ).

Area of one segment:

Area of one sector (angle $= 60°$ ):
$= \frac{60}{360} \times \frac{22}{7} \times 28^2 = \frac{1}{6} \times \frac{22}{7} \times 784 = \frac{1}{6} \times 2464 = \frac{2464}{6} = \frac{1232}{3} \text{ cm}^2$

Area of equilateral triangle (since $\theta = 60°$ , OA = OB = 28 cm, triangle is equilateral with side 28 cm):
$= \frac{\sqrt{3}}{4} \times 28^2 = \frac{1.7}{4} \times 784 = \frac{1332.8}{4} = 333.2 \text{ cm}^2$

Area of one segment:
$= \frac{1232}{3} - 333.2 = 410.67 - 333.2 = 77.47 \text{ cm}^2$

Total area of 6 designs:
$= 6 \times 77.47 = 464.82 \text{ cm}^2$

Cost of making the designs:
$= 464.82 \times 0.35 = ₹162.69$

Answer: Cost of making the designs $= ₹162.68$ (approximately)

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14Tick the correct answer in the following: Area of a sector of angle p (in degrees) of a circle with radius R is
(A) p/180 × 2πR
(B) p/180 × πR²
(C) p/360 × 2πR
(D) p/720 × 2πR²

Correct Answer: (D) $\dfrac{p}{720} \times 2\pi R^2$

Justification:
The standard formula for area of a sector is:
$\text{Area} = \frac{p}{360} \times \pi R^2$

This can be rewritten as:
$= \frac{p}{360} \times \pi R^2 = \frac{2p}{720} \times \pi R^2 = \frac{p}{720} \times 2\pi R^2$

Option (D) $\dfrac{p}{720} \times 2\pi R^2 = \dfrac{p}{360} \times \pi R^2$ , which matches the standard formula.

Options (A) and (C) give arc length (not area), and option (B) has the wrong denominator (180 instead of 360).

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