NCERT Solutions · Class 10 · Chapter 11

NCERT Solutions for Class 10 Maths Chapter 11: Areas Related to Circles

Length of an arc \(\dfrac{\theta}{360} \times 2\pi r\), area of a sector \(\dfrac{\theta}{360} \times \pi r^2\), and area of a segment = sector − triangle. Our own step-by-step solutions to every question in Exercise 11.1, set out for step marks.

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Exercise 11.1: Sectors and segments

What it tests. Sector of angle \(\theta\): area \(\dfrac{\theta}{360}\pi r^2\), arc \(\dfrac{\theta}{360}2\pi r\). Minor segment = sector − triangle \(\left(\tfrac12 r^2\sin\theta\right)\); major segment (or sector) = circle − minor one. Use the value of \(\pi\) the question gives; otherwise take \(\pi = \tfrac{22}{7}\), as the NCERT exercises do (no calculator: CBSE is a no-calculator exam).

Exercise 11.1, Question 1

Find the area of a sector of a circle of radius 6 cm with angle \(60^\circ\).

Show solution

\[\dfrac{60}{360} \times \dfrac{22}{7} \times 36 = \dfrac{132}{7}\]

Answer: \(\dfrac{132}{7}\ \text{cm}^2\) (about 18.86 cm²)

Practise this: Step 1, Secure the basics →

Exercise 11.1, Question 2

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

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\(2\pi r = 22 \Rightarrow r = \tfrac72\).
Quadrant \[= \tfrac14 \times \dfrac{22}{7} \times \dfrac{49}{4} = \dfrac{77}{8}\]

Answer: \(\dfrac{77}{8}\ \text{cm}^2\)

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Exercise 11.1, Question 3

The minute hand of a clock is 14 cm long. Find the area it sweeps in 5 minutes.

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5 minutes \[= \dfrac{5}{60} \times 360^\circ = 30^\circ\]
\[\dfrac{30}{360} \times \dfrac{22}{7} \times 196 = \dfrac{154}{3}\]

Answer: \(\dfrac{154}{3}\ \text{cm}^2\) (about 51.33 cm²)

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Exercise 11.1, Question 4

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of (use \(\pi = 3.14\)):

(i) the minor segment

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Quarter circle: \(\tfrac14 \times 3.14 \times 100 = 78.5\).
Triangle: \(\tfrac12 \times 10 \times 10 = 50\). Segment \(= 28.5\).

Answer: \(28.5\ \text{cm}^2\)

(ii) the major sector

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\(\tfrac34 \times 3.14 \times 100\).

Answer: \(235.5\ \text{cm}^2\)

Practise this: Step 2, Board standard →

Exercise 11.1, Question 5

In a circle of radius 21 cm, an arc subtends \(60^\circ\) at the centre. Find:

(i) the length of the arc

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\[\dfrac{60}{360} \times 2 \times \dfrac{22}{7} \times 21\]

Answer: 22 cm

(ii) the area of the sector formed by the arc

Show solution

\[\dfrac{60}{360} \times \dfrac{22}{7} \times 441\]

Answer: \(231\ \text{cm}^2\)

(iii) the area of the segment formed by the chord

Show solution

The triangle is equilateral (two radii, \(60^\circ\) between): area \(\dfrac{\sqrt3}{4} \times 441\).
Segment \(= 231 - \dfrac{441\sqrt3}{4}\).

Answer: \[\left(231 - \dfrac{441\sqrt3}{4}\right)\ \text{cm}^2\]

Practise this: Step 2, Board standard →

Exercise 11.1, Question 6

A chord of a circle of radius 15 cm subtends \(60^\circ\) at the centre. Find the areas of the minor and major segments (\(\pi = 3.14\), \(\sqrt3 = 1.73\)).

Show solution

Sector: \(\tfrac16 \times 3.14 \times 225 = 117.75\).
Triangle (equilateral): \(\tfrac{1.73}{4} \times 225 = 97.3125\). Minor segment \(= 20.4375\).
Circle \(= 706.5\); major segment \(= 706.5 - 20.4375 = 686.0625\).

Answer: Minor \(20.4375\ \text{cm}^2\), major \(686.0625\ \text{cm}^2\)

Practise this: Step 2, Board standard →

Exercise 11.1, Question 7

A chord of a circle of radius 12 cm subtends \(120^\circ\) at the centre. Find the area of the segment (\(\pi = 3.14\), \(\sqrt3 = 1.73\)).

Show solution

Sector: \(\tfrac13 \times 3.14 \times 144 = 150.72\).
Triangle: \[\begin{aligned}\tfrac12 \times 12^2 \times \sin 120^\circ &= 36\sqrt3 \\ &= 62.28\end{aligned}\]
Segment \(= 150.72 - 62.28\).

Answer: \(88.44\ \text{cm}^2\)

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Exercise 11.1, Question 8

A horse is tied to a peg at one corner of a square field of side 15 m by a 5 m rope (\(\pi = 3.14\)).

(i) Find the area it can graze.

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It grazes a quadrant of radius 5: \(\tfrac14 \times 3.14 \times 25\).

Answer: \(19.625\ \text{m}^2\)

(ii) Find the increase in grazing area if the rope were 10 m.

Show solution

\(\tfrac14 \times 3.14 \times 100 = 78.5\).
Increase \(= 78.5 - 19.625\).

Answer: \(58.875\ \text{m}^2\)

Practise this: Step 3, Full marks on long answers →

Exercise 11.1, Question 9

A brooch is made of silver wire: a circle of diameter 35 mm and 5 diameters dividing it into 10 equal sectors.

(i) Find the total length of wire.

Show solution

Circumference \(= \dfrac{22}{7} \times 35 = 110\); diameters \(= 5 \times 35 = 175\).

Answer: 285 mm

(ii) Find the area of each sector.

Show solution

\[\tfrac{1}{10} \times \dfrac{22}{7} \times \left(\tfrac{35}{2}\right)^2 = \dfrac{385}{4}\]

Answer: \(\dfrac{385}{4}\ \text{mm}^2\) (96.25 mm²)

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Exercise 11.1, Question 10

An umbrella has 8 equally spaced ribs; treat it as a flat circle of radius 45 cm. Find the area between two consecutive ribs.

Show solution

Each sector is \(\tfrac18\) of the circle: \[\tfrac18 \times \dfrac{22}{7} \times 2025 = \dfrac{22275}{28}\]

Answer: \(\dfrac{22275}{28}\ \text{cm}^2\) (about 795.54 cm²)

Practise this: Step 2, Board standard →

Exercise 11.1, Question 11

A car has two non-overlapping wipers, each with a 25 cm blade sweeping \(115^\circ\). Find the total area cleaned in each sweep.

Show solution

Each: \[\dfrac{115}{360} \times \dfrac{22}{7} \times 625\]
Two: \[2 \times \dfrac{158125}{252} = \dfrac{158125}{126}\]

Answer: \(\dfrac{158125}{126}\ \text{cm}^2\) (about 1254.96 cm²)

Practise this: Step 2, Board standard →

Exercise 11.1, Question 12

A lighthouse spreads red light over a sector of \(80^\circ\) to a distance of 16.5 km. Find the area of sea warned (\(\pi = 3.14\)).

Show solution

\[\dfrac{80}{360} \times 3.14 \times 16.5^2 = \tfrac29 \times 3.14 \times 272.25\]

Answer: \(189.97\ \text{km}^2\) (to 2 d.p.)

Practise this: Step 2, Board standard →

Exercise 11.1, Question 13

A round table cover has six equal designs (the segments cut off by a regular hexagon inscribed in a circle of radius 28 cm). Find the cost of the designs at ₹0.35 per cm² (\(\sqrt3 = 1.7\)).

Show solution

Each segment: sector \[\dfrac{60}{360} \times \dfrac{22}{7} \times 784 = \dfrac{1232}{3}\] minus equilateral triangle \(\dfrac{1.7}{4} \times 784 = 333.2\).
Six segments: \[\begin{aligned}6 \times \left(\dfrac{1232}{3} - 333.2\right) &= 2464 - 1999.2 \\ &= 464.8\ \text{cm}^2\end{aligned}\]
Cost \(= 464.8 \times 0.35\).

Answer: ₹162.68

Practise this: Step 4, 95+ stretch (HOTS) →

Exercise 11.1, Question 14

The area of a sector of angle \(p\) (degrees) of a circle of radius R is (A) \(\dfrac{p}{180} \times 2\pi R\) (B) \(\dfrac{p}{180} \times \pi R^2\) (C) \(\dfrac{p}{360} \times 2\pi R\) (D) \(\dfrac{p}{720} \times 2\pi R^2\)

Show solution

Area \[= \dfrac{p}{360}\pi R^2 = \dfrac{p}{720} \times 2\pi R^2\]

Answer: (D)

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Done the NCERT exercises? The board paper asks more

Areas Related to Circles has 37 original board-style questions (MCQ, assertion–reason, short and long answers, case studies) with step mark schemes, revision notes and a four-step Route to 95. Three sample questions are open to everyone; a free account opens the rest.

Also useful: free MCQs and case studies for Areas Related to Circles · Class 10 formula sheet · official CBSE board and sample papers · our sample papers with marking scheme · the Route to 95 plan

Textbook: NCERT Mathematics Class 10 (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in. Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log.