CBSE Class 10 · Chapter 11 · Mensuration · 2026-27

Areas Related to Circles Class 10: MCQ and case study questions

15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.

15 MCQs (1 mark each)
2 case studies (4 marks each)
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Multiple-choice questions

Tap an option to check it; the answer and a worked solution open. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.

Q1

·1 mark·Multiple choiceArea and arc of a sector

The area of a sector of a circle of radius $7$ cm with central angle $60^{\circ}$ is (Take $π = \frac{22}{7}$ .)

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Answer: (a)

\frac{77}{3} {cm}^{2}

Why: Area = (θ/360°) × πr² = (1/6) × (22/7) × 49.

$\frac{60}{360} \times \frac{22}{7} \times 7^{2} = \frac{1}{6} \times 154 = \frac{77}{3} {cm}^{2}$ .

Q2

·1 mark·Multiple choiceArea and arc of a sector

An arc of a circle of radius $28$ cm subtends $45^{\circ}$ at the centre. Its length is (Take $π = \frac{22}{7}$ .)

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Answer: (b)

22

Why: Arc = (θ/360°) × 2πr = (1/8) × 176.

$\frac{45}{360} \times 2 \times \frac{22}{7} \times 28 = \frac{1}{8} \times 176 = 22$ cm.

Q3

·1 mark·Multiple choiceArea and arc of a sector

The area of a sector is $\frac{3}{8}$ of the area of its circle. The central angle of the sector is

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Answer: (b)

135^{\circ}

Why: θ/360° = 3/8.

$\frac{θ}{360} \times π r^{2} = \frac{3}{8} π r^{2} \Rightarrow θ = \frac{3}{8} \times 360^{\circ} = 135^{\circ}$ .

Q4

·1 mark·Multiple choiceArea and arc of a sector

The minute hand of a clock is $10.5$ cm long. The area swept by it in $20$ minutes is (Take $π = \frac{22}{7}$ .)

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Answer: (c)

115.5 {cm}^{2}

Why: In 20 minutes the minute hand turns 120°, one-third of the circle.

Angle $= \frac{20}{60} \times 360^{\circ} = 120^{\circ}$ . Area $= \frac{1}{3} \times \frac{22}{7} \times {10.5}^{2} = \frac{1}{3} \times 346.5 = 115.5 {cm}^{2}$ .

Q5

·1 mark·Multiple choiceArea and arc of a sector

The arc of a sector of central angle $90^{\circ}$ is $22$ cm long. The radius of the circle is (Take $π = \frac{22}{7}$ .)

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Answer: (b)

14

Why: (90/360) × 2πr = πr/2 = 22.

$\frac{90}{360} \times 2 \times \frac{22}{7} \times r = \frac{11 r}{7} = 22 \Rightarrow r = 14$ cm.

Q6

·1 mark·Multiple choiceArea of a segment

A chord of a circle of radius $10$ cm subtends an angle of $60^{\circ}$ at the centre. The area of the minor segment is

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Answer: (a)

(\frac{50 π}{3} - 25 \sqrt{3}) {cm}^{2}

Why: Segment = sector − triangle; the triangle is equilateral with area (√3/4) × 100.

Sector $= \frac{60}{360} π (10)^{2} = \frac{50 π}{3}$ . The triangle is equilateral (two radii, angle $60^{\circ}$ ): area $= \frac{\sqrt{3}}{4} \times 100 = 25 \sqrt{3}$ . Segment $= \frac{50 π}{3} - 25 \sqrt{3} {cm}^{2}$ .

Q7

·1 mark·Multiple choiceArea and arc of a sector

Two sectors have the same central angle, but their radii are $3$ cm and $6$ cm. The ratio of their areas is

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Answer: (b)

1 : 4

Why: With the same angle, sector area is proportional to r².

$\frac{\frac{θ}{360} π (3)^{2}}{\frac{θ}{360} π (6)^{2}} = \frac{9}{36} = \frac{1}{4}$ , i.e. $1 : 4$ .

Q8

·1 mark·Multiple choiceArea and arc of a sector

A sector of a circle of radius $12$ cm has an arc of length $4 π$ cm. Its central angle is

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Answer: (c)

60^{\circ}

Why: (θ/360) × 24π = 4π gives θ/360 = 1/6.

$\frac{θ}{360} \times 2 π (12) = 4 π \Rightarrow \frac{θ}{360} = \frac{1}{6} \Rightarrow θ = 60^{\circ}$ .

Q9

·1 mark·Multiple choiceArea and arc of a sector

A sector cut from a circle of radius $21$ cm has a central angle of $60^{\circ}$ . Its perimeter (both radii and the arc) is (Take $π = \frac{22}{7}$ .)

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Answer: (d)

64

Why: Perimeter = two radii + arc = 42 + 22.

Arc $= \frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{6} \times 132 = 22$ cm. Perimeter $= 21 + 21 + 22 = 64$ cm. ( $43$ counts only one radius.)

Q10

·1 mark·Multiple choiceArea and arc of a sector

A sector of a circle of radius $6$ cm has an arc of length $5$ cm. The area of the sector is

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Answer: (a)

15 {cm}^{2}

Why: Sector area = ½ × arc length × radius.

Area $= \frac{θ}{360} π r^{2} = \frac{1}{2} r (\frac{θ}{360} \cdot 2 π r) = \frac{1}{2} \times 6 \times 5 = 15 {cm}^{2}$ .

Q11

·1 mark·Multiple choiceArea and arc of a sector

In a circle of radius $7$ cm, a minor sector has central angle $90^{\circ}$ . The area of the corresponding major sector is (Take $π = \frac{22}{7}$ .)

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Answer: (b)

115.5 {cm}^{2}

Why: The major sector has angle 360° − 90° = 270°.

$\frac{270}{360} \times \frac{22}{7} \times 49 = \frac{3}{4} \times 154 = 115.5 {cm}^{2}$ .

Q12

·1 mark·Multiple choiceArea and arc of a sector

The minute hand of a clock is $14$ cm long. The distance travelled by its tip in $15$ minutes is (Take $π = \frac{22}{7}$ .)

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Answer: (c)

22

Why: In 15 minutes the hand turns 90°: a quarter of the circumference.

$\frac{1}{4} \times 2 \times \frac{22}{7} \times 14 = \frac{1}{4} \times 88 = 22$ cm.

Q13

·1 mark·Multiple choiceArea and arc of a sector

The area of a circle is $616 {cm}^{2}$ . The area of a sector of this circle with central angle $45^{\circ}$ is

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Answer: (a)

77 {cm}^{2}

Why: A 45° sector is one-eighth of the circle.

$\frac{45}{360} \times 616 = \frac{616}{8} = 77 {cm}^{2}$ .

Q14

·1 mark·Multiple choiceArea of a segment

A chord of a circle of radius $6$ cm subtends an angle of $120^{\circ}$ at the centre. The area of the minor segment is

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Answer: (d)

(12 π - 9 \sqrt{3}) {cm}^{2}

Why: Sector 12π; the triangle has height 6 cos 60° = 3 on base 6√3, so its area is 9√3.

Sector $= \frac{120}{360} π (36) = 12 π$ . The triangle $O A B$ has height $6 \cos 60^{\circ} = 3$ on base $A B = 2 \times 6 \sin 60^{\circ} = 6 \sqrt{3}$ , so area $= \frac{1}{2} \times 6 \sqrt{3} \times 3 = 9 \sqrt{3}$ . Segment $= (12 π - 9 \sqrt{3}) {cm}^{2}$ .

Q15

·1 mark·Multiple choiceArea and arc of a sector

A sector of a circle of radius $r$ has the same area as a whole circle of radius $\frac{r}{2}$ . The central angle of the sector is

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Answer: (d)

90^{\circ}

Why: (θ/360)πr² = π(r/2)² = πr²/4 gives θ/360 = 1/4.

$\frac{θ}{360} π r^{2} = \frac{π r^{2}}{4} \Rightarrow θ = 90^{\circ}$ .

Case study questions

Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part. All Class 10 case study questions by chapter.

Case study 1: Lawn sprinkler (4 marks)

A gardener places a rotating sprinkler at the corner $O$ of a large lawn. The spray reaches $14$ m, and the sprinkler is set to turn through an angle of $90^{\circ}$ , so it waters a sector of radius $14$ m. (Take $π = \frac{22}{7}$ .)

(i) Find the area of lawn watered. [1 mark]

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Answer:

154 m^{2}

\frac{90}{360} \times \frac{22}{7} \times 14^{2} = \frac{1}{4} \times 616 = 154 m^{2}

. A1

(ii) Find the length of the curved edge of the watered region. [1 mark]

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Answer:

22

Arc

= \frac{1}{4} \times 2 \times \frac{22}{7} \times 14 = 22

m. A1

(iii) The sprinkler is reset to turn through

120^{\circ}

. How much more area does it now water? [2 marks]

Show answer

Answer:

\frac{154}{3} m^{2}

New area

= \frac{1}{3} \times 616 = \frac{616}{3} m^{2}

M1; increase

= \frac{616}{3} - 154 = \frac{154}{3} m^{2}

. A1

OR With the

90^{\circ}

setting, find the area of the watered region that lies beyond the straight line joining the two ends of the curved edge. [2 marks]

Show answer

Answer:

56 m^{2}

Segment

=

sector

-

right triangle

= 154 - \frac{1}{2} \times 14 \times 14

= 154 - 98 = 56 m^{2}

. A1

Case study 2: Pendulum in a science lab (4 marks)

In a physics practical, a pendulum with a thread $21$ cm long swings from one extreme position $A$ to the other extreme position $B$ , turning through an angle of $60^{\circ}$ about the fixed point $O$ . The bob traces the arc $A B$ . (Take $π = \frac{22}{7}$ .)

(i) Find the length of the arc

A B

. [1 mark]

Show answer

Answer:

22

\frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = \frac{1}{6} \times 132 = 22

cm. A1

(ii) Find the area of the sector

O A B

swept by the thread. [1 mark]

Show answer

Answer:

231 {cm}^{2}

\frac{1}{6} \times \frac{22}{7} \times 441 = 231 {cm}^{2}

. A1

(iii) Find the area of the region between the arc

A B

and the chord

A B

. Leave

\sqrt{3}

in your answer. [2 marks]

Show answer

Answer:

(231 - \frac{441 \sqrt{3}}{4}) {cm}^{2}

△ O A B

is equilateral with side

21

: area

= \frac{\sqrt{3}}{4} \times 441

M1. Segment

= 231 - \frac{441 \sqrt{3}}{4} {cm}^{2}

. A1

OR Find the perimeter of the region between the arc

A B

and the chord

A B

. [2 marks]

Show answer

Answer:

43

△ O A B

is equilateral, so chord

A B = 21

cm M1; perimeter

= 22 + 21 = 43

cm. A1

Next steps for Areas Related to Circles

This free set is separate from the chapter's question bank. On the Areas Related to Circles chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Pro adds all 37 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders, the full common-mistakes library and AI marking of your written working.

Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was checked twice, the second time by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.