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Areas Related to Circles: 3 marks Questions (CBSE Class 10)

30 different 3 marks questions on Areas Related to Circles from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (63)2 marks (12)3 marks (30)4 marks (13)5 marks (9)

Q273 marksShort AnswerAreas Related to Circles CBSE 2026 · Basic 430/4/1

A chord of a circle of radius 14 cm subtends a right angle at the centre. Find the area of the corresponding (i) minor segment (ii) major segment.

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Answer: (i) 56

c m^{2}

(ii) 560

c m^{2}

Area of sector = $\frac{90}{360} \times \frac{22}{7} \times 14 \times 14 = 154$ $c m^{2}$ .
Area of right triangle formed = $\frac{1}{2} \times 14 \times 14 = 98$ $c m^{2}$ .
(i) Minor segment = 154 – 98 = 56 $c m^{2}$ .
Area of circle = $\frac{22}{7} \times 14 \times 14 = 616$ $c m^{2}$ .
(ii) Major segment = 616 – 56 = 560 $c m^{2}$ .

Also asked in: 2026 Basic 430/4/2

Q283 marksShort AnswerAreas Related to Circles CBSE 2026 · Basic 430/4/3

A circle of radius 12 cm is divided into 11 identical sectors. Find the total area of 7 such sectors.

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Answer: 288

c m^{2}

Area of circle = $\frac{22}{7} \times 12 \times 12 = \frac{3168}{7}$ $c m^{2}$ .
Area of one sector = $\frac{1}{11} \times \frac{3168}{7} = \frac{288}{7}$ $c m^{2}$ .
Area of 7 sectors = $7 \times \frac{288}{7} = 288$ $c m^{2}$ .

Q313 marksShort AnswerAreas Related to Circles CBSE 2026 · Basic 430/5/1

Chord AB of a circle subtends an angle of $12 0^{\circ}$ at the centre O of the circle. Find the length of arc AB, if radius of the circle is 21 cm.

Diagram for CBSE 2026 Class 10 Maths question 31

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Answer: 44 cm

Arc length $= \frac{θ}{36 0 ^{\circ}} \times 2 π r = \frac{12 0 ^{\circ}}{36 0 ^{\circ}} \times 2 \times \frac{22}{7} \times 21$
$= \frac{1}{3} \times 132 = 44$ cm

Also asked in: 2026 Basic 430/5/2, 2026 Basic 430/5/3

Q303 marksShort AnswerAreas Related to Circles CBSE 2026 · Standard 30/2/1

Find the area of the sector of a circle of radius 42 cm and of central angle $3 0^{\circ}$ . Also, find the area of the corresponding major sector. [Use $π = \frac{22}{7}$ ]

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Answer: Minor sector = 462

c m^{2}

; major sector = 5082

c m^{2}

Area of sector $= \frac{30}{360} \times \frac{22}{7} \times 42 \times 42 = \frac{1}{12} \times 5544 = 462$ $c m^{2}$
Area of circle $= \frac{22}{7} \times 42 \times 42 = 5544$ $c m^{2}$
Area of major sector $= 5544 - 462 = 5082$ $c m^{2}$

Also asked in: 2026 Standard 30/2/3

Q273 marksShort AnswerAreas Related to Circles CBSE 2026 · Standard 30/2/2

Find the area of the segment AYB shown in the figure, if the radius of the circle is 21 cm and $∠ A O B = 12 0^{\circ}$ . [Use $π = \frac{22}{7}$ ]

Diagram for CBSE 2026 Class 10 Maths question 27

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Answer:

(462 - \frac{441 3}{4})

c m^{2}

(≈ 271.0

c m^{2}

)

Area of sector OAYB $= \frac{120}{360} \times \frac{22}{7} \times 21 \times 21 = 462$ $c m^{2}$
Area of $△ A O B = \frac{1}{2} r^{2} sin 12 0^{\circ} = \frac{1}{2} \times 441 \times \frac{3}{2} = \frac{441 3}{4}$ $c m^{2}$
(Draw OM $⊥$ AB: OM $= 21 cos 6 0^{\circ} = 10.5$ , AB $= 2 \times 21 sin 6 0^{\circ} = 213$ ; area $= \frac{1}{2} \times 213 \times 10.5$ .)
Area of segment $= 462 - \frac{441 3}{4}$ $c m^{2}$ ≈ 462 - 190.95 = 271.05 $c m^{2}$

Q273 marksShort AnswerAreas Related to Circles CBSE 2026 · Standard 30/3/1

In the given figure, chord AB subtends an angle of $12 0^{\circ}$ at the centre of the circle with radius 7 cm. Find (i) perimeter of major sector OACB, and (ii) area of the shaded segment, if area of $Δ$ OAB = 21.2 cm $^{2}$ .

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Answer: (i)

\frac{130}{3} \approx 43.33

cm (ii)

\approx 30.13

^{2}

(i) Major sector angle $= 36 0^{\circ} - 12 0^{\circ} = 24 0^{\circ}$
Arc ACB $= \frac{240}{360} \times 2 \times \frac{22}{7} \times 7 = \frac{88}{3}$ cm
Perimeter $= 7 + 7 + \frac{88}{3} = \frac{130}{3} \approx 43.33$ cm
(ii) Area of minor sector OAB $= \frac{120}{360} \times \frac{22}{7} \times 7^{2} = \frac{154}{3} \approx 51.33$ cm $^{2}$
Shaded segment $= 51.33 - 21.2 = 30.13$ cm $^{2}$ (approx.)

Also asked in: 2026 Standard 30/3/2, 2026 Standard 30/3/3

Q273 marksShort AnswerAreas Related to Circles CBSE 2026 · Standard 30/4/1

Chord AB of a circle with centre O and radius 21 mm subtends an angle of $12 0^{\circ}$ at the centre. Find the perimeters of the shaded region. (Use $3 = 1.73$ )

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Answer: 80.33 mm

The shaded region is the minor segment bounded by arc AB and chord AB.
Arc AB $= \frac{12 0 ^{\circ}}{36 0 ^{\circ}} \times 2 \times \frac{22}{7} \times 21 = 44$ mm
Draw $O M ⊥ A B$ : $∠ A O M = 6 0^{\circ}$ , $A M = 21 sin 6 0^{\circ} = \frac{21 3}{2}$
$A B = 213 = 21 \times 1.73 = 36.33$ mm
Perimeter $= 44 + 36.33 = 80.33$ mm

Q313 marksShort AnswerAreas Related to Circles CBSE 2026 · Standard 30/4/2

The perimeter of sector OAB of a circle with centre O and radius 5.6 cm, is 15.6 cm. Find length of the arc AB. Also find the value of $θ$ .

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Answer: Arc AB = 4.4 cm;

θ = 4 5^{\circ}

Perimeter = OA + OB + arc AB
$15.6 = 5.6 + 5.6 + arc A B \Rightarrow arc A B = 4.4$ cm
$\frac{θ}{36 0 ^{\circ}} \times 2 \times \frac{22}{7} \times 5.6 = 4.4 \Rightarrow \frac{θ}{36 0 ^{\circ}} \times 35.2 = 4.4$
$θ = 4 5^{\circ}$

Q303 marksShort AnswerAreas Related to Circles CBSE 2026 · Standard 30/4/3

Chord AB subtends an angle of $12 0^{\circ}$ at the centre O of the circle with radius $\frac{21}{2}$ cm. Find the perimeter of shaded segment ACB. (Use $3 = 1.7$ )

Diagram for CBSE 2026 Class 10 Maths question 30

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Answer: 39.85 cm

Arc ACB $= \frac{12 0 ^{\circ}}{36 0 ^{\circ}} \times 2 \times \frac{22}{7} \times \frac{21}{2} = 22$ cm
Draw $O M ⊥ A B$ ; $∠ A O M = 6 0^{\circ}$ , $A M = \frac{21}{2} sin 6 0^{\circ} = \frac{21 3}{4}$
$A B = \frac{21 3}{2} = 10.5 \times 1.7 = 17.85$ cm
Perimeter $= 22 + 17.85 = 39.85$ cm

Q263 marksShort AnswerAreas Related to Circles CBSE 2026 · Standard 30/5/1

A circle of diameter 20 cm is equally divided into five sectors. Find the area and perimeter of one of the sectors.

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Answer: Area

= \frac{440}{7}

c m^{2}

(≈ 62.86

c m^{2}

); perimeter

= \frac{228}{7}

cm (≈ 32.57 cm)

Radius $r = 10$ cm; each sector has angle $\frac{36 0 ^{\circ}}{5} = 7 2^{\circ}$ .
Area of one sector $= \frac{1}{5} \times π r^{2} = \frac{1}{5} \times \frac{22}{7} \times 100 = \frac{440}{7}$ $c m^{2}$ ≈ 62.86 $c m^{2}$
Arc length $= \frac{1}{5} \times 2 π r = \frac{1}{5} \times 2 \times \frac{22}{7} \times 10 = \frac{88}{7}$ cm
Perimeter $= 2 r + arc = 20 + \frac{88}{7} = \frac{228}{7}$ cm ≈ 32.57 cm

Also asked in: 2026 Standard 30/5/2

Q313 marksShort AnswerAreas Related to Circles CBSE 2026 · Standard 30/5/3

A chord of a circle, of radius 14 cm, subtends an angle of $6 0^{\circ}$ at the centre. Find the area of the smaller sector and perimeter of the smaller segment.

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Answer: Area of sector

= \frac{308}{3}

c m^{2}

(≈ 102.67

c m^{2}

); perimeter of segment

= \frac{86}{3}

cm (≈ 28.67 cm)

Area of smaller sector $= \frac{6 0 ^{\circ}}{36 0 ^{\circ}} \times \frac{22}{7} \times 14 \times 14 = \frac{616}{6} = \frac{308}{3}$ $c m^{2}$ ≈ 102.67 $c m^{2}$
Arc length $= \frac{6 0 ^{\circ}}{36 0 ^{\circ}} \times 2 \times \frac{22}{7} \times 14 = \frac{44}{3}$ cm
The triangle formed by the two radii and the chord is equilateral (angle $6 0^{\circ}$ , two sides 14 cm), so chord = 14 cm.
Perimeter of smaller segment $= \frac{44}{3} + 14 = \frac{86}{3}$ cm ≈ 28.67 cm

Q303 marksShort AnswerAreas Related to Circles CBSE 2025 · Basic 430/4/1

A chord of a circle of radius 10 cm subtends an angle of $6 0^{\circ}$ at the centre O. Find the area of the shaded region.
(Use $3 = 1.73, 2 = 1.41$ and $π = 3.14$ )

Diagram for CBSE 2025 Class 10 Maths question 30

Show answer & solution

Answer: 9.08 cm

^{2}

(approx.)

The shaded region is the minor segment cut off by chord AB.
Area of sector OAB $= \frac{60}{360} \times 3.14 \times 1 0^{2} = 52.33$ cm $^{2}$ .
$△ O A B$ is equilateral (OA = OB, $∠ A O B = 6 0^{\circ}$ ), so its area $= \frac{3}{4} \times 1 0^{2} = 43.25$ cm $^{2}$ .
Area of shaded segment $= 52.33 - 43.25 = 9.08$ cm $^{2}$ .

Also asked in: 2025 Basic 430/4/2

Q313 marksShort AnswerAreas Related to Circles CBSE 2025 · Basic 430/4/3

A chord of a circle of radius 14 cm subtends an angle of $9 0^{\circ}$ at the centre. Find perimeter of shaded region. (Use $2 = 1.41$ )

Diagram for CBSE 2025 Class 10 Maths question 31

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Answer: 41.74 cm

The shaded region is the minor segment bounded by chord AB and arc AB.
Arc AB $= \frac{90}{360} \times 2 \times \frac{22}{7} \times 14 = 22$ cm.
$△ O A B$ is right-angled at O, so chord $A B = 1 4^{2} + 1 4^{2} = 142 = 19.74$ cm.
Perimeter $= 22 + 19.74 = 41.74$ cm.

Q293 marksShort AnswerAreas Related to Circles CBSE 2025 · Basic 430/6/1

AB and CD are diameters of a circle with centre O and radius 7 cm. If $∠ B O D = 3 0^{\circ}$ , then find the area and perimeter of the shaded region.

Diagram for CBSE 2025 Class 10 Maths question 29

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Answer: Area

= \frac{77}{3}

^{2}

(about 25.67 cm

^{2}

); perimeter

= \frac{106}{3}

cm (about 35.33 cm)

The shaded region is two sectors BOD and AOC, each of angle $3 0^{\circ}$ ( $∠ A O C = ∠ B O D$ , vertically opposite).
Area $= 2 \times \frac{30}{360} \times \frac{22}{7} \times 7^{2} = 2 \times \frac{1}{12} \times 154 = \frac{77}{3}$ cm $^{2}$ .
Each arc $= \frac{30}{360} \times 2 \times \frac{22}{7} \times 7 = \frac{11}{3}$ cm.
Perimeter = 4 radii + 2 arcs $= 28 + \frac{22}{3} = \frac{106}{3}$ cm.

Also asked in: 2025 Basic 430/6/2, 2025 Basic 430/6/3

Q293 marksShort AnswerAreas Related to Circles CBSE 2025 · Standard 30/2/1

The length of the hour hand of a clock is 10 cm. Find the area of the minor sector swept by the hour hand of the clock between 5 a.m. to 8 a.m. Also, find the area of the major sector.

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Answer: Minor sector

= \frac{550}{7} \approx 78.57

^{2}

; major sector

= \frac{1650}{7} \approx 235.71

^{2}

The hour hand turns $3 0^{\circ}$ per hour; from 5 a.m. to 8 a.m. is 3 hours, so angle $= 9 0^{\circ}$
Minor sector $= \frac{90}{360} \times \frac{22}{7} \times 1 0^{2} = \frac{550}{7} \approx 78.57$ cm $^{2}$
Area of circle $= \frac{22}{7} \times 100 = \frac{2200}{7}$ cm $^{2}$
Major sector $= \frac{2200}{7} - \frac{550}{7} = \frac{1650}{7} \approx 235.71$ cm $^{2}$

Also asked in: 2025 Standard 30/2/2, 2025 Standard 30/2/3

Q293 marksShort AnswerAreas Related to Circles CBSE 2025 · Standard 30/3/1

A chord of a circle of radius 10 cm subtends a right angle at the centre of the circle. Find the area of the corresponding minor segment. [Use $π = 3.14$ ]

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Answer: 28.5 cm

^{2}

Area of sector $= \frac{90}{360} \times 3.14 \times 1 0^{2} = 78.5$ cm $^{2}$ .
Area of right triangle formed by the two radii $= \frac{1}{2} \times 10 \times 10 = 50$ cm $^{2}$ .
Area of minor segment $= 78.5 - 50 = 28.5$ cm $^{2}$ .

Also asked in: 2025 Standard 30/3/2, 2025 Standard 30/3/3

Q313 marksShort AnswerAreas Related to Circles CBSE 2024 · Basic 430/2/1

A car has two wipers which do not overlap. Each wiper has a blade of length 21 cm sweeping through an angle of $12 0^{\circ}$ . Find the area cleaned at each sweep of the blades.

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Answer: 924 cm

^{2}

Each blade sweeps a sector of radius 21 cm and angle $12 0^{\circ}$ .
Area of one sector $= \frac{120}{360} \times \frac{22}{7} \times 21 \times 21 = 462$ cm $^{2}$
The wipers do not overlap, so total area $= 2 \times 462 = 924$ cm $^{2}$

Also asked in: 2024 Basic 430/2/2, 2024 Basic 430/2/3

Q263 marksShort AnswerAreas Related to Circles CBSE 2024 · Basic 430/2/2

To warn ships for underwater rocks, a light house spreads a red coloured light over a sector of angle $8 0^{\circ}$ to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use $π = 3.14$ )

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Answer: 189.97 km

^{2}

Area of sector $= \frac{θ}{360} \times π r^{2}$
$= \frac{80}{360} \times 3.14 \times 16.5 \times 16.5$
$= \frac{2}{9} \times 3.14 \times 272.25 = 3.14 \times 60.5$
$= 189.97$ km $^{2}$

Q263 marksShort AnswerAreas Related to Circles CBSE 2024 · Basic 430/4/1

A chord of a circle of radius 10 cm subtends a right angle at the centre of the circle. Find the area of the corresponding (i) minor sector (ii) major sector. (Use $π = 3.14$ )

Show answer & solution

Answer: (i)

78.5

^{2}

(ii)

235.5

^{2}

Minor sector $= \frac{90}{360} \times 3.14 \times 1 0^{2} = \frac{314}{4} = 78.5$ cm $^{2}$ .
Area of circle $= 3.14 \times 100 = 314$ cm $^{2}$ .
Major sector $= 314 - 78.5 = 235.5$ cm $^{2}$ .

Also asked in: 2024 Basic 430/4/2, 2024 Basic 430/4/3

Q303 marksShort AnswerAreas Related to Circles CBSE 2024 · Basic 430/5/1

A horse is tied with a 14 m long rope at one corner of an equilateral triangular field having side 20 m. Find the area of the field where the horse cannot graze.

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Answer:

(1003 - \frac{308}{3})

^{2}

\approx 70.53

^{2}

Angle of an equilateral triangle $= 6 0^{\circ}$ , so the horse grazes a sector of radius 14 m and angle $6 0^{\circ}$ .
Area grazed $= \frac{60}{360} \times \frac{22}{7} \times 14 \times 14 = \frac{308}{3} \approx 102.67$ m $^{2}$ .
Area of field $= \frac{3}{4} \times 2 0^{2} = 1003 \approx 173.20$ m $^{2}$ .
Area not grazed $= 173.20 - 102.67 = 70.53$ m $^{2}$ (approx.).

Also asked in: 2024 Basic 430/5/2, 2024 Basic 430/5/3

Q30 (OR) (OR)3 marksShort AnswerAreas Related to Circles CBSE 2024 · Basic 430/5/1

The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand between 8:00 am and 8:05 am.

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Answer:

\frac{154}{3}

^{2}

\approx 51.33

^{2}

In 60 minutes the minute hand turns $36 0^{\circ}$ , so in 5 minutes it turns $3 0^{\circ}$ .
Area swept $= \frac{30}{360} \times \frac{22}{7} \times 14 \times 14$ .
$= \frac{1}{12} \times 616 = \frac{154}{3} \approx 51.33$ cm $^{2}$ .

Also asked in: 2024 Basic 430/5/2, 2024 Basic 430/5/3

Q313 marksShort AnswerAreas Related to Circles CBSE 2024 · Standard 30/5/1

An arc of a circle of radius 10 cm subtends a right angle at the centre of the circle. Find the area of the corresponding major sector. (Use $π = 3.14$ )

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Answer: 235.5 cm²

Angle of the major sector $= 36 0^{\circ} - 9 0^{\circ} = 27 0^{\circ}$ .
Area $= \frac{270}{360} \times 3.14 \times 1 0^{2} = \frac{3}{4} \times 314 = 235.5$ cm².

Also asked in: 2024 Standard 30/5/2

Q273 marksShort AnswerAreas Related to Circles CBSE 2024 · Standard 30/5/3

A sector is cut from a circle of radius 21 cm. The central angle of the sector is $15 0^{\circ}$ . Find the length of the arc of this sector and the area of the sector.

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Answer: Arc length = 55 cm; area = 577.5 cm²

Arc length $= \frac{150}{360} \times 2 \times \frac{22}{7} \times 21 = \frac{5}{12} \times 132 = 55$ cm.
Area $= \frac{150}{360} \times \frac{22}{7} \times 21 \times 21 = \frac{5}{12} \times 1386 = 577.5$ cm².

Q263 marksShort AnswerAreas Related to Circles CBSE 2023 · Basic 430/2/1

In the given figure, two concentric circles with centre O are shown. Radii of the circles are 2 cm and 5 cm respectively. Find the area of the shaded region.

Diagram for CBSE 2023 Class 10 Maths question 26

Show answer & solution

Answer: 11 cm

^{2}

The shaded region lies between the outer sector OAC and the inner sector OED, both with central angle $6 0^{\circ}$ .
Area of outer sector $= \frac{60}{360} \times \frac{22}{7} \times 5^{2}$
Area of inner sector $= \frac{60}{360} \times \frac{22}{7} \times 2^{2}$
Shaded area $= \frac{1}{6} \times \frac{22}{7} \times (25 - 4) = \frac{1}{6} \times \frac{22}{7} \times 21 = 11$ cm $^{2}$

Also asked in: 2023 Basic 430/2/2, 2023 Basic 430/2/3

Q303 marksShort AnswerAreas Related to Circles CBSE 2023 · Basic 430/4/1

Find the area of the sector of a circle of radius 7 cm and of central angle $9 0^{\circ}$ . Also, find the area of corresponding major sector.

Show answer & solution

Answer: Minor sector

= 38.5

^{2}

; major sector

= 115.5

^{2}

Area of sector $= \frac{90}{360} \times \frac{22}{7} \times 7 \times 7 = \frac{1}{4} \times 154 = 38.5$ cm $^{2}$ .
Area of circle $= \frac{22}{7} \times 49 = 154$ cm $^{2}$ .
Area of major sector $= 154 - 38.5 = 115.5$ cm $^{2}$ .

Also asked in: 2023 Basic 430/4/2, 2023 Basic 430/4/3

Q293 marksShort AnswerAreas Related to Circles CBSE 2023 · Standard 30/2/1

A car has two wipers which do not overlap. Each wiper has a blade of length 21 cm sweeping through an angle of $12 0^{\circ}$ . Find the total area cleaned at each sweep of the two blades.

Show answer & solution

Answer: 924 cm²

Area swept by one blade $= \frac{120}{360} \times \frac{22}{7} \times 21 \times 21 = 462$ cm²
Total for two blades $= 2 \times 462 = 924$ cm²

Q313 marksShort AnswerAreas Related to Circles CBSE 2023 · Standard 30/2/2

Reeti prepares a Rakhi for her brother Ronit. The Rakhi consists of a rectangle of length 8 cm and breadth 6 cm inscribed in a circle as shown in the figure. Find the area of the shaded region. (Use $π = 3.14$ )

Diagram for CBSE 2023 Class 10 Maths question 31

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Answer: 30.5 cm²

The diagonal of the rectangle is a diameter: $8^{2} + 6^{2} = 10$ cm, so $r = 5$ cm.
Area of circle $= 3.14 \times 5^{2} = 78.5$ cm²
Area of rectangle $= 8 \times 6 = 48$ cm²
Shaded area $= 78.5 - 48 = 30.5$ cm²

Q273 marksShort AnswerAreas Related to Circles CBSE 2023 · Standard 30/2/3

A chord of a circle of radius 14 cm subtends an angle of $6 0^{\circ}$ at the centre. Find the area of the corresponding minor segment of the circle.
(Use $π = 3.14$ and $3 = 1.73$ )

Show answer & solution

Answer: About 17.80 cm²

Area of sector $= \frac{60}{360} \times 3.14 \times 1 4^{2} = \frac{615.44}{6} \approx 102.57$ cm²
The triangle formed by the chord and the two radii is equilateral with side 14 cm.
Area of triangle $= \frac{3}{4} \times 1 4^{2} = 49 \times 1.73 = 84.77$ cm²
Area of minor segment $\approx 102.57 - 84.77 = 17.80$ cm²

Q313 marksShort AnswerAreas Related to Circles CBSE 2023 · Standard 30/6/1

In a circle of radius 21 cm, an arc subtends an angle of $6 0^{\circ}$ at the centre. Find the area of the sector formed by the arc. Also, find the length of the arc.

Show answer & solution

Answer: Area of sector = 231 cm

^{2}

; length of arc = 22 cm

Area of sector $= \frac{60}{360} \times \frac{22}{7} \times 21 \times 21 = 231$ cm $^{2}$
Length of arc $= \frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = 22$ cm

Also asked in: 2023 Standard 30/6/3

Q263 marksShort AnswerAreas Related to Circles CBSE 2023 · Standard 30/6/2

In the given figure, AB and CD are diameters of a circle with centre O perpendicular to each other. If OA = 7 cm, find the area of shaded region.

Show answer & solution

Answer: 28 cm

^{2}

r = 7 cm and $∠ B O C = ∠ C O A = 9 0^{\circ}$ .
Area of one segment = area of quadrant – area of right triangle
$= \frac{1}{4} \times \frac{22}{7} \times 7 \times 7 - \frac{1}{2} \times 7 \times 7 = 38.5 - 24.5 = 14$ cm $^{2}$
Shaded region = 2 segments = $2 \times 14 = 28$ cm $^{2}$

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