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Areas Related to Circles: 2 marks Questions (CBSE Class 10)

12 different 2 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)

Q252 marksVery Short AnswerAreas Related to Circles CBSE 2025 · Basic 430/1/1

A chord of a circle of diameter 20 cm subtends an angle of $6 0^{\circ}$ at the centre of the circle. Find the area of the corresponding minor segment of the circle. (Use $π = 3.14$ and $3 = 1.73$ )

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Answer: 9.08 cm² (approx.)

Radius $r = 10$ cm.
Area of sector $= \frac{60}{360} \times 3.14 \times 100 = 52.33$ cm².
The triangle formed is equilateral (two sides 10 cm, included angle $6 0^{\circ}$ ): area $= \frac{3}{4} \times 100 = \frac{1.73 \times 100}{4} = 43.25$ cm².
Area of minor segment $= 52.33 - 43.25 = 9.08$ cm².

Q232 marksVery Short AnswerAreas Related to Circles CBSE 2025 · Basic 430/1/2

In the given figure, two concentric circles with centre O and radii 2 cm and 3 cm are shown. Find the perimeter of the shaded region.

Diagram for CBSE 2025 Class 10 Maths question 23

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Answer:

\frac{152}{21}

cm ≈ 7.24 cm

The shaded region is bounded by an arc of the outer circle, an arc of the inner circle (both subtending $6 0^{\circ}$ ) and two radial segments each $3 - 2 = 1$ cm.
Outer arc $= \frac{60}{360} \times 2 π \times 3 = π$ cm; inner arc $= \frac{60}{360} \times 2 π \times 2 = \frac{2 π}{3}$ cm.
Perimeter $= π + \frac{2 π}{3} + 2 = \frac{5 π}{3} + 2 = \frac{5}{3} \times \frac{22}{7} + 2 = \frac{110}{21} + 2 = \frac{152}{21} \approx 7.24$ cm.

Q212 marksVery Short AnswerAreas Related to Circles CBSE 2025 · Basic 430/1/3

From a circular sheet of radius 70 cm, a quadrant is cut. Find the area of the remaining sheet.

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Answer: 11550 cm²

Area of the sheet $= π r^{2} = \frac{22}{7} \times 70 \times 70 = 15400$ cm².
Area of the quadrant $= \frac{1}{4} \times 15400 = 3850$ cm².
Remaining area $= 15400 - 3850 = 11550$ cm².

Q252 marksVery Short AnswerAreas Related to Circles CBSE 2025 · Basic 430/2/1

The area of a smaller circle is equal to the area of a sector of a larger circle with central angle $12 0^{\circ}$ . The radii of the smaller and larger circles are ‘r’ and ‘R’ respectively. Find r : R.

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Answer:

r : R = 1 : 3

Area of sector $= \frac{120}{360} π R^{2} = \frac{1}{3} π R^{2}$ .
$π r^{2} = \frac{1}{3} π R^{2} \Rightarrow r^{2} = \frac{R ^{2}}{3}$ .
$\frac{r}{R} = \frac{1}{3}$ , so $r : R = 1 : 3$ .

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

Q25 (OR) (OR)2 marksVery Short AnswerAreas Related to Circles CBSE 2025 · Basic 430/2/1

In the given figure, O is the centre of a circle of radius 7 cm. AB is a chord of the circle. Find the perimeter of the shaded region.

Diagram for CBSE 2025 Class 10 Maths question 25 (OR)

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Answer:

\frac{43}{3}

cm (about 14.33 cm)

In $△ O A B$ , OA = OB = 7 cm and $∠ A O B = 6 0^{\circ}$ , so the base angles are also $6 0^{\circ}$ : the triangle is equilateral and AB = 7 cm.
Length of arc AB $= \frac{60}{360} \times 2 \times \frac{22}{7} \times 7 = \frac{22}{3}$ cm.
Perimeter of shaded segment = AB + arc AB $= 7 + \frac{22}{3} = \frac{43}{3}$ cm $\approx 14.33$ cm.

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

Q252 marksVery Short AnswerAreas Related to Circles CBSE 2025 · Basic 430/3/1

In the given figure, OPRQ is a quadrant of a circle with centre O and radius 10 cm. If OP = 10 cm and OS = 6 cm, find the area of the shaded region. (Use $π = 3.14$ )

Diagram for CBSE 2025 Class 10 Maths question 25

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Answer: 48.5

cm^{2}

Area of quadrant OPRQ $= \frac{1}{4} \times 3.14 \times 1 0^{2} = 78.5$ $cm^{2}$ .
Area of right $△ P O S = \frac{1}{2} \times O P \times O S = \frac{1}{2} \times 10 \times 6 = 30$ $cm^{2}$ .
Shaded area $= 78.5 - 30 = 48.5$ $cm^{2}$ .

Q232 marksVery Short AnswerAreas Related to Circles CBSE 2025 · Basic 430/3/2

In the given figure, OABC is a quadrant of a circle with centre O and radius 8 cm. If OD = 6 cm, then find then perimeter of the shaded region.

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Answer:

\frac{172}{7}

\approx 24.57

The shaded region is bounded by arc ABC, CD and DA.
Arc ABC $= \frac{1}{4} \times 2 π r = \frac{1}{4} \times 2 \times \frac{22}{7} \times 8 = \frac{88}{7}$ cm.
$C D = O C - O D = 8 - 6 = 2$ cm.
$A D = O A^{2} + O D^{2} = 64 + 36 = 10$ cm.
Perimeter $= \frac{88}{7} + 2 + 10 = \frac{172}{7} \approx 24.57$ cm.

Q212 marksVery Short AnswerAreas Related to Circles CBSE 2025 · Basic 430/3/3

From a circular sheet of radius 10 cm, a quadrant is cut. Find the perimeter of the remaining sheet.

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Answer:

\frac{470}{7}

\approx 67.14

The remaining sheet is bounded by three-quarters of the circle and two radii.
Arc length $= \frac{3}{4} \times 2 π r = \frac{3}{4} \times 2 \times \frac{22}{7} \times 10 = \frac{330}{7}$ cm.
Perimeter $= \frac{330}{7} + 2 \times 10 = \frac{330}{7} + 20 = \frac{470}{7} \approx 67.14$ cm.

Q222 marksVery Short AnswerAreas Related to Circles CBSE 2025 · Standard 30/2/1

In the given figure, the shape of the top of a table is that of a sector of a circle with centre O and $∠ A O B = 9 0^{\circ}$ . If $A O = O B = 42$ cm, then find the perimeter of the top of the table.

Diagram for CBSE 2025 Class 10 Maths question 22

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Answer: 282 cm

The table top is the major sector, with angle $36 0^{\circ} - 9 0^{\circ} = 27 0^{\circ}$
Arc length $= \frac{270}{360} \times 2 \times \frac{22}{7} \times 42 = \frac{3}{4} \times 264 = 198$ cm
Perimeter $= 198 + 42 + 42 = 282$ cm

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

Q22 (OR) (OR)2 marksVery Short AnswerAreas Related to Circles CBSE 2025 · Standard 30/2/1

In the given figure, three sectors of a circle of radius 5 cm, making angles $3 5^{\circ}$ , $5 0^{\circ}$ and $9 5^{\circ}$ at the centre are shaded. Find the area of the shaded region. [Use $π = \frac{22}{7}$ ]

Diagram for CBSE 2025 Class 10 Maths question 22 (OR)

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Answer:

\frac{275}{7}

^{2}

\approx 39.29

^{2}

Total angle of shaded sectors $= 3 5^{\circ} + 5 0^{\circ} + 9 5^{\circ} = 18 0^{\circ}$
Shaded area $= \frac{180}{360} \times \frac{22}{7} \times 5^{2} = \frac{1}{2} \times \frac{550}{7} = \frac{275}{7}$ cm $^{2}$
$\approx 39.29$ cm $^{2}$

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

Q222 marksVery Short AnswerAreas Related to Circles CBSE 2024 · Standard 30/4/1

The minute hand of a clock is 14 cm long. Find the area on the face of the clock described by the minute hand in 5 minutes.

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Answer:

\frac{154}{3}

^{2}

\approx

51.33 cm

^{2}

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

Also asked in: 2024 Standard 30/4/2, 2024 Standard 30/4/3

Q22 (OR) (OR)2 marksVery Short AnswerAreas Related to Circles CBSE 2024 · Standard 30/4/1

Find the length of the arc of a circle which subtends an angle of $6 0^{\circ}$ at the centre of the circle of radius 42 cm.

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Answer: 44 cm

Arc length = $\frac{θ}{36 0 ^{\circ}} \times 2 π r$ .
$= \frac{60}{360} \times 2 \times \frac{22}{7} \times 42$ .
$= \frac{1}{6} \times 264 = 44$ cm.

Also asked in: 2024 Standard 30/4/2, 2024 Standard 30/4/3

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