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Areas Related to Circles: 5 marks Questions (CBSE Class 10)

9 different 5 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)

Q325 marksLong AnswerAreas Related to Circles CBSE 2024 · Basic 430/1/2

O is the centre of the circle. If AC = 28 cm, BC = 21 cm, $∠ B O D = 9 0^{\circ}$ and $∠ B O C = 3 0^{\circ}$ , then find the area of the shaded region given in the figure.

Diagram for CBSE 2024 Class 10 Maths question 32

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Answer: 427.875 cm

^{2}

(about 427.88 cm

^{2}

)

AB passes through O, so AB is a diameter and $∠ A C B = 9 0^{\circ}$ (angle in a semicircle).
$A B = 2 8^{2} + 2 1^{2} = 1225 = 35$ cm, so $r = \frac{35}{2}$ cm.
Area of semicircle on the side of C $= \frac{1}{2} \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} = 481.25$ cm $^{2}$ .
Area of $△ A C B = \frac{1}{2} \times 28 \times 21 = 294$ cm $^{2}$ , so the two shaded segments $= 481.25 - 294 = 187.25$ cm $^{2}$ .
Area of quadrant BOD $= \frac{90}{360} \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} = 240.625$ cm $^{2}$ .
Shaded area $= 187.25 + 240.625 = 427.875$ cm $^{2}$ .

Q355 marksLong AnswerAreas Related to Circles CBSE 2024 · Basic 430/1/3

A chord of a circle of radius 14 cm subtends an angle of $9 0^{\circ}$ at the centre. Find the area of the corresponding minor and major segments of the circle.

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Answer: Minor segment

= 56

^{2}

; major segment

= 560

^{2}

Area of the minor sector $= \frac{90}{360} \times \frac{22}{7} \times 14 \times 14 = 154$ cm $^{2}$ .
Area of the triangle formed by the two radii $= \frac{1}{2} \times 14 \times 14 = 98$ cm $^{2}$ .
Minor segment $= 154 - 98 = 56$ cm $^{2}$ .
Area of circle $= \frac{22}{7} \times 196 = 616$ cm $^{2}$ , so major segment $= 616 - 56 = 560$ cm $^{2}$ .

Q325 marksLong AnswerAreas Related to Circles CBSE 2024 · Standard 30/1/1

An arc of a circle of radius 21 cm subtends an angle of $6 0^{\circ}$ at the centre. Find :
(i) the length of the arc.
(ii) the area of the minor segment of the circle made by the corresponding chord.

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Answer: (i) 22 cm (ii)

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

cm^{2}

\approx 40.05

cm^{2}

(i) Arc length $= \frac{60}{360} \times 2 \times \frac{22}{7} \times 21 = 22$ cm.
(ii) Area of sector $= \frac{60}{360} \times \frac{22}{7} \times 2 1^{2} = 231$ $cm^{2}$ .
The triangle formed by the two radii and the chord is equilateral (angle $6 0^{\circ}$ , two equal sides), so its area $= \frac{3}{4} \times 2 1^{2} = \frac{441 3}{4}$ $cm^{2}$ .
Area of minor segment $= 231 - \frac{441 3}{4} \approx 231 - 190.95 = 40.05$ $cm^{2}$ .

Q355 marksLong AnswerAreas Related to Circles CBSE 2024 · Standard 30/1/3

In the given figure, diameters AC and BD of the circle intersect at O. If $∠ A O B = 6 0^{\circ}$ and OA = 10 cm, then :
(i) find the length of the chord AB.
(ii) find the area of shaded region.
(Take $π = 3.14$ and $3 = 1.73$ )

Diagram for CBSE 2024 Class 10 Maths question 35

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Answer: (i) AB = 10 cm (ii) 113.75

cm^{2}

(i) OA = OB = 10 cm and $∠ A O B = 6 0^{\circ}$ , so $△ A O B$ is equilateral and AB = 10 cm.
(ii) The shaded region is the minor segment APB together with sector BOC.
$∠ B O C = 18 0^{\circ} - 6 0^{\circ} = 12 0^{\circ}$ . Area of sector BOC $= \frac{120}{360} \times 3.14 \times 100 = 104.67$ $cm^{2}$ (approx).
Area of segment APB = sector AOB $-$ $△ A O B = \frac{60}{360} \times 3.14 \times 100 - \frac{1.73}{4} \times 100 = 52.33 - 43.25 = 9.08$ $cm^{2}$ (approx).
Shaded area $= \frac{314}{3} + \frac{314}{6} - 43.25 = 157 - 43.25 = 113.75$ $cm^{2}$ .

Q355 marksLong AnswerAreas Related to Circles CBSE 2024 · Standard 30/2/1

The perimeter of a certain sector of a circle of radius 5.6 m is 20.0 m. Find the area of the sector.

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Answer: 24.64 m

^{2}

Perimeter of sector $= 2 r + l$ , where $l$ is the arc length.
$2 \times 5.6 + l = 20$ , so $l = 8.8$ m.
Area of sector $= \frac{1}{2} \times l \times r = \frac{1}{2} \times 8.8 \times 5.6 = 24.64$ m $^{2}$
(Check: $\frac{θ}{36 0 ^{\circ}} \times 2 π r = 8.8$ gives $θ = 9 0^{\circ}$ , and $\frac{1}{4} \times \frac{22}{7} \times 5. 6^{2} = 24.64$ .)

Also asked in: 2024 Standard 30/2/3

Q335 marksLong AnswerAreas Related to Circles CBSE 2023 · Basic 430/5/1

In the given figure, AB is a chord of a circle of radius 7 cm and centred at O. Find the area of the shaded region if $∠ A O B = 9 0^{\circ}$ . Also, find length of minor arc AB.

Diagram for CBSE 2023 Class 10 Maths question 33

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Answer: Area of shaded region = 14 cm

^{2}

; minor arc AB = 11 cm

Area of sector OAB $= \frac{90}{360} \times \frac{22}{7} \times 7 \times 7 = 38.5$ cm $^{2}$ .
Area of $△ A O B = \frac{1}{2} \times 7 \times 7 = 24.5$ cm $^{2}$ .
Shaded segment = 38.5 − 24.5 = 14 cm $^{2}$ .
Minor arc AB $= \frac{90}{360} \times 2 \times \frac{22}{7} \times 7 = 11$ cm.

Also asked in: 2023 Basic 430/5/2, 2023 Basic 430/5/3

Q33 (OR) (OR)5 marksLong AnswerAreas Related to Circles CBSE 2023 · Basic 430/5/1

AB and CD are arcs of two concentric circles of radii 3.5 cm and 10.5 cm respectively and centred at O. Find the area of the shaded region if $∠ A O B = 6 0^{\circ}$ . Also, find the length of arc CD.

Diagram for CBSE 2023 Class 10 Maths question 33 (OR)

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Answer: Area of shaded region

= \frac{154}{3} \approx 51.33

^{2}

; arc CD = 11 cm

Shaded area = sector OCD $-$ sector OAB $= \frac{60}{360} \times \frac{22}{7} \times (10. 5^{2} - 3. 5^{2})$ .
$= \frac{1}{6} \times \frac{22}{7} \times (110.25 - 12.25) = \frac{1}{6} \times \frac{22}{7} \times 98 = \frac{154}{3} \approx 51.33$ cm $^{2}$ .
Arc CD $= \frac{60}{360} \times 2 \times \frac{22}{7} \times 10.5 = 11$ cm.

Also asked in: 2023 Basic 430/5/2, 2023 Basic 430/5/3

Q345 marksLong AnswerAreas Related to Circles CBSE 2023 · Standard 30/1/1

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. Also find the area of the major segment of the circle.

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Answer: Minor segment

= (\frac{308}{3} - 493)

^{2}

≈ 17.89 cm

^{2}

; major segment

= (\frac{1540}{3} + 493)

^{2}

≈ 598.11 cm

^{2}

Area of sector $= \frac{60}{360} \times \frac{22}{7} \times 14 \times 14 = \frac{308}{3}$ cm $^{2}$ ≈ 102.67 cm $^{2}$ .
The triangle formed is equilateral (two radii with $6 0^{\circ}$ between them): area $= \frac{3}{4} \times 1 4^{2} = 493$ cm $^{2}$ ≈ 84.87 cm $^{2}$ .
Minor segment $= \frac{308}{3} - 493$ ≈ 17.89 cm $^{2}$ .
Area of circle $= \frac{22}{7} \times 1 4^{2} = 616$ cm $^{2}$ .
Major segment $= 616 - (\frac{308}{3} - 493) = \frac{1540}{3} + 493$ ≈ 598.11 cm $^{2}$ .

Also asked in: 2023 Standard 30/1/2, 2023 Standard 30/1/3

Q355 marksLong AnswerAreas Related to Circles CBSE 2023 · Standard 30/5/1

A 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 the area of that part of the field in which the horse can graze. Also, find the increase in grazing area if length of rope is increased to 10 m. (Use $π = 3.14$ )

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Answer: 19.625 m

^{2}

; increase = 58.875 m

^{2}

The horse grazes a quadrant (angle $9 0^{\circ}$ ) of radius equal to the rope.
With 5 m rope: area $= \frac{1}{4} \times 3.14 \times 5^{2} = 19.625$ m $^{2}$ .
With 10 m rope: area $= \frac{1}{4} \times 3.14 \times 1 0^{2} = 78.5$ m $^{2}$ .
Increase = 78.5 – 19.625 = 58.875 m $^{2}$ .

Also asked in: 2023 Standard 30/5/2, 2023 Standard 30/5/3

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