CBSE Class 10 Mathematics Basic 2025: question 37
Set in the CBSE Class 10 board exam of 2025, paper 430/5/1, section E, question 37, for 4 marks. Try it first; the hints and our step-by-step solution are below.
Q37 · 4 marks · Case-basedCompetency-basedAreas Related to Circles
A farmer has put up a decorative windmill in his farm in which there are eight blades of equal width and equally placed in a circular arrangement. A circular wire goes through them. The diagram shows two blades OAB and OPQ in a quarter circle with centre O. , cm, cm. O is the centre of both the circles.
- (i)1 mark
Determine the measure of .
- (ii)1 mark
Find length of arc CD.
- (iii)2 marks
(a) Find the area of region CABD.
OR
(iii)2 marks(b) Find perimeter of region CABD.
Figure in the paper: Left: a windmill with eight equal blades arranged in a circle around a central hub, joined by a circular wire. Right: concentric circles with centre O. Radii OA (outer, 28 cm, pointing down) and OR (to the right). Points A, B on outer circle and C, D on inner circle along the radius OA-side; OB, OD and OP, OQ are drawn. Angles 30° are marked between OA and OB, and between OP and OQ (with OR as reference). Points on outer arc: A, B, P, Q, R; inner circle points: C, D. See it in the official paper.
Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 430/5/1.
Hints
Open one hint at a time and try again before the next.
Hint 1
The quarter circle AOR is 90°. Subtract the two blade angles and share what is left equally between the two gaps.
Hint 2
Arc length = (θ/360°) × 2πr, using the radius of the circle on which the arc lies.
Hint 3
Region CABD is a sector of radius 28 cm minus a sector of radius 21 cm. Its boundary has two arcs and two straight pieces, each 28 − 21 cm.
Step-by-step solution
Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.
- 1 mark
(i) (quarter circle). The blades OAB and OPQ take up . The blades are equally placed, so the remaining splits into two equal gaps, and . Hence .
- 1 mark
(ii) Arc CD lies on the inner circle ( cm) and subtends at O. Length of arc cm.
- 1 mark
(iii)(a) Area of region CABD = area of sector OAB − area of sector OCD
- 1 mark
(iii)(a)
- 1 mark
(iii)(b) Length of arc cm.
- 1 mark
(iii)(b) Perimeter of CABD = arc AB + arc CD + CA + DB. Here cm, so the perimeter cm.
Answer: (i) 15°; (ii) 11 cm; (iii)(a) ≈ 89.83 cm²; (iii)(b) ≈ 39.67 cm
Where the marks are
- (i) ∠BOP = ½(90° − 60°) = 15° (1 mark)
- (ii) Arc CD = (30/360) × 2 × (22/7) × 21 = 11 cm (1 mark)
- (iii)(a) Writes area as difference of two sectors: (30/360) × (22/7) × (28² − 21²) (1 mark)
- (iii)(a) Area ≈ 89.8 cm² (1 mark)
- (iii)(b) Arc AB = 44/3 ≈ 14.67 cm (1 mark)
- (iii)(b) Perimeter = 14.67 + 11 + 2 × 7 = 39.67 cm (1 mark)
Revise the chapter
- Chapter: Areas Related to Circles (notes, formulas, where marks are lost and every past question)
- Topic: Area of a Sector of a Circle
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