MathsCBSE Class 10 · Basic · Board exam 2026, set 430/5/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Basic 2026: question 38

Set in the CBSE Class 10 board exam of 2026, paper 430/5/1, section E, question 38, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q38 · 4 marks · Case-basedCompetency-basedAreas Related to Circles

In a circular museum hall of radius 14 m, some statues are displayed. Statues are kept inside the inner concentric circle of radius 7 m. One such statue lying in sector OAB, is fenced along line segments OA, AP, PB and BO where P is a point on outer circle.

Based on above information, answer the following questions:

(i)
Find $m ∠ A O P$ .
1 mark
(ii)
Prove that $△ O A P ≅ △ O B P$ .
1 mark
(iii)(a)
Find the length of fencing required to protect the statue. (Take $3 = 1.73$ )
2 marks
OR
(iii)(b)
Find area of quadrilateral OAPB. (Take $3 = 1.73$ )
2 marks

Figure in the paper: Photograph of a circular museum hall with statues, and a figure: an outer circle of radius 14 m and an inner concentric circle of radius 7 m, both centred at O. A and B lie on the inner circle; P lies on the outer circle. Segments OA, AP, PB and BO are drawn, with OP as a dashed line from O through the inner region to P. The region OAPB is the fenced area. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 430/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

PA touches the inner circle at A, so what is the angle between OA and AP?

Hint 2

In right triangle OAP you know the side next to $∠ A O P$ (7 m) and the hypotenuse (14 m). Pick the trigonometric ratio that links them.

Hint 3

The two triangles OAP and OBP are congruent, so the fencing is $2 (O A + A P)$ and the area is twice the area of one triangle.

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.

(i) Here $O A = 7$ m (radius of the inner circle) and $O P = 14$ m (radius of the outer circle). PA is a tangent to the inner circle at A, so $∠ O A P = 9 0^{\circ}$ [tangent ⟂ radius].
In right $△ O A P$ : $cos ∠ A O P = \frac{O A}{O P} = \frac{7}{14} = \frac{1}{2}$ , so $m ∠ A O P = 6 0^{\circ}$ .
1 mark
(ii) In $△ O A P$ and $△ O B P$ :
$∠ O A P = ∠ O B P = 9 0^{\circ}$ [radius ⟂ tangent at the point of contact]
$O P = O P$ [common hypotenuse]
$O A = O B$ [radii of the inner circle]
Therefore $△ O A P ≅ △ O B P$ [RHS congruence].
1 mark
(iii)(a) In right $△ O A P$ : $sin 6 0^{\circ} = \frac{A P}{O P} \Rightarrow \frac{3}{2} = \frac{A P}{14} \Rightarrow A P = 73 = 7 \times 1.73 = 12.11$ m.
1 mark
$P B = A P$ (CPCT) and $O B = O A$ .
Length of fencing $= O A + A P + P B + B O = 2 (O A + A P) = 2 (7 + 12.11) = 38.22$ m.
1 mark
(iii)(b) As in (iii)(a), $sin 6 0^{\circ} = \frac{A P}{14} \Rightarrow A P = 73 = 12.11$ m.
1 mark
Area of OAPB $= 2 \times ar (△ O A P)$ [the triangles are congruent] $= 2 \times \frac{1}{2} \times O A \times A P = 7 \times 12.11 = 84.77 m^{2}$ .
1 mark

Answer: (i) $6 0^{\circ}$ ; (ii) $△ O A P ≅ △ O B P$ by RHS; (iii)(a) 38.22 m; (iii)(b) 84.77 m²

Where the marks are

Uses $cos ∠ A O P = 7/14$ to get $∠ A O P = 6 0^{\circ}$ (1 mark)
Proves the congruence by RHS with correct reasons (right angle, common hypotenuse, equal radii) (1 mark)
(iii)(a) Finds $A P = 73 = 12.11$ m (1 mark)
(iii)(a) Fencing $= 2 (O A + A P) = 38.22$ m (1 mark)
(iii)(b) Finds $A P = 12.11$ m (1 mark)
(iii)(b) Area $= 2 \times \frac{1}{2} \times O A \times A P = 84.77$ m² (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

Practise this chapterBoard-style questions from Areas Related to Circles, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 430/5/1 (Basic): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Basic 2026: question 38

Hints

Step-by-step solution

Revise the chapter

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