MathsCBSE Class 10 · Standard · Board exam 2026, set 30/4/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard 2026: question 36

Set in the CBSE Class 10 board exam of 2026, paper 30/4/1, section E, question 36, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q36 · 4 marks · Case-basedCompetency-basedTriangles

Carom board is a very popular game. The board is a square of side length 65 cm. It has circular pockets in each corner.

Ansh strikes a disc, kept at position P with a striker. The disc, hits the boundary of the board at R and goes straight to pocket at corner C. It is given that PS = 9 cm, PQ = 35 cm, BR = $x$ , $∠ P R Q = α$ and $∠ C R B = θ$ .

Based on the above information, answer the following questions :

(i)
Using law of reflection i.e. $∠ P R T = ∠ C R T$ , prove that $θ = α$ .
1 mark
(ii)
Prove that $△ P QR \sim △ C B R$ given that PQ is perpendicular to AB.
1 mark
(iii)(a)
Find the value of $x$ using similarity of triangles.
2 marks
OR
(iii)(b)
If $\frac{Area △ P QR}{Area △ C B R} = \frac{P Q ^{2}}{C B ^{2}}$ , then find the value of $x$ .
2 marks

Figure in the paper: Photo of a carom board plus a diagram: square board with corners C (top-left), B (top-right), A (bottom-right), D (bottom-left); side CB = 65 cm. Pockets (circles) at each corner. R lies on side AB at distance x from B. Q is on AB below R, with PQ horizontal = 35 cm, PQ perpendicular to AB. P is a point inside; PS = 9 cm is the perpendicular from P to side DA (bottom side) with S on DA. T is on a dashed horizontal line through R extended into the board (normal at R). Line CR is drawn from C to R; line PR from P to R. Angle PRQ = α, angle CRB = θ. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 30/4/1.

Hints

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

Hint 1

The normal TR makes 90° with side AB, so angles on each side of TR add up to 90°.

Hint 2

For similarity, look for a pair of equal angles at R and a right angle in each triangle.

Hint 3

QR = 65 − AQ − BR, and AQ equals PS = 9 cm; then compare corresponding sides or areas.

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) TR is the normal at R, so TR ⟂ AB. Hence $∠ T R Q = ∠ T R B = 9 0^{\circ}$ .
$∴ α + ∠ P R T = 9 0^{\circ} = θ + ∠ C R T$
Since $∠ P R T = ∠ C R T$ [law of reflection], we get $α = θ$ .
1 mark
(ii) In $△ P QR$ and $△ C B R$ :
- $∠ P R Q = ∠ C R B$ [ $α = θ$ , from (i)]
- $∠ P QR = ∠ C B R = 9 0^{\circ}$ [PQ ⟂ AB, and the board is a square]
$∴ △ P QR \sim △ C B R$ [AA similarity].
1 mark
(iii)(a) Since $△ P QR \sim △ C B R$ , $\frac{P Q}{C B} = \frac{QR}{B R}$ .
½ mark
We have $Q A = P S = 9$ cm, so $QR = A B - Q A - B R = 65 - 9 - x = 56 - x$ .
$\Rightarrow \frac{35}{65} = \frac{56 - x}{x}$
1 mark
$\Rightarrow 35 x = 65 (56 - x) \Rightarrow 100 x = 3640 \Rightarrow x = 36.4$ cm
½ mark
(iii)(b) $\frac{ar ( △ P QR )}{ar ( △ C B R )} = \frac{\frac{1}{2} \times P Q \times QR}{\frac{1}{2} \times C B \times B R} = \frac{P Q ^{2}}{C B ^{2}}$
$\Rightarrow \frac{\frac{1}{2} \times 35 \times ( 65 - 9 - x )}{\frac{1}{2} \times 65 \times x} = \frac{35 \times 35}{65 \times 65}$
1 mark
$\Rightarrow \frac{56 - x}{x} = \frac{35}{65}$
½ mark
$\Rightarrow 100 x = 3640 \Rightarrow x = 36.4$ cm
½ mark

Answer: (i) θ = α proved; (ii) ΔPQR ~ ΔCBR by AA; (iii)(a) x = 36.4 cm; (iii)(b) x = 36.4 cm

Where the marks are

(i) Uses TR ⟂ AB to write α + ∠PRT = θ + ∠CRT and concludes α = θ (1 mark)
(ii) Equal angles α = θ and two right angles, giving similarity by AA (1 mark)
(iii)(a) Writes PQ/CB = QR/BR (½ mark)
(iii)(a) Substitutes QR = 56 − x to get 35/65 = (56 − x)/x (1 mark)
(iii)(a) x = 36.4 cm (½ mark)
(iii)(b) Writes the area ratio using ½ × base × height and equates it to 35²/65² (1 mark)
(iii)(b) Simplifies to (56 − x)/x = 35/65 (½ mark)
(iii)(b) x = 36.4 cm (½ mark)

Revise the chapter

Chapter: Triangles (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Triangles, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 30/4/1 (Standard): 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 Standard 2026: question 36

Hints

Step-by-step solution

Revise the chapter

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