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Mathematics · 2026 · 4 marks
CBSE 2026 · Region 4 · Set 1 · Q36
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…
Carom board is a very popular game. The board is a square of side length $\displaystyle 65$ 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 $\displaystyle \mathrm{PS}=9 \mathrm{~cm}, \mathrm{PQ}=35 \mathrm{~cm}, \mathrm{BR}=x, \angle \mathrm{PRQ}=\alpha$ and $\displaystyle \angle \mathrm{CRB}=\theta$. Based on the above information, answer the following questions :
(i)
Using law of reflection i.e. $\displaystyle \angle \mathrm{PRT}=\angle \mathrm{CRT}$, prove that $\displaystyle \theta=\alpha$.
(ii)
Prove that $\displaystyle \triangle \mathrm{PQR} \sim \triangle \mathrm{CBR}$ given that PQ is perpendicular to AB.
(iii)
(a)
Find the value of $\displaystyle x$ using similarity of triangles.
OR
(b)
If $\displaystyle \frac{\text { Area } \triangle \mathrm{PQR}}{\text { Area } \triangle \mathrm{CBR}}=\frac{\mathrm{PQ}^{2}}{\mathrm{CB}^{2}}$, then find the value of $\displaystyle x$.
Marking-scheme solution
(i)
\(\displaystyle \mathrm{TR} \perp \mathrm{AB}\)
\(\displaystyle \therefore \alpha+\angle \mathrm{PRT}=\theta+\angle \mathrm{TRC}\)
As \(\displaystyle \angle \mathrm{PRT}=\angle \mathrm{TRC}\), so \(\displaystyle \alpha=\theta\)
(ii)
As \(\displaystyle \theta=\alpha\), so \(\displaystyle \angle \mathrm{PRQ}=\angle \mathrm{CRB}\)
\[\text { and } \angle \mathrm{PQR}=\angle \mathrm{CBR}=90^{0}
\]
\[\therefore \triangle \mathrm{PQR} \sim \triangle \mathrm{CBR}
\]
(iii)(a)
\(\displaystyle \triangle \mathrm{PQR} \sim \triangle \mathrm{CBR}\)
\[\therefore \frac{P Q}{\mathrm{CB}}=\frac{Q R}{\mathrm{BR}}
\]
\[\begin{aligned}
\Rightarrow & \frac{35}{65}=\frac{65-9-x}{x} \\
& \Rightarrow 35 x=65(56-x)
\end{aligned}
\]
\[\Longrightarrow x=36.4 \mathrm{~cm}
\]
(b)
\(\displaystyle \frac{A r . \triangle P Q R}{A r . \triangle C B R}=\frac{P Q^{2}}{B C^{2}}\)
\[\Rightarrow \frac{\dfrac{1}{2} \times 35 \times(65-x-9)}{\dfrac{1}{2} \times 65 \times x}=\frac{35 \times 35}{65 \times 65}
\]
\[\Rightarrow \frac{56-x}{x}=\frac{35}{65}
\]
\(\displaystyle \Rightarrow \mathrm{x}=36.4 \mathrm{~cm}\)
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CBSE Class 10 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.