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Mathematics · 2026 · 4 marks
CBSE 2026 · Region 4 · Set 2 · Q38
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)Find the value of $\displaystyle x$ using similarity of triangles.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 $\displaystyle x$. $\displaystyle 1171$-$\displaystyle 2$ { } of $\displaystyle 24$
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)
Find the value of $\displaystyle x$ using similarity of triangles.
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 $\displaystyle x$. $\displaystyle 1171$-$\displaystyle 2$ { } of $\displaystyle 24$
Marking-scheme solution
(i) \(\displaystyle \mathrm{TR} \perp \mathrm{AB}\)
\[\therefore \alpha+\angle \mathrm{PRT}=\theta+\angle \mathrm{TRC}
\]
As \(\displaystyle \angle \mathrm{PRT}=\angle \mathrm{TRC}\), so \(\displaystyle \alpha=\theta\)
(ii) As \(\displaystyle \theta=\alpha\), so \(\displaystyle \angle \mathrm{PRQ}=\angle \mathrm{CRB}\)
\[\text { and } \angle \mathrm{PQR}=\angle \mathrm{CBR}=90^{\circ}
\]
\[\therefore \triangle \mathrm{PQR} \sim \triangle \mathrm{CBR}
\]
(iii) (a) \(\displaystyle \triangle \mathrm{PQR} \sim \triangle \mathrm{CBR}\)
\[\therefore \frac{P Q}{\mathrm{CB}}=\frac{Q R}{\mathrm{BR}}
\]
\[\Rightarrow \frac{35}{65}=\frac{65-9-x}{x}
\]
\[\Rightarrow 35 x=65(56-x)
\]
\[\Rightarrow x=36.4 \mathrm{~cm}
\]
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CBSE Class 10 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.