✓ Board-verified
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 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) \(\displaystyle \mathrm{TR} \perp \mathrm{AB}\)
\[\therefore \alpha+\angle \mathrm{PRT}=\theta+\angle \mathrm{TRC}
\]
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) \(\displaystyle \triangle \mathrm{PQR} \sim \triangle \mathrm{CBR}\)
\[\therefore \frac{P Q}{\mathrm{CB}}=\frac{Q R}{\mathrm{BR}}
\]
\[\Rightarrow \frac{35}{65}=\frac{65-9-x}{x}
\]
\[\Rightarrow 35 x=65(56-x)
\]
\[\Rightarrow x=36.4 \mathrm{~cm}
\]
All Introduction to Trigonometry questions2026 question paperNCERT solutions: Introduction to TrigonometryPractice Introduction to Trigonometry →
More from Introduction to Trigonometry
- Prove that: (1+ cot θ-cosec θ)(1+ tan θ+ sec θ)=22026
- Evaluate: (5 cos^2 60^°+4 sec^2 30^°- tan^2 45^°)/(sin^2 30^°+ cos^2 30^°) OR Prove that: 1+(cot^2 α)/(1+cosec α)=cosec α2026
- [5/8 sec^2 60^°- tan^2 60^°+ cos^2 45^°] is equal to2023
- If sin θ+ cos θ=√3, then prove that tan θ+ cot θ=1 OR Prove that: ( sin A+ sec A)^2+( cos A+cosec A)^2=(1+ sec A cosec A)^22026
- The hour-hand of a clock is 6 cm long. The angle swept by it between 7:20 a.m. and 7:55 a.m. is:2023
- If x=a cos θ and y=b sin θ, then the value of b^2 x^2+a^2 y^2 is:2024
- Assertion: For an acute angle θ, value of cosec θ cannot be 1/(√2). Reason (R): cosec θ ≥ 1 for 0^° ≤ θ ≤ 90^°2025
- Prove that (cosec A- sin A)( sec A- cos A)=1/(cot A+ tan A).2023
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.