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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, $BR = x$, $\angle PRQ = \alpha$ and $\angle CRB = \theta$.
Based on the above information, answer the following questions:
(i) Using law of reflection i.e. $\angle PRT = \angle CRT$, prove that $\theta = \alpha$.
(ii) Prove that $\triangle PQR \sim \triangle CBR$ given that $PQ$ is perpendicular to $AB$.
(iii) (a) Find the value of $x$ using similarity of triangles.
OR
(b) If $\frac{\text{Area } \triangle PQR}{\text{Area } \triangle CBR} = \frac{PQ^2}{CB^2}$, then find the value of $x$.
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, $BR = x$, $\angle PRQ = \alpha$ and $\angle CRB = \theta$.
Based on the above information, answer the following questions:
(i) Using law of reflection i.e. $\angle PRT = \angle CRT$, prove that $\theta = \alpha$.
(ii) Prove that $\triangle PQR \sim \triangle CBR$ given that $PQ$ is perpendicular to $AB$.
(iii) (a) Find the value of $x$ using similarity of triangles.
OR
(b) If $\frac{\text{Area } \triangle PQR}{\text{Area } \triangle CBR} = \frac{PQ^2}{CB^2}$, then find the value of $x$.
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Sol. (i) $TR \perp AB$
$\therefore \alpha + \angle PRT = \theta + \angle TRC$
As $\angle PRT = \angle TRC$, so $\alpha = \theta$ (I) (1 Mark)
(ii) As $\theta = \alpha$, so $\angle PRQ = \angle CRB$
and $\angle PQR = \angle CBR = 90^\circ$
$\therefore \triangle PQR \sim \triangle CBR$ (I) (1 Mark)
(iii) (a) $\triangle PQR \sim \triangle CBR$
$\therefore \frac{PQ}{CB} = \frac{QR}{BR}$ (I) ($\frac{1}{2}$ Mark)
$\Rightarrow \frac{35}{65} = \frac{65-9-x}{x}$ (II) (1 Mark)
$\Rightarrow 35x = 65 (56 - x)$
$\Rightarrow x = 36.4$ cm (III) ($\frac{1}{2}$ Mark)
OR
(b) $\frac{\text{Ar. } \triangle PQR}{\text{Ar. } \triangle CBR} = \frac{PQ^2}{BC^2}$ (I) (1 Mark)
$\Rightarrow \frac{\frac{1}{2} \times 35 \times (65-x-9)}{\frac{1}{2} \times 65 \times x} = \frac{35 \times 35}{65 \times 65}$ (II) ($\frac{1}{2}$ Mark)
$\Rightarrow \frac{56-x}{x} = \frac{35}{65}$
$\Rightarrow x = 36.4$ cm (III) ($\frac{1}{2}$ Mark)
$\therefore \alpha + \angle PRT = \theta + \angle TRC$
As $\angle PRT = \angle TRC$, so $\alpha = \theta$ (I) (1 Mark)
(ii) As $\theta = \alpha$, so $\angle PRQ = \angle CRB$
and $\angle PQR = \angle CBR = 90^\circ$
$\therefore \triangle PQR \sim \triangle CBR$ (I) (1 Mark)
(iii) (a) $\triangle PQR \sim \triangle CBR$
$\therefore \frac{PQ}{CB} = \frac{QR}{BR}$ (I) ($\frac{1}{2}$ Mark)
$\Rightarrow \frac{35}{65} = \frac{65-9-x}{x}$ (II) (1 Mark)
$\Rightarrow 35x = 65 (56 - x)$
$\Rightarrow x = 36.4$ cm (III) ($\frac{1}{2}$ Mark)
OR
(b) $\frac{\text{Ar. } \triangle PQR}{\text{Ar. } \triangle CBR} = \frac{PQ^2}{BC^2}$ (I) (1 Mark)
$\Rightarrow \frac{\frac{1}{2} \times 35 \times (65-x-9)}{\frac{1}{2} \times 65 \times x} = \frac{35 \times 35}{65 \times 65}$ (II) ($\frac{1}{2}$ Mark)
$\Rightarrow \frac{56-x}{x} = \frac{35}{65}$
$\Rightarrow x = 36.4$ cm (III) ($\frac{1}{2}$ Mark)