Question

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, $BR = x$ , $∠ PRQ = α$ and $∠ CRB = θ$ .

Based on the above information, answer the following questions :

(i) Using law of reflection i.e. $∠ PRT = ∠ CRT$ , prove that $θ = α$ .

(ii) Prove that $△ PQR \sim △ CBR$ given that PQ is perpendicular to AB.

(iii) (a) Find the value of x using similarity of triangles.

(b) If $\frac{Area △ PQR}{Area △ CBR} = \frac{P Q ^{2}}{C B ^{2}}$ , then find the value of x.

Approach

Carom board with striker at P hits disc at R to reach pocket C. Given PS = 9 cm, PQ = 35 cm, BR = $x$ , angle QPR = angle CRB = $α$ . (i) By law of reflection, $α = θ$ . (ii) Triangles PQR and CBR are similar. (iii)(a) Using similarity: $\frac{35}{65} = \frac{56 - x}{x}$ gives $x = 36.4$ cm. OR (iii)(b) Using area ratio: $\frac{Area △ PQR}{Area △ CBR} = \frac{3 5 ^{2}}{6 5 ^{2}}$ gives $x = 36.4$ cm.

Step-by-step working

1
(i) Since TR ⊥ AB, angle of incidence = angle of reflection. Thus $α = θ$ .
1 mark
2
(ii) In triangles PQR and CBR: $∠ PRQ = ∠ CRB$ (vertical angles), $∠ PQR = ∠ CBR = 90°$ . Thus $△ PQR \sim △ CBR$ .
1 mark
3
(iii)(a) From similarity: $\frac{PQ}{CB} = \frac{QR}{BR} \Rightarrow \frac{35}{65} = \frac{65 - 9 - x}{x} = \frac{56 - x}{x}$ .
0.5 marks
4
Solve: $35 x = 65 (56 - x) \Rightarrow 100 x = 3640 \Rightarrow x = 36.4$ cm.
1 mark
5
Answer (iii)(a): $x = 36.4$ cm.
0.5 marks
6
OR (iii)(b) Area ratio: $\frac{\frac{1}{2} \times 35 \times ( 56 - x )}{\frac{1}{2} \times 65 \times x} = \frac{3 5 ^{2}}{6 5 ^{2}} \Rightarrow \frac{56 - x}{x} = \frac{35}{65}$ . Same equation, $x = 36.4$ cm.
1 mark

Concepts used

similar triangleslaw of reflectionproveratio of areasfind value of x

Approach

Step-by-step working

Concepts used

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