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 $\angle CRB = \beta$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:21 · grounding stimulus
Model Answer
(i) RT is the boundary AB (the line of reflection).
∠PRQ = α (given) and ∠PRQ = ∠PRT (since Q lies on AB, PQ ⊥ AB)
By law of reflection, ∠PRT = ∠CRT = α
Now, ∠CRT + ∠CRB = 180° and ∠PRT + ∠PRQ = 180° — but since ∠CRB = β is vertically opposite to ...
Actually: ∠PRT = α, and ∠CRB = β. Since ∠PRT and ∠CRB are vertically opposite angles, α = β. ∎
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(ii) In △PQR and △CBR:
- ∠PQR = ∠CBR = 90° (PQ ⊥ AB, BC ⊥ AB)
- ∠PRQ = ∠CRB (i.e. α = β, proved above)
By AA similarity criterion, △PQR ~ △CBR. ∎
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(iii) Since △PQR ~ △CBR:
$$\frac{PQ}{CB} = \frac{QR}{BR}$$
Here PQ = 35 cm, CB = 65 cm, BR = x.
QR = AB − QA = AB − PS = 65 − 9 = 56 cm
$$\frac{35}{65} = \frac{56}{x}$$
$$x = \frac{56 \times 65}{35} = \frac{3640}{35} = \mathbf{104 \ cm}$$
Source: Triangles (Similarity of Triangles), CBSE Class 10 Mathematics
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Explanation
- (i) The key is recognising that ∠PRT = ∠CRB as vertically opposite angles, and the law of reflection gives ∠PRT = α, so β = α.
- (ii) Both right angles + equal angles from reflection → AA similarity. Examiners expect you to name the criterion explicitly.
- (iii) QR = 65 − 9 = 56 cm (since QA = PS = 9 cm on the square board). Set up the ratio correctly from the similarity and solve for x. Many students lose marks by using the wrong side-correspondence — always match sides opposite equal angles.