Q) 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 ∼ △CBR given that PQ is perpendicular to AB.
(iii) Find the value of x using similarity of triangles.
(iv) If Area △ PQR/ Area △ CBR = PQ2/ CB2, then find the value of x.

(Q 38 – 30/4/2 – CBSE 2026 Question Paper)

Ans:

(i) prove that θ = α:

In the diagram, we have TR Ʇ AB

∴ ∠ TRA = ∠ TRB = 90 ⁰

Since it is given that ∠ PRT = ∠ CRT

Since, ∠ PRT = 90 ⁰ – α and ∠ CRT = 90 ⁰ – θ

∴ (90 ⁰ – α) = (90 ⁰ – θ)

∴ – α = – θ

∴ θ = α

Hence Proved!

(ii) Prove that △ PQR ∼ △CBR:

Let’s compare △ PQR and △CBR:

∠ PQR = ∠ CBR = 90 ⁰
(Given that PQ is Ʇ AB and square board’s corners are 90)

∠ PRQ = ∠ CRB    [∵ θ = α from part (i)]

∴ By AA Similarity criterion,

△ PQR ∼ △CBR

Hence Proved !

(iii) Value of x:

Since △ PQR ∼ △CBR

∴

Since ABCD is a square, ∴ AB = CB = 65 cm

∵ BR = x

∴ AR = 65 – x

∴ QR = AR – AQ = AR – PS = (65 – x) – 9 = (56 – x) cm

∵

∴

∴ 7 x = 13 (56 – x)

∴ 7 x = 728 – 13 x

∴ 20 x = 728

∴ x = = 36.4 cm

Therefore the value of x is 36.4 cm

(iv) Value of x if Area △ PQR/ Area △ CBR = PQ2/ CB2:

Since △ PQR and △ CBR are right angled triangles,

Area △ PQR = x PQ x QR

Area △ CBR = x CB x BR

Given that Area △ PQR / Area △ CBR =

∴

∴

∴

∴

∴

∴ 7 x = 13 (56 – x)

∴ 7 x = 728 – 13 x

∴ 20 x = 728

∴ x = = 36.4 cm

Therefore the value of x is 36.4 cm

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