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Carom cm. It board has 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. - Mathematics

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Question

Carom cm. It board has 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.   [1]

(ii) Prove that ΔPQR ~ ΔCBR given that PQ is perpendicular to АВ.   [1]

(ii) (a) Find the value of x using similarity of triangles.   [2]

OR

(b) If `(Area  ΔPQR)/(Area  ΔCBR) = (PQ^2)/(CB^2)`, then find the value of x.

Case Study
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Solution

(i) In the diagram, let line RT be the normal to the boundary AB at point R. 

By the law of reflection: ∠PRT = ∠CRT.

The line RT is perpendicular to the side of the carom board, so ∠QRT = ∠BRT = 90°.

Now, ∠PRQ = α = 90° – ∠PRT.

And ∠CRB = θ = 90° – ∠CRT.

Since ∠PRT = ∠CRT, their complements must also be equal:

90° – ∠PRT = 90° – ∠CRT

α = θ

Hence, θ = α.

(ii) In ΔPQR and ΔCBR

∠PQR = ∠CBR = 90° (Given PQ ⊥ AB and the corner of the square carom board is 90°).

 ∠PRQ = ∠CRB (Proved in part (i) as α = θ).

Therefore, by AA Similarity Criterion: Two triangles are similar if two of their corresponding angles are equal.

ΔPQR ∼ ΔCBR

Hence proved.

(iii) (a) Side of square board = 65 cm. Thus, CB = 65.

Given PQ = 35.

 From diagram, S is on AD and PQ is perpendicular to AB. PS = 9 cm represents the distance of Q from corner A.

So, AQ = 9.

Since AB = 65, the length QB = 65 – 9 = 56.

We are given BR = x. Since R is on the segment QB, QR = QB – BR = 56 – x.

 From ΔPQR ∼ ΔCBR

`(PQ)/(CB) = (QR)/(BR)`

`35/65 = (56 - x)/x`

`7/13 = (56 - x)/x`

7x = 13(56 – x)

7x = 728 – 13x

20x = 728

`x = 728/20`

x = 36.4 cm

The value of x is 36.4 cm.

(iii) (b) ΔPQR ∼ ΔCBR

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

`(PQ)/(CB) = (QR)/(BR)`

`35/65 = (56 - x)/x`

⇒ x = 36.4 cm

The value of x is 36.4 cm.

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2025-2026 (March) Standard - 30/4/2
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