SSC CGL 19 Oct 2014 Shift 2 Paper

Question : 142 of 200

Marks: +1, -0

In ΔPQR, S and T are points on sides PR and PQ respectively such that ∠PQR = ∠PST. If PT = 5 cm, PS = 3 cm and TQ = 3 cm, then length of SR is

Solution:

∠ PQR = ∠ PST
∠ P = ∠ P
∴ Δ PST and ΔPQR are similar,
∴

{PQ}/{PS}

{PR}/{PT}

⇒

8/3

{PR}/5

⇒ PR =

{8×5}/3

{40}/3

∴ SR = PR – PS =

{40}/3

- 3
=

{40-9}/3

{31}/3

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