Question 142
In ∠PQR, 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
Let SR be 'x'.
In $$\triangle{PTS}$$ and $$\triangle{PQR}$$,
$$\angle{PST}=\angle{PQR}$$.
and $$\angle{TPS}=\angle{RPS}$$.
$$\therefore \triangle{PTS}$$ is similar to $$\triangle{PRQ}$$.
$$\therefore \frac{PT}{PR}=\frac{PS}{PQ}$$.
$$\frac{5}{3+x}=\frac{3}{3+5}$$.
$$x+3=\frac{40}{3}$$.
$$x=\frac{31}{3}$$.
Hence, Option C is correct.
Create a FREE account and get:
- Free SSC Study Material - 18000 Questions
- 230+ SSC previous papers with solutions PDF
- 100+ SSC Online Tests for Free
