Question

In the given figure, in a triangle PQR, ST $||$ QR and $\frac{PS}{SQ}=\frac{3}{5}$ and $PR=28$cm, find PT.

Answer 1

$ST\parallel QR$ [ Given ]

$\therefore$ $\frac{PS}{PQ}=\frac{PT}{PR}$ [ By BPT ]

$\Rightarrow$ $\frac{PS}{SQ}=\frac{3}{5}$ and $PR=28cm$ [ Given ]

$\Rightarrow$ $\frac{SQ}{PS}=\frac{5}{3}$

$\Rightarrow$ $\frac{SQ+PS}{PS}=\frac{5+3}{3}$

$\therefore$ $\frac{PQ}{PS}=\frac{8}{3}$

$\Rightarrow$ $\frac{PS}{PQ}=\frac{3}{8}$

$\Rightarrow$ $\frac{PS}{PQ}=\frac{PT}{PR}$

$\Rightarrow$ $\frac{3}{8}=\frac{PT}{28}$

$\Rightarrow$ $PT=\frac{28\times 3}{8}$

$\therefore$ $PT=10.5cm$