Question

In $△$XYZ, P is any point on XY and PQ $\perp$ XZ. If XP $=4$ cm, XY $=16$ cm and XZ $=24$ cm, find XQ.

Answer 1

In $△XQP$ and $△XYZ,$

$\angle XQP=\angle XYZ={90}^{\circ }$

$\angle QXP=\angle YXZ$ (Common angle)

$\therefore △XQP\sim △XYZ$

$\frac{XP}{XZ}=\frac{XQ}{XY}$ (By properties of similar triangle)

$\Rightarrow \frac{4cm}{24cm}=\frac{XQ}{16cm}$

$\Rightarrow XQ=\frac{16}{6}$ cm

$\Rightarrow XQ=\frac{8}{3}$ cm