Question

PQR is a right-angle triangle right angled at Q. XY is parallel to QR. PQ = 6 cm and PX : XQ = 1: 2. Calculate the lengths of PR and QR.

Answer 1

Given,

$\frac{PX}{XQ}=\frac{1}{2}$

$\frac{XQ}{PX}=2$

$\frac{PX+XQ}{PX}=2+1$

$\frac{PQ}{PX}=3$

In $△PQR$ and $△PXY$,
$\angle XPY=\angle QPR$ (Common angle)
$\angle PXY=\angle PQR={90}^{\circ }$ (XY II QR)
$\angle XYP=\angle QRP$ (third angle)
hence, $△PQR\sim △PXY$ (AAA rule)

Hence, $\frac{PQ}{PX}=\frac{QR}{XY}=\frac{PR}{PY}$ (Corresponding sides)
$\frac{PR}{PY}=3$

therefore, $PR=3\times PR=3\times 4=12cm$
Now, In $△PQR$,
$P{Q}^2+Q{R}^2=P{R}^2$ (Pythagoras theorem)
$6^2+Q{R}^2={12}^2$
$Q{R}^2=144-36$
$QR=\sqrt{108}$
$QR=10.392cm$