Question

In the figure below. $PQR$ is a right-angle triangle right angled at $Q$. $XY$ is parallel to $QR$. $PQ=6$ cm, $PY=4$ cm and $PX:XQ=1:2$. Calculate the lengths of $PR$ and $QR$.

• A. $PR=12$ cm; $QR=10.392$ cm
• B. $PR=13$ cm; $QR=11.392$ cm
• C. $PR=11$ cm; $QR=12.392$ cm
• D. none of the above

Answer 1

The correct option is A $PR=12$ cm; $QR=10.392$ cm

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

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

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

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

In $△PQR$ and $△PXY$, we have

$\angle XPY=\angle QPR$ (Common angle)

$\angle PXY=\angle PQR={90}^{\circ }$ ($XY\parallel QR$)

$\angle XYP=\angle QRP$ (third angle)

Hence, $△PQR\sim △PXY$ (AAA rule)

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

$\Rightarrow \frac{PR}{PY}=3$

Therefore, $PR=3\times PR=3\times 4=12$ cm

Now, In $△PQR$, we have

$P{Q}^2+Q{R}^2=P{R}^2$ (Pythagoras theorem)

$\Rightarrow 6^2+Q{R}^2={12}^2$

$\Rightarrow Q{R}^2=144-36$

$\Rightarrow QR=\sqrt{108}$

$\Rightarrow QR=10.392$ cm