Question

$\Delta PQR$ is a right triangle with $PR$ as the hypotenuse. If $PQ=4m,QR=3mandQS\perp PR$, find the area of the triangle $PQR$. Also find the length of $QS$.

• A. $6m^2,2.4m$
• B. $6m^2,2.5m$
• C. $5m^2,2.4m$
• D. $6m^2,1.4m$

Answer 1

Answer: (A)

Given: In $△PQR,PQ=4m,QR=3m$ and $\angle PQR={90}^{\circ }$

Therefore, using, Pythagoras Theorem,

$\Rightarrow P{Q}^2+Q{R}^2=P{R}^2$

$\Rightarrow 4^2+3^2=P{R}^2$

$\Rightarrow 16+9=P{R}^2$

$\Rightarrow 25=P{R}^2$

$\Rightarrow 5^2=P{R}^2$

$\Rightarrow PR=5m$

Now, area of $△PQR=\frac{1}{2}\times PQ\times QR=\frac{1}{2}\times 4\times 3=6m^2$...(1)

Also, area of $△PQR=\frac{1}{2}\times QS\times PR$ [as $QS$ is also height/altitude of the triangle]..(2)

From (1) and (2), we can say that,

$\Rightarrow \frac{1}{2}\times QS\times PR=6$

$\Rightarrow \frac{1}{2}\times QS\times 5=6$

$\Rightarrow QS=\frac{6\times 2}{5}=2.4m$

Hence, option A. $6m^2,2.4m$ is correct.