Question

$P(2,1),Q(2,5)$ and $R(k,5)$ are the three vertices of $△PQR$ in which $\angle Q={90}^o$. If the area of $△PQR$ is $10{unit}^2$, the value of $k$ is:

• A. $5$
• B. $6$
• C. $7$
• D. $8$

Answer 1

The correct option is C $7$

$\frac{1}{2}PQ\times QR=10$

$\sqrt{(2-2{)}^2+(5-1{)}^2}=PQ$

$QR=\sqrt{(k-2{)}^2+(5-5{)}^2}$

$\Rightarrow \sqrt{4^2}\times \sqrt{(k-2{)}^2}=20$

$\Rightarrow (k-2)=5\Rightarrow k=7$