Question

Let $PQR$ be a triangle in which $PQ=3$. From the vertex $R$, draw the altitude $RS$ to meet $PQ$ at $S$. Assume that $RS=\sqrt{3}$ and $PS=QR$. Then $PR$ equals

• A. $\sqrt{5}$
• B. $\sqrt{6}$
• C. $\sqrt{7}$
• D. $\sqrt{8}$

Answer 1

The correct option is C $\sqrt{7}$

Given :
$RS=\sqrt{3}$, $PS=QR$ and $PS+SQ=3$

Let $PS=x=RQ$
$\therefore SQ=3-x$
$\text{In}△RSQ\Rightarrow R{S}^2+S{Q}^2=R{Q}^2\Rightarrow 3+(3-x{)}^2=x^2\Rightarrow 3+9+x^2-6x=x^2\Rightarrow x=2$

Now,
$PR=\sqrt{(x{)}^2+(\sqrt{3}{)}^2}=\sqrt{7}$