Question

P is the vertex of cuboid and Q, R and S are three points on the adjacent edges passing through p as shown. PQ = PR = 2 cm and PS = 1 cm. Then the area of $\Delta QRS\left(inc{m}^2\right)$ is:

• A. $\frac{\sqrt{15}}{4}$
• B. $\frac{5}{2}$
• C. $\sqrt{6}$
• D. $2\sqrt{2}$

Answer 1

The correct option is C $\sqrt{6}$

Since $\Delta PQS$ is a right angled $\Delta$
$\therefore Q{S}^2=P{Q}^2+P{S}^2$
$\Rightarrow QS=\sqrt{(2{)}^2+(1{)}^2}=\sqrt{5}$
In the same way $QR=\sqrt{8},RS=\sqrt{5}$

$S=\frac{\sqrt{5}+\sqrt{5}+\sqrt{8}}{2}$
$=\frac{2\sqrt{5}+2\sqrt{2}}{2}=\sqrt{5}+\sqrt{2}$
$\therefore$ By using Heron’s Formula
$A=\sqrt{(\sqrt{5}+\sqrt{2})(\sqrt{5}+\sqrt{2}-\sqrt{5})(\sqrt{5}+\sqrt{2}-\sqrt{5})(\sqrt{5}+\sqrt{2}-2\sqrt{2})}$
$=\sqrt{(\sqrt{5}+\sqrt{2})\left(\sqrt{2}\right)\left(\sqrt{2}\right)(\sqrt{5}-\sqrt{2})}$
$=\sqrt{2\times (5-2)}=\sqrt{6}$