A rectangular plank 2 metre wide is placed symmetrically on the diagonal of a square of side 8 metres as shown in the figure. The area of the plank is :

A rectangular plank $$\sqrt{2}$$ metre wide is placed symmetrically on the diagonal of a square of side 8 metres as shown in the figure. The area of the plank is :

Answer: Option B

Solution

Answer: Option B

Solution:

$$\begin{array}{rl}& \text{ Let }AP=AQ=x\text{ metres}\\ & \text{Then,}\\ & \Rightarrow x^2+x^2={\left(\sqrt{2}\right)}^2\\ & \Rightarrow 2x^2=2\\ & \Rightarrow x^2=1\\ & \Rightarrow x=1m\\ & So,△PAQ\text{ is isosceles}\text{.}\\ & \therefore PT=QT=\left(\frac{\sqrt{2}}{2}\right)m=\left(\frac{1}{\sqrt{2}}\right)m\\ & \text{In }△PTA\text{, we have : }\angle PTA=90^{\circ }\\ & \therefore A{T}^2=A{P}^2-P{T}^2\\ & =1^2-{\left(\frac{1}{\sqrt{2}}\right)}^2\\ & =1-\frac{1}{2}=\frac{1}{2}\\ & Or,AT=\left(\frac{1}{\sqrt{2}}\right)m\\ & \text{Similarly,}\\ & CX=\left(\frac{1}{\sqrt{2}}\right)m\\ & \therefore PS=QR=XT\\ & =AC-2\times AT\\ & =\left[8\sqrt{2}-\left(2\times \frac{1}{\sqrt{2}}\right)\right]m\\ & =\left(8\sqrt{2}-\frac{2}{\sqrt{2}}\right)m\\ & =\frac{14}{\sqrt{2}}m\\ & \text{Area of the plank :}\\ & =\left(\frac{14}{\sqrt{2}}\times \sqrt{2}\right)m^2\\ & =14m^2\end{array}$$