Subject: General Aptitude · Chapter: Area · Exam: · Marks: · Difficulty:
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 $7 \sqrt{2}$ sq. m
(b)14 sq. m
(c)98 sq. m
(d)$(16 \sqrt{2}-3)$ sq. m
Answer
Answer (as printed): B
Explanation
Let $A P=A Q=x$ metres. Then, $x^{2}+x^{2}=(\sqrt{2})^{2} \Rightarrow 2 x^{2}=2 \Rightarrow x^{2}=1 \Rightarrow x=1 \mathrm{m}$. So, $\triangle P A Q$ is isosceles. $$\therefore P T=Q T=\left(\frac{\sqrt{2}}{2}\right) \mathrm{m}=\left(\frac{1}{\sqrt{2}}\right) \mathrm{m} .$$ In $\triangle P T A$, we have: $\angle P T A=90^{\circ}$. $$\therefore A T^{2}=A P^{2}-P T^{2}=1^{2}-\left(\frac{1}{\sqrt{2}}\right)^{2}=1-\frac{1}{2}=\frac{1}{2}$$ or $A T=\left(\frac{1}{\sqrt{2}}\right) \mathrm{m}$. Similarly, $C X=\left(\frac{1}{\sqrt{2}}\right) \mathrm{m}$.$$\begin{aligned} & \therefore P S=Q R=X T=A C-2 \times A T=\left[8 \sqrt{2}-\left(2 \times \frac{1}{\sqrt{2}}\right)\right] \mathrm{m} \\ & =\left(8 \sqrt{2}-\frac{2}{\sqrt{2}}\right) \mathrm{m}=\frac{14}{\sqrt{2}} \mathrm{m} . \end{aligned}$$ Area of the plank $=\left(\frac{14}{\sqrt{2}} \times \sqrt{2}\right) \mathrm{m}^{2}=14 \mathrm{m}^{2}$.