Question

A rectangular plank $\sqrt{2}$ m wide is placed symmetrically on the diagonal of a square of side $8$ metres. What is the area of the plank?

• A. $(16\sqrt{2}-3)$ sq. m
• B. $7\sqrt{2}$ sq. m
• C. $98$ sq. m
• D. $14$ sq. m

Answer 1

The correct option is D $14$ sq. m

Let $AF=AE=a$

$\Rightarrow A{F}^2+A{E}^2=E{F}^2$ ...( By Pythagoras theorem)

$\Rightarrow a^2+a^2={\left(\sqrt{2}\right)}^2$

$\Rightarrow 2a^2=2$

$\Rightarrow a=1$

$\therefore AF=CG=1M$

$\Rightarrow FB=BG=8m-1m=7m$

In $\Delta FGB,$

$F{G}^2=F{B}^2+B{G}^2$ ...( Pythagoras Theorem)

$\Rightarrow F{G}^2=7^2+7^2=98$

$\Rightarrow FG=7\sqrt{2}$

$\therefore$ Area of the plank $=FG\times FE$

$=(7\sqrt{2}\times \sqrt{2})m^2=14m^2$