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1. 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 :

A. $7 \sqrt{2}$ sq. m

B. 14 sq. m

C. 98 sq. m

D. ( $16 \sqrt{2}$ - 3) sq. m

Solution:

\begin{aligned} Let A P = A Q = x metres \\ Then, \\ \Rightarrow x^{2} + x^{2} = {(\sqrt{2})}^{2} \\ \Rightarrow 2 x^{2} = 2 \\ \Rightarrow x^{2} = 1 \\ \Rightarrow x = 1 m \\ S o, △ P A Q is isosceles . \\ ∴ P T = Q T = (\frac{\sqrt{2}}{2}) m = (\frac{1}{\sqrt{2}}) m \\ In △ P T A, we have : ∠ P T A = 90^{\circ} \\ ∴ A T^{2} = A P^{2} - P T^{2} \\ = 1^{2} - {(\frac{1}{\sqrt{2}})}^{2} \\ = 1 - \frac{1}{2} = \frac{1}{2} \\ O r, A T = (\frac{1}{\sqrt{2}}) m \\ Similarly, \\ C X = (\frac{1}{\sqrt{2}}) m \\ ∴ P S = Q R = X T \\ = A C - 2 \times A T \\ = [8 \sqrt{2} - (2 \times \frac{1}{\sqrt{2}})] m \\ = (8 \sqrt{2} - \frac{2}{\sqrt{2}}) m \\ = \frac{14}{\sqrt{2}} m \\ Area of the plank : \\ = (\frac{14}{\sqrt{2}} \times \sqrt{2}) m^{2} \\ = 14 m^{2} \end{aligned}

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