In the figure below, PQRS is a square. If RS = 4 units, then the area of the shaded portion (in sq. units) is:

2 (\sqrt{2 π} - 1)

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2 \sqrt{2} (π - 1)

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4 (\sqrt{2} π - 1)

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(2 π - 4)

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Solution

The correct option is D $(2 π - 4)$
$Δ Q S R$ is a right-angled isosceles triangle.
RS = 4 units
So, $Q S = 4 \sqrt{2}$ units, $O S = 2 \sqrt{2}$
Hence, shaded area $= \frac{1}{2} (\frac{π (2 \sqrt{2})^{2}}{2} - 8) \Rightarrow (2 π - 4) s q . u n i t$

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