Question : ABCD is a square inscribed in a circle of radius $r$. Then the total area (in square units) of the portions of the circle lying outside the square is:
Option 1: $\pi (r^{2}-4)$
Option 2: $2\pi (r ^{2}-1)$
Option 3: $\pi^{2} r(r-7)$
Option 4: $r^{2}(\pi -2)$
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Correct Answer: $r^{2}(\pi -2)$
Solution :
Let the radius of the circle be $r$ units.
Area of circle = $\pi r^2$
Diagonal of ABCD = BD = $2r$ units
Area of square = $\frac{1}{2}\times \text{(BD)}^2$
= $\frac{1}{2}\times 4r^2$
= $2r^2$
Required difference = Area of the circle – Area of the square
= $\pi r^2 - 2r^2$
= $r^2 (\pi- 2)$ sq. units
Hence, the correct answer is $r^{2}(\pi -2)$.
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