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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