Bunuel wrote:
In the figure above, a circle is inscribed in a square with side b and a square with side a is inscribed in the circle. What is the area oF the square with side b ?
(1) a = 4
(2) The radius of the circle is \(2\sqrt{2}\)
Notes:
Diagonal of a square = \(Side*\sqrt{2}\)
Radius of a circle = \(\frac{Diameter}{2}\)
(1) a = 4. So, Diagonal of small square = Diameter of the circle = Side of large square = \(b = 4\sqrt{2}\). Area of large square is: \(b^2=(4\sqrt{2})^2=(16)(2)=32\), sufficient.
(2) The radius of the circle is \(2\sqrt{2}\). Then, Diameter of circle = \(2*Radius\) = \(2*(2\sqrt{2})\) = Side of large square = \(b = 4\sqrt{2}\). Area of large square is: \(b^2=(4\sqrt{2})^2=(16)(2)=32\), sufficient.
(D) is the answer.