Correct option is (c) 2 : 5
Let ABCD be the square inscribed in the semi-circle with center O, and side CD the diameter of the semi-circle.
Let the point M be the mid-point of AB.
Now, join OB and OM to get the right △OMB with OB as the hypotenuse.
Therefore, OM2 + MB2 = OB2
OM = side of the square inscribed in the semi-circle,
MB = half of the side; and
OB = radius
Let the side of the square be x.
\(x^2 + (\frac x2)^2 = r^2\)
\(x^2 = \frac{4r^2}5\)
Hence, the area of the semi-circle = \(\frac{4r^2}5\)
Now diagonal of the square inscribed in the circle =2r
Therefore, its area = \(\frac{(2r)^2}{2} = 2r^2\)
Area of the square = \(\frac{\text{diagonal}^2}2\)
Hence, the required ratio = \(\frac{4r^2}5 : 2r^2\)
\(= \frac 25 : 1\)
\(= 2:5\)