Let ABCD be the square inscribed in the semi-circle with center 0 , 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.
x2+()2=r2 x2= Hence, the area of the semi-circle
= Now diagonal of the square inscribed in the circle
=2r Therefore, its area
==2r2 Area of the square
= Hence, the required ratio
=:2r2=:1=2:5