Last Funky Circle Area Problem

I had to cheat on this one a little, ACG....I wasn't sure about how to  construct a square in a semi-circle without a little online "help"....LOL!!!!...oh, well..."any port in a storm "

Look at the following :

Without a loss of generality, let us construct  rectangle ABEF where AB  = 2  and AF  = 4.....the area of this is 8....so by symmetry, the square ABCD has an area of 4.....

Next....draw diagonal BF and construct a circle with G as a center and GB as a radius.....then square  ABCD will be the largest square that we can inscribe  in the upper half of the circle....and it will have an area of AB^2  = 2^2  = 4

Because GC  = (1/2 )AB  = 1   and BC  = AB  = 2, then, by the Pythagorean Theorem,  the radius of the circle, GB, is a hypotenuse of a right triangle and  will be equal to  √[ GC^2 + BC^2]  = √[ 1^2 + 2^2]  = √5

So  we can construct square HIJK  where  H  = (√5/√2, √5/√2)  and  I  = (- √5/√2, √5/√2)

So....HI   will be the side of the square   and its length is  just  √  ( √5/√2 -  -√5/√2)^2  =

(√5/√2 + √5/√2)   =  2√5/√2  = (2/√2)*√5 =   √2*√5  =  √10

So the area of this square  is just  HI^2  = (√10)^2  = 10

So.....since any construction will yield the same ratio....the ratio of the area of a square inscribed in a semicircle with radius r to the area of a square inscribed in a circle with radius r =   4/10   =  2/5

The area of the small square ABCD = 2 x 2 =4
Hypotenuse of the small triangle BCG=Radius of the semi-circle
H=sqrt(5) = The radius of the semi-circle
The diagonal of the large square IHJK =2sqrt(5)
Sin(45) x 2sqrt(5) =sqrt(10) = One side of large square IHJK
So, the ratio of the area of ABCD / area of IHJK =4 / sqrt(10)^2 =2/5

Thanks everyone!