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