Two side-by-side squares are inscribed in a semicircle as shown above.

Side or larger square = a, side of smaller square = b, OX = c

r^2 = a^2 + (a-c)^2
r^2 = b^2 + (b+c)^2

=> a^2 + (a-c)^2 = b^2 + (b+c)^2
=> 2*a^2 - 2ac = 2*b^2 + 2bc
=> a^2 - b^2 = c(a+b)
=> a-b = c

Substitute in any equation, r^2 = a^2 + (a-(a-b))^2 = a^2 + b^2 => a^2 + b^2 = 100

or,

we can obtain the result in a generalized manner. since no specific ratio has been given with respect to the sides of the squares, let's consider them to be of equal size.

So the common side will be perpendicular at origin.
Let side of each square be x.

r^2 = x^2 + x^2 = 2*x^2

2*x^2 is nothing but the combined area of the squares, which is equal to 100.

Ans: C

Explanation;

let small Square with side X
Large Square with side Y

By using Pythagoras theorem we can deduce to below formula

Area : X^2 + Y^2 = R^2
100

IMO-C

Since the radius is 10. The line segment of diameter = 20

The maximum side of a square is 10. And since other square is smaller. Then its side to the maximum can be 9
Since the maximum of square will never touch the corners.
The side of both square will be less than 9
=> Let Sum of area of square with side 8 & 7 = 64+49 = 113 (Not among option)

=> Let Sum of area of square with side 8 & 6 = 64+36 = 100
The possible biggest area possible in semicircular sphere.

IMO C

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Everyone has already posted a great solution, so I would just add a different approach that I used.

Assume both the squares to be of same height, this condition will combine the them into a rectangle.
Calculating the area of the rectangle is quite trivial, the length and breadth will come out to be 10√2 and 5√2 respectively.
Area = length * breadth = 100

Provide small square with side a; big square with size b;
Two triangles with r as hypotenuses
Prove that the two triangles are the same size: same corners
Then we have a^2+b^2=r^2

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