What is the ratio of the area of a semi-circle to the area of an
inscribed square? (Picture a half circle enclosing as big a square as
can fit inside it, i.e., one side of the square shares the straight
line of the half-circle shape while the two opposite corners of the
square just touch the arc of the half circle.) (The method is more
important than the answer.)

We know from basic geometry that the area of a semicircle is
calculated by (pi)r^2/2 and that the area of a square is determined by
squaring the length of its side.

By inscribing a square within a semi circle, drawing r from the center
of the semi circle to each of the corners of the square to comprise
two right triangles within the square, and using the Pythagorean
Theorem, we learn that r^2 = a^2 + b^2, where b is the length of a
side of the square, r is the radius of the circle, and a = b/2.  Each
a, b, and r comprise a right triangle.

Therefore, b^2 = r^2 - a^2.  Replacing a with b/2 yields b^2 = r^2 -
b^2/4 or r^2 = 5b^2/4.

The desired ratio of the area of the semi-circle to that of the square
= (pi)r^2/2 * 1/b^2.  Substituting 5b^2/4 for r^2 yields (pi)5b^2/8b^2
or 5/8(pi).

Sincerely,

Wonko