Assume a circle of diameter $d$. Inscribe a square $A$ centred in the circle with its diagonal equal to the diameter of the circle. Now escribe a square $B$ with the sides equal to the diameter of the circle. Show how to obtain the ratio of the area of square $A$ to the area of square $B$.
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This can be done by a computation. The outer square $B$ has area $d^2$. Let the side of the inner square $A$ be $s$. Then by the Pythagorean theorem, $s^2+s^2=d^2$. But $s^2$ is the area if the inner square, and we are finished.
But there is a neater way! Rotate the inner square $A$ about the centre of the circle, until the corners of the inner square are the midpoints of the sides of the outer square. (Sorry that I cannot draw a picture: I hope these words are enough for you to do it.)
Now draw the two diagonals of the inner square. As a check on the correctness of your picture, the diagonals of the inner square are parallel to the sides of the outer square. We have divided the outer square into $8$ congruent isosceles right triangles. And the inner square is made up of $8$ of these triangles. So the outer square $B$ has twice the area of the inner square.