# Riddler Express, November 15 2019

Thought I’d see if I can get back into the groove of doing these, at least sometimes. This week‘s is geometry. I admit I got a bit of fear when I looked at it; geometry is something I haven’t used much since school, so it doesn’t come back so easily.

Anyway, we have to calculate the area of the darkly-shaded area, the smaller semicircle, given that the lightly-shaded areas have a total area of seven:

This only requires us to make a right-angled triangle:

The lines BE and DE are both radiuses of the small semi-circle. The line BD is a radius of the larger semi-circle. From Pythagoras we can see that it is √2 times as long as the the radius of the small semi-circle.

Semi-circles, like circles, have areas that vary with the square of their radius. So the larger semi-circle has an area that is √2 ×√2 as large as the smaller semi-circle. That is, it’s twice as large. So the smaller semi-circle is exactly half the size of the larger one, and the lightly-shaded area is the same size as the darkly-shaded area: seven units.