@klaus said in Puzzle time: Geometry:
Here's a solution that doesn't involve any angles. Radius of upper circle is a and of lower circle is b and of the outer circle r.
[image: 1612553719437-57550d93-1bc7-4811-8fe9-72c8b580c5bd-image.png]
The segment marked with "b-a" has length b-a by the intersecting chords theorem.
Now, by Pythagoras, (2a)^+(2b)^2=(2r)^2, or a^2+b^2=r^2.
The sum of the areas of the semi-circles is 1/2 pi a^2 + 1/2 pi b^2, or pi/2 (a^2+b^2).
Substituting in the equation from above, we get that the combined area is
pi/2 r^2.
Since r^2*pi = 1, we get that the area is 1/2.
Yeap, that's the solution that requires Thales theorem for you to establish that the "hypothenuse" has to cut through the middle of the circle and thus has the length 2r. Without this you would not be able to relate "r" to "a" and 'b".