# All in the square

Another Catriona Shearer puzzle.
11.
All in the Square

Spoiler!
This is cute. Never mind the big diagram, just consider a circle inscribed in a right triangle with hypotenuse $h$:

With the vertices and contact points joined to the incenter, you can see the triangle is divided into an $r\times r$ square and four triangles, two of which can be rearranged: to make an $h\times r$ rectangle. So the triangle’s area equals $r \times (r+h)$. Solving, $r = (\sqrt{h^2+4A}-h)/2$ and with $h=10$, $A=24$, we get $r=2$; the circle’s area is $4\pi$.