All in the square

Another Catriona Shearer puzzle.
All in the Square

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.

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