Consider the construction below (it is self-evident, no tricks or misleading drawing)
What is the area enclosed by the curve that goes through the points H,J,M,L ?
I solved the problem using integration after rotating everything by $\pi/4$ and scaling length by a factor $\sqrt{2}$:
here's what I did
After scaling back areas by a factor $\frac{1}{2}$, the area is $$\frac{1}{2} \int_{-\sqrt{7}/4}^{\sqrt{7}/4} \sqrt{\frac{1}{2}-x^2}-\sqrt{2-x^2}+1 dx=\frac{1}{8} \left( \sqrt{7}-8sin^{-1}(\frac{\sqrt{7}}{4\sqrt{2}})+2tan^{-1}(\sqrt{7}) \right) \simeq 0.146$$
Now, is there a proof that would not resort to a heavy use of calculus ?