Problem: square(ABCD) is a regular square, and a circle touches internally in the square. Also, arc(BD) divides the square. Then calculate the area of the colored region.
This question is easily solved when you use analytical geometry. But I can't solve it by only using axiomatic geometry. How to solve this by not using analytical geometry?
I'm referring to the following figure:
The area in question can be viewed as the formal sum of two circular sectors minus two triangles.
Note that $$b={a\over2}, \quad c={a\over\sqrt{2}}\ ,$$ and the cosine theorem allows to compute the angles $\alpha$ and $\beta$: $$\cos\alpha={b^2+c^2-a^2\over 2 bc}=-{\sqrt{2}\over4},\qquad \cos\beta={a^2+c^2-b^2\over 2ac}={5\sqrt{2}\over 8}\ .$$ The areas of the two circular sectors are then given by $$A_\alpha={b^2\over2}\cdot2\alpha={\alpha\over4}a^2,\qquad A_\beta={a^2\over2}\cdot 2\beta=\beta a^2\,$$ and the area of one triangle is given by $$A_\triangle={1\over2}ac\>\sin\beta={\sqrt{7}\over16}a^2\ .$$ Therefore the shaded area comes to $$A=A_\alpha+A_\beta-2A_\triangle=\left({1\over4}\arccos{-\sqrt{2}\over4}+\arccos{5\sqrt{2}\over8}-{\sqrt{7}\over8}\right)a^2\doteq0.639\>a^2\ .$$