Cambridge IGCSE Additional Mathematics Challenge Q This is a challenge question from my Cambridge IGCSE Additional Maths textbook. Bear with me on the drawing. The drawing consists of a square, circle, and quarter circle. The only measurement give is that the side length of the square is $10$cm. Can someone help me find the area of the shaded region? I am looking for an explanation of the answer as well.

 A: Hint. Let the $x$-axis run along the diagonal from bottom left to top right. Then the equation of the small circle is $x^2+y^2=5^2$ and that of the big circle $x^2+(y+\sqrt{50})^2=10^2.$ The two intersect at the points $$\left(\pm\frac{5\sqrt 7}{2\sqrt2},\frac{5}{2\sqrt2}\right).$$
Thus the area is given by $$2\int_0^{5\sqrt 7/2\sqrt 2}\left(\sqrt{5^2-x^2}-\sqrt{10^2-x^2}+5\sqrt{2}\right)\mathrm dx.$$
Can you now proceed?

Based on the fact that OP may not know calculus, as hinted at in the comments, I add that the integral evaluates to $$25\left(\alpha-4\beta+\frac{\sqrt 7}{2}\right),$$ where $\cos\alpha=1/2\sqrt 2,\,\cos\beta=5/4\sqrt 2,$ and the acute angles $\alpha,\,\beta$ are in radians.
A: Sketch. Here is an elementary way to obtain the area of the lune. Join the points of intersection of the two arcs, which gives a common chord $C$ for the two involved circles. Thus the area we seek is the difference in area of the segment of the small circle and the big circle, cut off by $C.$ Let these areas respectively be $S$ and $T.$ Then we want $S-T.$ Now to get each of these, we subtract the area of the isosceles triangle defined by radii of the involved circle and $C$ from the area of the sector formed by this triangle and the segment. It follows that we need the length of $C,$ which I'll call $2y,$ and the angles subtended by the given arcs at their respective centres. Let the one for the small circle be $2\phi,$ and the other $2\psi.$ Finally let $x$ be the distance from the centre of the small circle to the line segment $C.$ If you represent all of this information on a diagram, you get a triangle defined by a half-diagonal of the given square, a radius of the small circle, and a radius of the large circle, with sides $5\sqrt2,5$ and $10$ respectively. The angles opposite these sides are an unnamed unknown (not needed to solve the problem), the angle $\psi,$ and the angle $180°-\phi.$ [All angles are measured in degrees.]
Thus applying the cosine rule to this triangle gives us that $$\cos\psi=\frac{5}{4\sqrt 2}.$$ Thus we obtain $$\sin\psi=\frac{\sqrt 7}{4\sqrt 2}.$$ Then using the sine rule gives us that $\sin\phi=2\sin\psi=\frac{\sqrt 7}{2\sqrt 2}.$ Thus we obtain that $\cos\phi=\frac{1}{2\sqrt 2}.$ This gives us $$x=5\cos\phi=\frac{5}{2\sqrt 2}$$ and $$y=\frac{5\sqrt 7}{2\sqrt 2}.$$ 
Hence we have that the area of the small triangle is $$xy=\frac{25}{8}\sqrt 7$$ and the area of the large triangle is $$(x+5\sqrt 2)y=xy+5y\sqrt 2=\frac{125}{8}\sqrt 7.$$ Therefore we have that the area $S$ of the small segment is given by $$\frac{2\phi}{360°}×π×5^2-xy=\frac54\left(\frac{π\phi}{9}-\frac58\sqrt 7\right)$$ and similarly that $$T=\frac{2\psi}{360°}×π×10^2-\frac{125}{8}\sqrt 7=5\left (\frac{π\psi}{9}-\frac{25}{8}\sqrt 7\right).$$
Therefore the area needed is given by $$S-T=\frac{5π}{9}\left(\frac{\phi}{4}-\psi\right)+\frac{425}{32}\sqrt 7,$$ where $$\cos\phi=\frac14\sqrt 2,\,\cos\psi=\frac58\sqrt 2$$ and $\phi,\,\psi$ are in degrees.
