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.
On
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.
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.