A geometry problem by A Former Brilliant Member

By pythagorean theorem, $BD=\sqrt{10^2+10^2}=10\sqrt{2}$ . It follows that $OD=\dfrac{10\sqrt{2}}{2}=5\sqrt{2}$ . Let $x$ be the area of segment $AD$ . Since segment $AD$ equals segment $AB$ , we have

$2x=\dfrac{1}{2}(\pi)(5\sqrt{2})^2-\dfrac{1}{2}(10^2)=25\pi-50$

Let $y$ be the area of segement $DO$ . By symmentry segment $DO$ is equal to segment $OB$ . Let $z$ be the area of the region in the diagram. By symmetry, $z=2y$ .

$y=\dfrac{1}{4}(\pi)(5^2)-\dfrac{1}{2}(5)(5)=\dfrac{25}{4}\pi-\dfrac{25}{4}$

The area of the shaded region is

$A_{shaded}=2x+2y+z=2x+2y+2y=2x+4y$

$A_{shaded}=25\pi - 50 + 4\left(\dfrac{25}{4}\pi - \dfrac{25}{2}\right)= 25\pi-50+25\pi-50=$ $\boxed{50\pi-100}$

Note:

1. $2x$ is equal to the area of the semicircle with diameter of $DB$ (or radius of $OD$ ) minus the area of triangle $ABD$ .

2. $y$ is a segment of a circle. It is equal to the area of quarter circle with radius $5$ minus area of triangle $DOE$ .

3. If we combine segment $DO$ and segment $BO$ , it forms region $z$ .