A geometry problem by Yahia El Haw

The area of a circle is $\pi r^2$ , so we have

$25\pi=\pi r^2$

$25=r^2$

$r=5$

By cosine law on $\triangle AEB$ , we have

$(5\sqrt{2})^2=5^2+5^2-2(5)(5)(\cos \angle AEB)$

$\angle AEB=90^\circ$

Therefore, $\angle BEC=\angle CED = 45^\circ$ .

The area of the shaded region is therefore, $\dfrac{1}{2}(5)(5)+2\left(\dfrac{1}{2}\right)(5)(5)(\sin 45) \approx \boxed{30.178}$ .