, 2025

$\triangle ABC$ is
right-angled at $B$, and so by the
Pythagorean Theorem, we get $BC^2=AC^2-AB^2$ or $BC^2=40^2-20^2=1600-400=1200$, and so
$BC=\sqrt{1200}$ (since $BC>0$).

The radius of the semi-circle is $\dfrac12BC=\dfrac{\sqrt{1200}}{2}$, and so
its area is $\dfrac12\pi\left(\dfrac{\sqrt{1200}}{2}\right)^2=\dfrac12\pi\left(\dfrac{1200}{4}\right)$,
which when simplified is $\dfrac12\pi(300)=150\pi$.