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Geometry Difficulty 2.5 Junior Find the answer Canada

In the diagram, $\triangle ABC$ is right-angled at $B$ and the semi-circle has diameter $BC$ . If $AB=20$ and $AC=40$ , what is the area of the
semi-circle?

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Solution

$\triangle ABC$ 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 $\dfrac12\pi(300)=150\pi$ .

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.