Track / Stage 3 / 67 of 260 #67 of 1964

Problem 67

AMC 10/12, early questions

Geometry Difficulty 3.2 Multiple choice

A square of area 40 is inscribed in a semicircle as shown. What is the area of the semicircle?

Figure (Asymptote source)

defaultpen(linewidth(0.8)); size(100); real r=sqrt(50), s=sqrt(10); draw(Arc(origin, r, 0, 180)); draw((r,0)--(-r,0), dashed); draw((-s,0)--(s,0)--(s,2*s)--(-s,2*s)--cycle);

Pick one

Official solution

Since the area of the square is $40$ , the length of a side is $\sqrt{40}=2\sqrt{10}$ . The distance between the center of the semicircle and one of the bottom vertices of the square is half the length of the side, which is $\sqrt{10}$ .
Using the Pythagorean Theorem to find the radius $r$ of the semicircle, $r^2 = (2\sqrt{10})^2 + (\sqrt{10})^2 = 50$ . So, the area of the semicircle is $\frac{1}{2}\cdot \pi \cdot 50 = \boxed{\textbf{(B) }25\pi}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.