AMC8 2010

AMC8 2010 · Q23

AMC8 2010 · Q23. It mainly tests Pythagorean theorem, Circle theorems.

Semicircles $POQ$ and $ROS$ pass through the center of circle $O$. What is the ratio of the combined areas of the two semicircles to the area of the circle $O$?

半圆 $POQ$ 和 $ROS$ 都经过圆心 $O$ 的圆 $O$ 的圆心。两个半圆的总面积与圆 $O$ 的面积之比是多少？

(A) $\frac{\sqrt{2}}{4}$ $\frac{\sqrt{2}}{4}$

(B) $\frac{1}{2}$ $\frac{1}{2}$

(D) $\frac{2}{3}$ $\frac{2}{3}$

(E) $\frac{\sqrt{2}}{2}$ $\frac{\sqrt{2}}{2}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

By the Pythagorean Theorem, the radius of the larger circle turns out to be $\sqrt{1^2 + 1^2} = \sqrt{2}$. Therefore, the area of the larger circle is $(\sqrt{2})^2\pi = 2\pi$. Using the coordinate plane given, we find that the radius of each of the two semicircles to be 1. So, the area of the two semicircles is $1^2\pi=\pi$. Finally, the ratio of the combined areas of the two semicircles to the area of circle $O$ is $\boxed{\textbf{(B)}\ \frac{1}{2}}$.

由勾股定理，大圆的半径为 $\sqrt{1^2 + 1^2} = \sqrt{2}$，面积为 $(\sqrt{2})^2\pi = 2\pi$。由坐标图，两个半圆每个半径为 1，总面积为 $1^2\pi=\pi$。因此，两个半圆总面积与圆 $O$ 面积之比为 $\boxed{\textbf{(B)}\ \frac{1}{2}}$。

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