2006 AMC 10B 第 8 题

Question

如图，一个面积为

40

的正方形内接于一个半圆。这个半圆的面积是多少？

A square of area

40

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

20\pi

25\pi

30\pi

40\pi

50\pi

设正方形边长为

s

，满足

s^2=40

，且底边位于直径中央。

Let the square have side

s

with

s^2=40,

base centered on the diameter

半径满足

r^2=\left(\tfrac{s}{2}\right)^2+s^2

。

The radius satisfies

r^2=\left(\tfrac{s}{2}\right)^2+s^2

设正方形边长为

s

，则

s^2=40

。它的底边在直径上居中，而上方的一个顶点

\left(\tfrac{s}{2},s\right)

位于圆上。

因此

r^2=\left(\tfrac{s}{2}\right)^2+s^2=\tfrac{40}{4}+40=50

，半圆面积为

\tfrac12\pi r^2=\tfrac12\pi(50)=25\pi

。

所以正确答案是 B。

Let the square have side

s,

so

s^2=40.

Its base lies centered on the diameter, and a top corner at

\left(\tfrac{s}{2},s\right)

lies on the circle.

Then

r^2=\left(\tfrac{s}{2}\right)^2+s^2=\tfrac{40}{4}+40=50.

The semicircle area is

\tfrac12\pi r^2=\tfrac12\pi(50)=25\pi.

Thus, the correct answer is B.

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