Question

图中的两个圆有相同的圆心 $C$ 。弦 $\overline{AD}$ 在 $B$ 点与内圆相切， $AC$ 长 $10$ ，且弦 $\overline{AD}$ 长 $16$ 。两个圆之间区域的面积是多少？

The two circles pictured have the same center $C.$ Chord $\overline{AD}$ is tangent to the inner circle at $B,$ $AC$ is $10,$ and chord $\overline{AD}$ has length $16.$ What is the area between the two circles?

$\ 36 \pi$

$\ 49 \pi$

$\ 64 \pi$

$\ 81 \pi$

$\ 100 \pi$

答案：C

知识点：圆面积弦勾股定理

难度评级：1390

解答：

两圆之间的面积等于大圆面积减去小圆面积，即由勾股定理，因此只需求 $AB^2 \pi$ 。 $(AC)^2\pi - (CB)^2 \pi$ $= \pi(AC^2 - CB^2).$ $AC^2 - CB^2 = AB^2.$

因为 $AB$ 是 $AD$ 的一半，所以 $AB = 8$ 。于是 $AB^2 \pi = 64 \pi$ 。

所以正确答案是 C。

The area between the two circles is the area of the larger circle minus the area of the smaller circle. This would be $(AC)^2\pi - (CB)^2 \pi$ $= \pi(AC^2 - CB^2).$ By the Pythagorean Theorem, we can get $AC^2 - CB^2 = AB^2.$ Therefore, we need to find $AB^2 \pi.$

Since $AB$ is half of $AD,$ we get $AB = 8.$ This makes $AB^2 \pi = 64 \pi.$

Thus, the answer is C .

2010 AMC 8 第 19 题

答案：C

难度评级：1390

解答：

其他年份的第 19 题