AMC8 2010

AMC8 2010 · Q19

AMC8 2010 · Q19. It mainly tests Pythagorean theorem, Circle theorems.

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?

图中两个圆心相同，为点 $C$。弦 $\overline{AD}$ 在点 $B$ 处与内圆相切，$AC = 10$，弦 $\overline{AD}$ 长为 16。两个圆之间的面积是多少？

(A) $36\pi$ $36\pi$

(B) $49\pi$ $49\pi$

(D) $81\pi$ $81\pi$

(E) $100\pi$ $100\pi$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Since $\triangle ACD$ is isosceles, $CB$ bisects $AD$. Thus $AB=BD=8$. From the Pythagorean Theorem, $CB=6$. Thus the area between the two circles is $100\pi - 36\pi=64\pi$ $\boxed{\textbf{(C)}\ 64\pi}$ Note: The length $AC$ is necessary information, as this tells us the radius of the larger circle. The area of the annulus is $\pi(AC^2-BC^2)=\pi AB^2=64\pi$.

由于 $\triangle ACD$ 是等腰三角形，$CB$ 平分 $AD$。因此 $AB = BD = 8$。由勾股定理，$CB = 6$。两个圆之间的面积为 $100\pi - 36\pi = 64\pi$ $\boxed{\textbf{(C)}\ 64\pi}$ 注：$AC$ 的长度是必要信息，它给出了大圆的半径。环形面积为 $\pi(AC^2 - BC^2) = \pi AB^2 = 64\pi$。

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