AMC12 2004 B

AMC12 2004 B · Q10

AMC12 2004 B · Q10. It mainly tests Pythagorean theorem, Circle theorems.

An annulus is the region between two concentric circles. The concentric circles in the figure have radii $b$ and $c$, with $b>c$. Let $OX$ be a radius of the larger circle, let $XZ$ be tangent to the smaller circle at $Z$, and let $OY$ be the radius of the larger circle that contains $Z$. Let $a=XZ$, $d=YZ$, and $e=XY$. What is the area of the annulus?

环形区域（annulus）是两个同心圆之间的区域。图中的同心圆半径分别为 $b$ 和 $c$，且 $b>c$。设 $OX$ 为大圆的一条半径，$XZ$ 在 $Z$ 点与小圆相切，$OY$ 为经过 $Z$ 的大圆半径。令 $a=XZ$，$d=YZ$，$e=XY$。该环形区域的面积是多少？

(A) \pi a^2 \pi a^2

(B) \pi b^2 \pi b^2

(D) \pi d^2 \pi d^2

(E) \pi e^2 \pi e^2

Answer

Correct choice: (A)

正确答案：（A）

Solution

The area of the large circle is $\pi b^2$, the area of the small one is $\pi c^2$, hence the shaded area is $\pi(b^2-c^2)$. From the Pythagorean Theorem for the right triangle $OXZ$ we have $a^2 + c^2 = b^2$, hence $b^2-c^2=a^2$ and thus the shaded area is $\boxed{\mathrm{(A)\ }\pi a^2}$.

大圆面积为 $\pi b^2$，小圆面积为 $\pi c^2$，因此阴影面积为 $\pi(b^2-c^2)$。对直角三角形 $OXZ$ 使用勾股定理得 $a^2 + c^2 = b^2$，因此 $b^2-c^2=a^2$，从而阴影面积为 $\boxed{\mathrm{(A)\ }\pi a^2}$。

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