AMC10 2006 B

AMC10 2006 B · Q19

AMC10 2006 B · Q19. It mainly tests Circle theorems, Area & perimeter.

A circle of radius 2 is centered at O. Square OABC has side length 1. Sides AB and CB are extended past B to meet the circle at D and E, respectively. What is the area of the shaded region in the figure, which is bounded by BD, BE, and the minor arc connecting D and E?

圆心为 O，半径为 2。正方形 OABC 边长为 1。边 AB 和 CB 分别向 B 外延长，与圆相交于 D 和 E。图中阴影区域由 BD、BE 和连接 D 和 E 的短弧所围成，其面积是多少？

(A) $\frac{\pi}{3} + 1 - \sqrt{3}$ $\frac{\pi}{3} + 1 - \sqrt{3}$

(B) $\frac{\pi}{2}(2 - \sqrt{3})$ $\frac{\pi}{2}(2 - \sqrt{3})$

(D) $\frac{\pi}{6} + \frac{\sqrt{3} - 1}{2}$ $\frac{\pi}{6} + \frac{\sqrt{3} - 1}{2}$

(E) $\frac{\pi}{3} - 1 + \sqrt{3}$ $\frac{\pi}{3} - 1 + \sqrt{3}$

Answer

Correct choice: (A)

正确答案：（A）

Solution

(A) Since $OC=1$ and $OE=2$, it follows that $\angle EOC=60^\circ$ and $\angle EOA=30^\circ$. The area of the shaded region is the area of the $30^\circ$ sector $DOE$ minus the area of congruent triangles $OBD$ and $OBE$. First note that $Area(\text{Sector }DOE)=\frac{1}{12}(4\pi)=\frac{\pi}{3}.$ In right triangle $OCE$, we have $CE=\sqrt{3}$, so $BE=\sqrt{3}-1$. Therefore $Area(\triangle OBE)=\frac{1}{2}(\sqrt{3}-1)(1).$ The required area is consequently $\frac{\pi}{3}-2\left(\frac{\sqrt{3}-1}{2}\right)=\frac{\pi}{3}+1-\sqrt{3}.$

（A）由于 $OC=1$ 且 $OE=2$，可得 $\angle EOC=60^\circ$，$\angle EOA=30^\circ$。阴影部分的面积等于 $30^\circ$ 扇形 $DOE$ 的面积减去全等三角形 $OBD$ 和 $OBE$ 的面积。先注意到 $Area(\text{Sector }DOE)=\frac{1}{12}(4\pi)=\frac{\pi}{3}.$ 在直角三角形 $OCE$ 中，$CE=\sqrt{3}$，因此 $BE=\sqrt{3}-1$。所以 $Area(\triangle OBE)=\frac{1}{2}(\sqrt{3}-1)(1).$ 因此所求面积为 $\frac{\pi}{3}-2\left(\frac{\sqrt{3}-1}{2}\right)=\frac{\pi}{3}+1-\sqrt{3}.$

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