AMC10 2006 B

AMC10 2006 B · Q24

AMC10 2006 B · Q24. It mainly tests Circle theorems, Area & perimeter.

Circles with centers at O and P have radii 2 and 4, respectively, and are externally tangent. Points A and B on the circle with center O and points C and D on the circle with center P are such that AD and BC are common external tangents to the circles. What is the area of the concave hexagon AOBCPD?

中心分别为O和P的圆，半径分别为2和4，且外部相切。在以O为中心的圆上的点A和B，以及以P为中心的圆上的点C和D，使得AD和BC是圆的公共外部切线。凹六边形AOBCPD的面积是多少？

(A) $18\sqrt{3}$ $18\sqrt{3}$

(B) $24\sqrt{2}$ $24\sqrt{2}$

(D) $24\sqrt{3}$ $24\sqrt{3}$

(E) $32\sqrt{2}$ $32\sqrt{2}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) Through $O$ draw a line parallel to $\overline{AD}$ intersecting $\overline{PD}$ at $F$. Then $AOFD$ is a rectangle and $OPF$ is a right triangle. Thus $DF=2$, $FP=2$, and $OF=4\sqrt{2}$. The area of trapezoid $AOPD$ is $12\sqrt{2}$, and the area of hexagon $AOBCPD$ is $2\cdot 12\sqrt{2}=24\sqrt{2}$.

（B）过 $O$ 作一条与 $\overline{AD}$ 平行的直线，与 $\overline{PD}$ 交于 $F$。则 $AOFD$ 是一个矩形，$OPF$ 是一个直角三角形。因此 $DF=2$，$FP=2$，且 $OF=4\sqrt{2}$。梯形 $AOPD$ 的面积为 $12\sqrt{2}$，六边形 $AOBCPD$ 的面积为 $2\cdot 12\sqrt{2}=24\sqrt{2}$。

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