AMC12 2006 B

AMC12 2006 B · Q16

AMC12 2006 B · Q16. It mainly tests Triangles (properties), Circle theorems.

Regular hexagon $ABCDEF$ has vertices $A$ and $C$ at $(0,0)$ and $(7,1)$, respectively. What is its area?

正六边形 $ABCDEF$ 的顶点 $A$ 和 $C$ 分别位于 $(0,0)$ 和 $(7,1)$。它的面积是多少？

(A) $20\sqrt{3}$ $20\sqrt{3}$

(B) $22\sqrt{3}$ $22\sqrt{3}$

(D) $27\sqrt{3}$ $27\sqrt{3}$

(E) 50 50

Answer

Correct choice: (C)

正确答案：（C）

Solution

To find the area of the regular hexagon, we only need to calculate the side length. a distance of $\sqrt{7^2+1^2} = \sqrt{50} = 5\sqrt{2}$ apart. Half of this distance is the length of the longer leg of the right triangles. Therefore, the side length of the hexagon is $\frac{5\sqrt{2}}{2}\cdot\frac{1}{\sqrt{3}}\cdot2 = \frac{5\sqrt{6}}{3}$. The apothem is thus $\frac{1}{2}\cdot\frac{5\sqrt{6}}{3}\cdot\sqrt{3} = \frac{5\sqrt{2}}{2}$, yielding an area of $\frac{1}{2}\cdot10\sqrt{6}\cdot\frac{5\sqrt{2}}{2}=25\sqrt{3} \implies \mathrm{(C)}$.

要找正六边形的面积，只需计算边长。 $\overline{AC}$ 的距离为 $\sqrt{7^2+1^2} = \sqrt{50} = 5\sqrt{2}$。这个距离的一半是直角三角形较长的直角边的长度。因此，六边形的边长为 $\frac{5\sqrt{2}}{2}\cdot\frac{1}{\sqrt{3}}\cdot2 = \frac{5\sqrt{6}}{3}$。因此，内切圆半径（apothem）为 $\frac{1}{2}\cdot\frac{5\sqrt{6}}{3}\cdot\sqrt{3} = \frac{5\sqrt{2}}{2}$，从而面积为 $\frac{1}{2}\cdot10\sqrt{6}\cdot\frac{5\sqrt{2}}{2}=25\sqrt{3} \implies \mathrm{(C)}$.

Topics