Question

如图，考虑 $12$ 边形 $ABCDEFGHIJKL$ ，每条边长为 $4$ ，且每两条相邻边成直角。设 $\overline{AG}$ 与 $\overline{CH}$ 交于 $M$ 。四边形 $ABCM$ 的面积是多少？

Consider the $12$ -sided polygon $ABCDEFGHIJKL,$ as shown. Each of its sides has length $4,$ and each two consecutive sides form a right angle. Suppose that $\overline{AG}$ and $\overline{CH}$ meet at $M.$ What is the area of quadrilateral $ABCM?$

$\dfrac{44}{3}$

$16$

$\dfrac{88}{5}$

$20$

$\dfrac{62}{3}$

答案：C

知识点：坐标几何鞋带公式

难度评级：1790

解答：

将图形放在坐标系中，取 $A = (-2, 6)$ 、 $B = (2, 6)$ 、 $C = (2, 2)$ 、 $G = (2, -6)$ 、 $H = (-2, -6)$ 。

直线 $AG$ 为 $y = -3x$ ，直线 $CH$ 为 $y = 2x - 2$ 。

它们的交点为 $M = \left(\tfrac25, -\tfrac65\right)$ 。

对 $A, B, C, M$ 使用鞋带公式，面积为 $\begin{aligned} [ABCM] &=\dfrac12\left|{-8}-\dfrac{136}{5}\right|\\ &=\dfrac{88}{5}. \end{aligned}$

所以正确答案是 C。

Put the figure on coordinates with $A = (-2, 6),$ $B = (2, 6),$ $C = (2, 2),$ $G = (2, -6),$ and $H = (-2, -6).$

Line $AG$ is $y = -3x,$ and line $CH$ is $y = 2x - 2.$

Their intersection is $M = \left(\tfrac25, -\tfrac65\right).$

Applying the shoelace formula to $A, B, C, M$ gives $\begin{aligned} [ABCM] &=\dfrac12\left|{-8}-\dfrac{136}{5}\right|\\ &=\dfrac{88}{5}. \end{aligned}$

Thus, the correct answer is C.

2007 AMC 10A 第 18 题

答案：C

难度评级：1790

解答：

其他年份的第 18 题