Question

下图由三个单位正方形和两条线段组成。求 $\triangle ABC$ 的面积。

Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC?$

$\dfrac16$

$\dfrac15$

$\dfrac29$

$\dfrac13$

$\dfrac{\sqrt{2}}{4}$

答案：B

知识点：坐标几何三角形面积

难度评级：1480

解答：

可以用坐标几何求两条直线的交点。

令 $A$ 为原点， $B = (1, 0)$ 。经过 $A$ 的直线斜率为 $-\dfrac{1}{2}$ ，方程为经过 $B$ 的直线斜率为 $2$ ， $y$ 轴截距为 $-2$ ，所以方程为 $y = -\dfrac{1}{2}x.$ $y = 2x - 2.$

联立两式，得到 $2x - 2 = -\dfrac{1}{2}x$ $x = \dfrac{4}{5}.$

因此点 $C$ 的 $y$ 坐标为 $-\dfrac{1}{2} \cdot \dfrac{4}{5} = -\dfrac{2}{5}.$

三角形 $ABC$ 的面积为 $\dfrac{1}{2} \cdot 1 \cdot \dfrac{2}{5} = \dfrac{1}{5}.$

所以正确答案是 B。

We can use coordinate geometry to figure out where the intersection of the two lines occurs.

Let $A$ be the origin and $B = (1, 0).$ Then the slope of the line through $A$ is $-\dfrac{1}{2},$ which makes the equation of the line $y = -\dfrac{1}{2}x.$ The slope of the line through $B$ is $2.$ The $y$ -intercept is $-2.$ This makes the equation of this line $y = 2x - 2.$

Equating the equations, we get $2x - 2 = -\dfrac{1}{2}x$ $x = \dfrac{4}{5}.$

This makes the $y$ -coordinate of $C$ $-\dfrac{1}{2} \cdot \dfrac{4}{5} = -\dfrac{2}{5}.$

The area of triangle $ABC$ is then $\dfrac{1}{2} \cdot 1 \cdot \dfrac{2}{5} = \dfrac{1}{5}.$

Thus, B is the correct answer.

2012 AMC 10A 第 15 题

答案：B

难度评级：1480

解答：

其他年份的第 15 题