2012 AMC 10A 第 15 题

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

小提示：

将

A

和

B

放在坐标轴上。

Place

A

and

B

on coordinate axes

大提示：

求出两条所画直线的交点。

Find the intersection of the two drawn lines

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

y = -\dfrac{1}{2}x\text{。}

经过

B

的直线斜率为

2

，

y

轴截距为

-2

，所以方程为

y = 2x - 2\text{。}

联立两式，得到

2x - 2 = -\dfrac{1}{2}x

x = \dfrac{4}{5}\text{。}

因此点

C

的

y

坐标为

-\dfrac{1}{2} \cdot \dfrac{4}{5} = -\dfrac{2}{5}\text{。}

三角形

ABC

的面积为

\dfrac{1}{2} \cdot 1 \cdot \dfrac{2}{5} = \dfrac{1}{5}\text{。}

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

小提示：

大提示：

解答：

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