AMC10 2012 A

AMC10 2012 A · Q15

AMC10 2012 A · Q15. It mainly tests Area & perimeter, Coordinate geometry.

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

给出三个单位正方形和连接两对顶点的两条线段。求 $\triangle ABC$ 的面积。

(A) \frac{1}{6} \frac{1}{6}

(B) \frac{1}{5} \frac{1}{5}

(D) \frac{1}{3} \frac{1}{3}

(E) \frac{\sqrt{2}}{4} \frac{\sqrt{2}}{4}

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Place the figure on the coordinate plane with $A$ at the origin, $B$ on the positive $x$-axis, and label the other points as shown. Then the equation of line $AE$ is $y=-\frac{1}{2}x$, and the equation of line $BF$ is $y=2x-2$. Solving the simultaneous equations shows that $C=\left(\frac{4}{5},-\frac{2}{5}\right)$. Therefore $\triangle ABC$ has base $AB=1$ and altitude $\frac{2}{5}$, so its area is $\frac{1}{5}$.

答案（B）：将图形放在坐标平面上，使 $A$ 在原点，$B$ 在正 $x$ 轴上，并按图所示标出其他点。则直线 $AE$ 的方程为 $y=-\frac{1}{2}x$，直线 $BF$ 的方程为 $y=2x-2$。联立方程可得 $C=\left(\frac{4}{5},-\frac{2}{5}\right)$。因此 $\triangle ABC$ 的底边 $AB=1$，高为 $\frac{2}{5}$，所以面积为 $\frac{1}{5}$。

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