AMC8 2008

AMC8 2008 · Q23

AMC8 2008 · Q23. It mainly tests Area & perimeter, Coordinate geometry.

In square $ABCE$, $AF = 2FE$ and $CD = 2DE$. What is the ratio of the area of $\triangle BFD$ to the area of square $ABCE$?

在正方形 $ABCE$ 中，$AF = 2FE$ 且 $CD = 2DE$。$ riangle BFD$ 的面积与正方形 $ABCE$ 的面积之比是多少？

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

(B) \frac{2}{9} \frac{2}{9}

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

(E) \frac{7}{20} \frac{7}{20}

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Because the answer is a ratio, it does not depend on the side length of the square. Let $AF=2$ and $FE=1$. That means square $ABCE$ has side length $3$ and area $3^2=9$ square units. The area of $\triangle BAF$ is equal to the area of $\triangle BCD=\frac{1}{2}\cdot 3\cdot 2=3$ square units. Triangle $DEF$ is an isosceles right triangle with leg lengths $DE=FE=1$. The area of $\triangle DEF$ is $\frac{1}{2}\cdot 1\cdot 1=\frac{1}{2}$ square units. The area of $\triangle BFD$ is equal to the area of the square minus the areas of the three right triangles: $9-(3+3+\frac{1}{2})=\frac{5}{2}$. So the ratio of the area of $\triangle BFD$ to the area of square $ABCE$ is $\frac{\frac{5}{2}}{9}=\frac{5}{18}$.

答案（C）：因为答案是一个比值，所以它不依赖于正方形的边长。令 $AF=2$ 且 $FE=1$。这意味着正方形 $ABCE$ 的边长为 $3$，面积为 $3^2=9$ 平方单位。$\triangle BAF$ 的面积等于 $\triangle BCD$ 的面积，即 $\frac{1}{2}\cdot 3\cdot 2=3$ 平方单位。三角形 $DEF$ 是等腰直角三角形，两条直角边长为 $DE=FE=1$。$\triangle DEF$ 的面积为 $\frac{1}{2}\cdot 1\cdot 1=\frac{1}{2}$ 平方单位。$\triangle BFD$ 的面积等于正方形的面积减去三个直角三角形的面积：$9-(3+3+\frac{1}{2})=\frac{5}{2}$。因此，$\triangle BFD$ 的面积与正方形 $ABCE$ 的面积之比为 $\frac{\frac{5}{2}}{9}=\frac{5}{18}$。

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