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AMC8 2018

AMC8 2018 · Q20

AMC8 2018 · Q20. It mainly tests Similarity, Area & perimeter.

In $\triangle ABC$, point $E$ is on $\overline{AB}$ with $AE = 1$ and $EB = 2$. Point $D$ is on $\overline{AC}$ so that $\overline{DE} \parallel \overline{BC}$ and point $F$ is on $\overline{BC}$ so that $\overline{EF} \parallel \overline{AC}$. What is the ratio of the area of $\triangle CDEF$ to the area of $\triangle ABC$?
在 $\triangle ABC$ 中,点 $E$ 在 $\overline{AB}$ 上,$AE = 1$,$EB = 2$。点 $D$ 在 $\overline{AC}$ 上,使得 $\overline{DE} \parallel \overline{BC}$,点 $F$ 在 $\overline{BC}$ 上,使得 $\overline{EF} \parallel \overline{AC}$。$ riangle CDEF$ 的面积与 $\triangle ABC$ 的面积之比是多少?
stem
(A) $\frac{4}{9}$ $\frac{4}{9}$
(B) $\frac{1}{2}$ $\frac{1}{2}$
(C) $\frac{5}{9}$ $\frac{5}{9}$
(D) $\frac{3}{5}$ $\frac{3}{5}$
(E) $\frac{2}{3}$ $\frac{2}{3}$
Answer
Correct choice: (A)
正确答案:(A)
Solution
Answer (A): Triangles $AED$, $EBF$, and $ABC$ are similar with their sides in the ratio of $1:2:3$. Therefore their areas are in the ratio of $1:4:9$. The combined areas of $\triangle AED$ and $\triangle EBF$ constitute $\frac{5}{9}$ of the area of $\triangle ABC$, so the area of $CDEF$ is $\frac{4}{9}$ of the area of $\triangle ABC$.
答案(A):三角形 $AED$、$EBF$ 和 $ABC$ 相似,它们的边长比为 $1:2:3$,因此面积比为 $1:4:9$。$\triangle AED$ 与 $\triangle EBF$ 的面积之和占 $\triangle ABC$ 面积的 $\frac{5}{9}$,所以 $CDEF$ 的面积占 $\triangle ABC$ 面积的 $\frac{4}{9}$。
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