AMC10 2009 A

AMC10 2009 A · Q17

AMC10 2009 A · Q17. It mainly tests Triangles (properties), Similarity.

Rectangle $ABCD$ has $AB = 4$ and $BC = 3$. Segment $EF$ is constructed through $B$ so that $EF \perp DB$, and $A$ and $C$ lie on $DE$ and $DF$, respectively. What is $EF$?

矩形 $ABCD$ 有 $AB = 4$ 和 $BC = 3$。通过 $B$ 构造线段 $EF$ 使得 $EF \perp DB$，且 $A$ 和 $C$ 分别位于 $DE$ 和 $DF$ 上。$EF$ 等于多少？

(A) 9 9

(B) 10 10

(D) $\frac{103}{9}$ $\frac{103}{9}$

(E) 12 12

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Note that $DB=5$ and $\triangle EBA$, $\triangle DBC$, and $\triangle BFC$ are all similar. Therefore $\frac{4}{EB}=\frac{3}{5}$, so $EB=\frac{20}{3}$. Similarly, $\frac{3}{BF}=\frac{4}{5}$, so $BF=\frac{15}{4}$. Thus \[ EF=EB+BF=\frac{20}{3}+\frac{15}{4}=\frac{125}{12}. \]

答案（C）：注意 $DB=5$，且 $\triangle EBA$、$\triangle DBC$ 和 $\triangle BFC$ 都相似。因此 $\frac{4}{EB}=\frac{3}{5}$，所以 $EB=\frac{20}{3}$。同理，$\frac{3}{BF}=\frac{4}{5}$，所以 $BF=\frac{15}{4}$。因此 \[ EF=EB+BF=\frac{20}{3}+\frac{15}{4}=\frac{125}{12}. \]

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