Question

长方形 $ABCD$ 中， $AB = 4$ ， $BC = 3$ 。过 $B$ 作线段 $EF$ ，使得 $EF \perp DB$ ，并且 $A$ 和 $C$ 分别在 $DE$ 和 $DF$ 上。求 $EF$ ？

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 $EF?$

$9$

$10$

$\dfrac{125}{12}$

$\dfrac{103}{9}$

$12$

答案：C

知识点：相似直角三角形勾股定理

难度评级：1580

解答：

对角线 $DB = \sqrt{4^2 + 3^2} = 5$ 。

直角三角形 $EBA$ 、 $DBC$ 和 $BFC$ 都与 $\triangle DBA$ 相似。由 $\triangle EBA$ ， $\begin{aligned} \dfrac{EB}{AB} &= \dfrac{DB}{BC} \\ &\implies \dfrac{EB}{4} = \dfrac{5}{3} \\ &\implies EB = \dfrac{20}{3}. \end{aligned}$

由 $\triangle BFC$ ， $\begin{aligned} \dfrac{BF}{BC} &= \dfrac{DB}{AB} \\ &\implies \dfrac{BF}{3} = \dfrac{5}{4} \\ &\implies BF = \dfrac{15}{4}. \end{aligned}$

因此 $\begin{aligned} EF &= EB + BF \\ &= \dfrac{20}{3} + \dfrac{15}{4} \\ &= \dfrac{125}{12}. \end{aligned}$

所以正确答案是 C。

The diagonal is $DB = \sqrt{4^2 + 3^2} = 5.$

Right triangles $EBA,$ $DBC,$ and $BFC$ are all similar to $\triangle DBA.$ From $\triangle EBA,$ $\begin{aligned} \dfrac{EB}{AB} &= \dfrac{DB}{BC} \\ &\implies \dfrac{EB}{4} = \dfrac{5}{3} \\ &\implies EB = \dfrac{20}{3}. \end{aligned}$

From $\triangle BFC,$ $\begin{aligned} \dfrac{BF}{BC} &= \dfrac{DB}{AB} \\ &\implies \dfrac{BF}{3} = \dfrac{5}{4} \\ &\implies BF = \dfrac{15}{4}. \end{aligned}$

Therefore $\begin{aligned} EF &= EB + BF \\ &= \dfrac{20}{3} + \dfrac{15}{4} \\ &= \dfrac{125}{12}. \end{aligned}$

Thus, the correct answer is C.

2009 AMC 10A 第 17 题

答案：C

难度评级：1580

解答：

其他年份的第 17 题