Question

El rectángulo $ABCD$ tiene $AB = 4$ y $BC = 3.$ El segmento $EF$ se construye pasando por $B$ de modo que $EF \perp DB,$ y $A$ y $C$ están sobre $DE$ y $DF,$ respectivamente. ¿Cuánto vale $EF$ ?

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 $EF?$

$9$

$10$

$\dfrac{125}{12}$

$\dfrac{103}{9}$

$12$

Solución:

La diagonal es $DB = \sqrt{4^2 + 3^2} = 5$ .

Los triángulos rectángulos $EBA,$ $DBC,$ y $BFC$ son todos semejantes a $\triangle DBA.$ A partir de $\triangle EBA,$ $\begin{aligned} \dfrac{EB}{AB} &= \dfrac{DB}{BC} \\ &\implies \dfrac{EB}{4} = \dfrac{5}{3} \\ &\implies EB = \dfrac{20}{3}. \end{aligned}$

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

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

Así, la respuesta correcta es 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 Problema 17

Solución:

El Problema 17 en otros años