Question

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 $EF?$

$9$

$10$

$\dfrac{125}{12}$

$\dfrac{103}{9}$

$12$

Answer: C

Concepts:similarity right triangle Pythagorean Theorem

Difficulty rating: 1580

Solution:

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 Problem 17

Answer: C

Difficulty rating: 1580

Solution:

Problem 17 in Other Years