2009 AMC 10A Problem 17

Problem 17 of 25IntermediateGeometry

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?

Show solution

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.

More practice

Concepts: similarity · right triangle · Pythagorean Theorem

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.