2009 AMC 10A Problems/Problem 17

Problem

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

$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 10 \qquad \mathrm{(C)}\ \frac {125}{12} \qquad \mathrm{(D)}\ \frac {103}{9} \qquad \mathrm{(E)}\ 12$

Solution

The situation is shown in the picture below.

$[asy] unitsize(0.6cm); defaultpen(0.8); pair A=(0,0), B=(4,0), C=(4,3), D=(0,3); pair EF=rotate(90)*(D-B); pair E=intersectionpoint( (0,-100)--(0,100), (B-100*EF)--(B+100*EF) ); pair F=intersectionpoint( (-100,3)--(100,3), (B-100*EF)--(B+100*EF) ); draw(A--B--C--D--cycle); draw(B--D, dashed); draw(E--F); draw(A--E, dashed); draw(C--F, dashed); label("$A$",A,W); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,NW); label("$E$",E,SW); label("$F$",F,NE); label("$3$",A--D,W); label("$4$",C--D,N); [/asy]$

Obviously, from the Pythagorean theorem we have $BD=5$ .

Triangle $EAB$ is similar to $ABD$ , as they have the same angles. Hence $BE/AB = DB/AD$ , and therefore $BE = AB\cdot DB/AD = 20/3$ .

Also triangle $CBF$ is similar to $ABD$ . Hence $BF/BC = DB/AB$ , and therefore $BF=BC\cdot DB / AB = 15/4$ .

We then have $EF = EB+BF = \frac{20}3 + \frac{15}4 = \frac{80 + 45}{12} = \boxed{\frac{125}{12}}$ .

2009 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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2009 AMC 10A Problems/Problem 17

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