2009 AMC 10A Problems/Problem 17

Revision as of 20:21, 30 October 2017 by Lj2016 (talk | contribs) (Solution 2)


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 1

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]

From the Pythagorean theorem we have $BD=5$.

Triangle $EAB$ is similar to $DAB$, as they have the same angles. Segment $BA$ is perpendicular to $DA$, meaning that angle $DAB$ and $BAE$ are right angles and congruent. Also, angle $DBE$ is a right angle. Because it is a rectangle, angle $BDC$ is congruent to $DBA$ and angle $ADC$ is also a right angle. By the transitive property:

$mADB + mBDC = mDBA + mABE$

$mBDC = mDBA$

$mADB + mBDC = mBDC + mABE$

$mADB = mABE$

Next, because every triangle has a degree measure of 180, angle $E$ and angle $DBA$ are congruent.

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

Solution 2

Since $BD$ is the altitude from $B$ to $EF$, we can use the equation $BD^2 = EB\cdot BF$.

Looking at the angles, we see that triangle $BDE$ is similar to $DCB$. Because of this, $\frac{AB}{CB} = \frac{EB}{DB}$. From the given information and the Pythagorean theorem, $AB=4$, $CB=3$, and $DB=5$. Solving gives $EB=20/3$.

We can use the above formula to solve for $BF$. $BD^2 = 20/3\cdot BF$. Solve to obtain $BF=15/4$.

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

Solution 3 There is a better solution where we find EF directly instead of in parts (use similarity). The strategy is similar.

See Also

2009 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 10 Problems and Solutions

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