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 $EF$ ?

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

Solutions

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 $BAD$ , 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 $BEA$ and angle $DBA$ are similar.

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 $D$ 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(Coordinate Bash)

To keep things simple, we will use coordinates in only the first quadrant. The picture will look like the diagram above reflected over the $x$ -axis.It is also worth noting the $F$ will lie on the $x$ axis and $E$ on the $y$ . Let $D$ be the origin, $A(3,0)$ , $C(4,0)$ , and $B(4,3)$ . We can express segment $DB$ as the line $y=\frac{3x}{4}$ . Since $EF$ is perpendicular to $DB$ , and we know that $(4,3)$ lies on it, we can use this information to find that segment $EF$ is on the line $y=\frac{-4x}{3}+\frac{25}{3}$ . Since $E$ and $F$ are on the $y$ and $x$ axis, respectively, we plug in $0$ for $x$ and $y$ , we find that point $E$ is at $(0,\frac{25}{3})$ , and point $F$ is at $(\frac{25}{4},0)$ . Applying the distance formula, we obtain that $EF$ = $\boxed{\frac{125}{12}}$ .

2009 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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