2005 AMC 10B Problems/Problem 23

Problem

In trapezoid $ABCD$ we have $\overline{AB}$ parallel to $\overline{DC}$ , $E$ as the midpoint of $\overline{BC}$ , and $F$ as the midpoint of $\overline{DA}$ . The area of $ABEF$ is twice the area of $FECD$ . What is $AB/DC$ ?

$\mathrm{(A)} 2 \qquad \mathrm{(B)} 3 \qquad \mathrm{(C)} 5 \qquad \mathrm{(D)} 6 \qquad \mathrm{(E)} 8$

Solution

Since the height of both trapezoids are equal, and the area of $ABEF$ is twice the area of $FECD$ ,

$\frac{AB+EF}{2}=2\left(\frac{CD+EF}{2}\right)$ .

$\frac{AB+EF}{2}=CD+EF$ , so

$AB+EF=2CD+2EF$ .

$EF$ is exactly halfway between $AB$ and $CD$ , so $EF=\frac{AB+CD}{2}$ .

$AB+\frac{AB+CD}{2}=2CD+AB+CD$ , so

$\frac{3}{2}AB+\frac{1}{2}CD=3CD+AB$ , and

$\frac{1}{2}AB=\frac{5}{2}CD$ .

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2005 AMC 10B Problems/Problem 23

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