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 1

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

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

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

$AB+EF=2DC+2EF$ .

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

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

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

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

$\frac{AB}{DC} = \boxed{5}$ .

Solution 2

Mark $DC=z$ , $AB=x$ , and $FE=y.$ Note that the heights of trapezoids $ABEF$ & $FECD$ are the same. Mark the height to be $h$ .

Then, we have that $\tfrac{x+y}{2}\cdot h=2(\tfrac{y+z}{2} \cdot h)$ .

From this, we get that $x=2z+y$ .

We also get that $\tfrac{x+z}{2} \cdot h= 3(\tfrac{y+z}{2} \cdot h)$ .

Simplifying, we get that $2x=z+3y$

Notice that we want $\tfrac{AB}{DC}=\tfrac{x}{z}$ .

Dividing the first equation by $z$ , we get that $\tfrac{x}{z}=2+\tfrac{y}{z}\implies 3(\tfrac{x}{z})=6+3(\tfrac{y}{z})$ .

Dividing the second equation by $z$ , we get that $2(\tfrac{x}{z})=1+3(\tfrac{y}{z})$ .

Now, when we subtract the top equation from the bottom, we get that $\tfrac{x}{z}=5$

Hence, the answer is $\boxed{5}$

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

Contents

Problem

Solution 1

Solution 2

See Also