2005 AMC 10B Problems/Problem 23

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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 $AB/DC$?

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

Solution 1

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


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


$EF$ 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


$\frac{AB}{DC} = \boxed{\textbf{(C) }5}$.

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

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 $\tfrac{x}{z}=\boxed{\textbf{(C) }5}$

Solution 3

Since the bases of the trapezoids along with the height are the same, the only thing that matters is the second base. Denote the length of the bigger trapezoid $x$. The area of the smaller trapezoid is $A$ = $h \cdot \frac {b_1 + b_2}{2}$. The area of the larger trapezoid is $A$ = $h \cdot \frac {b_1 + x}{2}$. Since this problem asks for proportions, assume that $b_1 = 1$ and $b_2 = 2$.

The smaller trapezoid has area $h$ while the larger trapezoid must have area $5h$. We have the equation $\frac {x}{2} = 5$. $x$ = 10, and our answer is $\boxed{\textbf{(C) }5}$


Video Solution

https://www.youtube.com/watch?v=fsNJbC3hGtk ~David

See Also

2005 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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All AMC 10 Problems and Solutions

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