Difference between revisions of "2005 AMC 10B Problems/Problem 23"

(Solution)
Line 25: Line 25:
 
== See Also ==
 
== See Also ==
 
{{AMC10 box|year=2005|ab=B|num-b=22|num-a=24}}
 
{{AMC10 box|year=2005|ab=B|num-b=22|num-a=24}}
 +
{{MAA Notice}}

Revision as of 11:15, 4 July 2013

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

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

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

See Also

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png