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

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== Solution ==
 
== Solution ==
The length of <math>EF</math> is the arithmetic mean of the two parallel bases.  Since the base EF is shared between the two smaller trapezoids, you want AB/DC = \boxed{2}$.
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The length of <math>EF</math> is the arithmetic mean of the two parallel bases.  Since the base EF is shared between the two smaller trapezoids, you want AB/DC = /boxed{2}$.
  
 
== 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}}

Revision as of 20:56, 23 August 2011

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

The length of $EF$ is the arithmetic mean of the two parallel bases. Since the base EF is shared between the two smaller trapezoids, you want AB/DC = /boxed{2}$.

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