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

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== Solution ==
 
== Solution ==
 
Since the height of both trapezoids are equal, and the area of <math>ABEF</math> is twice the area of <math>FECD</math>,  
 
Since the height of both trapezoids are equal, and the area of <math>ABEF</math> is twice the area of <math>FECD</math>,  
 +
 
<math>\frac{AB+EF}{2}=2\left(\frac{CD+EF}{2}\right)</math>.
 
<math>\frac{AB+EF}{2}=2\left(\frac{CD+EF}{2}\right)</math>.
 +
 
<math>\frac{AB+EF}{2}=CD+EF</math>, so
 
<math>\frac{AB+EF}{2}=CD+EF</math>, so
 +
 
<math>AB+EF=2CD+2EF</math>.
 
<math>AB+EF=2CD+2EF</math>.
 +
 
<math>EF</math> is exactly halfway between <math>AB</math> and <math>CD</math>, so <math>EF=\frac{AB+CD}{2}</math>. <math>AB+\frac{AB+CD}{2}=2CD+AB+CD</math>, so
 
<math>EF</math> is exactly halfway between <math>AB</math> and <math>CD</math>, so <math>EF=\frac{AB+CD}{2}</math>. <math>AB+\frac{AB+CD}{2}=2CD+AB+CD</math>, so
 +
 
<math>\frac{3}{2}AB+\frac{1}{2}CD=3CD+AB</math>, and <math>\frac{1}{2}AB=\frac{5}{2}CD</math>. <math>AB/DC = \boxed{5}</math>.
 
<math>\frac{3}{2}AB+\frac{1}{2}CD=3CD+AB</math>, and <math>\frac{1}{2}AB=\frac{5}{2}CD</math>. <math>AB/DC = \boxed{5}</math>.
  
 
== 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 22:36, 24 May 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
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All AMC 10 Problems and Solutions