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Difference between revisions of "2005 AMC 10B Problems/Problem 23"
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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{DC+EF}{2}\right)</math>.
  
−
<math>\frac{AB+EF}{2}=CD+EF</math>, so
+
<math>\frac{AB+EF}{2}=DC+EF</math>, so
  
−
<math>AB+EF=2CD+2EF</math>.
+
<math>AB+EF=2DC+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>EF</math> is exactly halfway between <math>AB</math> and <math>DC</math>, so <math>EF=\frac{AB+DC}{2}</math>.  
  
−
<math>AB+\frac{AB+CD}{2}=2CD+AB+CD</math>, so
+
<math>AB+\frac{AB+DC}{2}=2DC+AB+DC</math>, so
  
−
<math>\frac{3}{2}AB+\frac{1}{2}CD=3CD+AB</math>, and  
+
<math>\frac{3}{2}AB+\frac{1}{2}DC=3DC+AB</math>, and  
  
−
<math>\frac{1}{2}AB=\frac{5}{2}CD</math>.
+
<math>\frac{1}{2}AB=\frac{5}{2}DC</math>.
  
 
  <math>AB/DC = \boxed{5}</math>.
 
  <math>AB/DC = \boxed{5}</math>.

Revision as of 19:52, 5 January 2016

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 $FECD$,

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

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

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

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

See Also

2005 AMC 10B (Problems • Answer Key • Resources)
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

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