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

Revision as of 22:37, 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

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

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

@@ Line 13: / Line 13: @@
 <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>\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>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>.
 == See Also ==
 {{AMC10 box|year=2005|ab=B|num-b=22|num-a=24}}

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

Revision as of 22:37, 24 May 2013

Problem

Solution

See Also