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>, | |
| + | <math>\frac{AB+EF}{2}=2\left(\frac{CD+EF}{2}\right)</math>. | ||
| + | <math>\frac{AB+EF}{2}=CD+EF</math>, so | ||
| + | <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>\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 15:30, 6 December 2011
Problem
In trapezoid 
we have 
parallel to 
, 
as the midpoint of 
, and 
as the midpoint of 
. The area of 
is twice the area of 
. What is 
?
Solution
Since the height of both trapezoids are equal, and the area of is twice the area of ,
.
, so
.
is exactly halfway between 
and 
, so 
. 
, so
, and 
. 
.
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 | ||
