Difference between revisions of "2005 AMC 10B Problems/Problem 23"
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<math>\mathrm{(A)} 2 \qquad \mathrm{(B)} 3 \qquad \mathrm{(C)} 5 \qquad \mathrm{(D)} 6 \qquad \mathrm{(E)} 8 </math> | <math>\mathrm{(A)} 2 \qquad \mathrm{(B)} 3 \qquad \mathrm{(C)} 5 \qquad \mathrm{(D)} 6 \qquad \mathrm{(E)} 8 </math> | ||
− | == Solution == | + | == Solution 1== |
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>, | ||
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<math>\frac{1}{2}AB=\frac{5}{2}DC</math>. | <math>\frac{1}{2}AB=\frac{5}{2}DC</math>. | ||
− | <math>AB | + | <math>\frac{AB}{DC} = \boxed{5}</math>. |
==Solution 2== | ==Solution 2== |
Revision as of 10:17, 25 November 2018
Contents
[hide]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 1
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
.
.
Solution 2
Mark , , and Note that the heights of trapezoids & are the same. Mark the height to be .
Then, we have that .
From this, we get that .
We also get that .
Simplifying, we get that
Notice that we want .
Dividing the first equation by , we get that .
Dividing the second equation by , we get that .
Now, when we subtract the top equation from the bottom, we get that
Hence, the answer is
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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