Difference between revisions of "2005 AMC 10B Problems/Problem 23"
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== Problem == | == Problem == | ||
− | == Solution == | + | In trapezoid <math>ABCD</math> we have <math>\overline{AB}</math> parallel to <math>\overline{DC}</math>, <math>E</math> as the midpoint of <math>\overline{BC}</math>, and <math>F</math> as the midpoint of <math>\overline{DA}</math>. The area of <math>ABEF</math> is twice the area of <math>FECD</math>. What is <math>AB/DC</math>? |
+ | |||
+ | <math>\mathrm{(A)} 2 \qquad \mathrm{(B)} 3 \qquad \mathrm{(C)} 5 \qquad \mathrm{(D)} 6 \qquad \mathrm{(E)} 8 </math> | ||
+ | |||
+ | == 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>, | ||
+ | |||
+ | <math>\frac{AB+EF}{2}=2\left(\frac{DC+EF}{2}\right)</math>. | ||
+ | |||
+ | <math>\frac{AB+EF}{2}=DC+EF</math>, so | ||
+ | |||
+ | <math>AB+EF=2DC+2EF</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+DC}{2}=2DC+AB+DC</math>, so | ||
+ | |||
+ | <math>\frac{3}{2}AB+\frac{1}{2}DC=3DC+AB</math>, and | ||
+ | |||
+ | <math>\frac{1}{2}AB=\frac{5}{2}DC</math>. | ||
+ | |||
+ | <math>\frac{AB}{DC} = \boxed{5}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Mark <math>DC=z</math>, <math>AB=x</math>, and <math>FE=y.</math> | ||
+ | Note that the heights of trapezoids <math>ABEF</math> & <math>FECD</math> are the same. Mark the height to be <math>h</math>. | ||
+ | |||
+ | Then, we have that <math>\tfrac{x+y}{2}\cdot h=2(\tfrac{y+z}{2} \cdot h)</math>. | ||
+ | |||
+ | From this, we get that <math>x=2z+y</math>. | ||
+ | |||
+ | We also get that <math>\tfrac{x+z}{2} \cdot h= 3(\tfrac{y+z}{2} \cdot h)</math>. | ||
+ | |||
+ | Simplifying, we get that <math>2x=z+3y</math> | ||
+ | |||
+ | Notice that we want <math>\tfrac{AB}{DC}=\tfrac{x}{z}</math>. | ||
+ | |||
+ | Dividing the first equation by <math>z</math>, we get that <math>\tfrac{x}{z}=2+\tfrac{y}{z}\implies 3(\tfrac{x}{z})=6+3(\tfrac{y}{z})</math>. | ||
+ | |||
+ | Dividing the second equation by <math>z</math>, we get that <math>2(\tfrac{x}{z})=1+3(\tfrac{y}{z})</math>. | ||
+ | |||
+ | Now, when we subtract the top equation from the bottom, we get that <math>\tfrac{x}{z}=5</math> | ||
+ | |||
+ | Hence, the answer is <math>\boxed{5}</math> | ||
+ | |||
== See Also == | == See Also == | ||
− | + | {{AMC10 box|year=2005|ab=B|num-b=22|num-a=24}} | |
+ | {{MAA Notice}} |
Latest revision as of 20:22, 26 December 2020
Contents
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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