Difference between revisions of "2005 AMC 10B Problems/Problem 23"
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From this, we get that <math>x=2z+y</math>. | From this, we get that <math>x=2z+y</math>. | ||
− | We also get that <math>\frac{x+z}{2} \cdot h | + | We also get that <math>\frac{x+z}{2} \cdot h= 3(\frac{y+z}{2} \cdot h)</math>. |
− | Simplifying, we get that < | + | Simplifying, we get that <math>2x=z+3y</math> |
− | Notice that we want < | + | Notice that we want <math>\frac{AB}{DC}=\frac{x}{z}</math>. |
− | Dividing the first equation by < | + | Dividing the first equation by <math>z</math>, we get that <math>\frac{x}{z}=2+\frac{y}{z}\implies 3(\frac{x}{z})=6+3(\frac{y}{z})</math>. |
− | Dividing the second equation by < | + | Dividing the second equation by <math>z</math>, we get that <math>2(\frac{x}{z})=1+3(\frac{y}{z})</math>. |
− | Now, when we subtract the top equation from the bottom, we get that < | + | Now, when we subtract the top equation from the bottom, we get that <math>\frac{x}{z}=5</math> |
− | Hence, the answer is < | + | Hence, the answer is <math>\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}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:34, 16 June 2017
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.