Difference between revisions of "2005 AMC 10B Problems/Problem 23"
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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 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 </math>z<math>, we get that </math>2(\frac{x}{z}=1+3(\frac{y}{z}<math>. | + | 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 </math>\frac{x}{z}=5<math> | Now, when we subtract the top equation from the bottom, we get that </math>\frac{x}{z}=5<math> | ||
Hence, the answer is </math>\boxed{5}$ | Hence, the answer is </math>\boxed{5}$ | ||
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== 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 12:33, 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 = 3(\frac{y+z}{2} \cdot h)$.
Simplifying, we get that$ (Error compiling LaTeX. Unknown error_msg)2x=z+3y\frac{AB}{DC}=\frac{x}{z}$.
Dividing the first equation by$ (Error compiling LaTeX. Unknown error_msg)z\frac{x}{z}=2+\frac{y}{z}\implies 3(\frac{x}{z})=6+3(\frac{y}{z})$.
Dividing the second equation by$ (Error compiling LaTeX. Unknown error_msg)z2(\frac{x}{z})=1+3(\frac{y}{z})$.
Now, when we subtract the top equation from the bottom, we get that$ (Error compiling LaTeX. Unknown error_msg)\frac{x}{z}=5\boxed{5}$
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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