Difference between revisions of "2005 AMC 10B Problems/Problem 23"
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== Solution 1== | == Solution 1== | ||
− | Since the | + | Since the heights 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}=2\left(\frac{DC+EF}{2}\right)</math>. | ||
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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>\tfrac{x+z}{2} \cdot | + | We also get that <math>\tfrac{x+z}{2} \cdot 2h= 3(\tfrac{y+z}{2} \cdot h)</math>. |
Simplifying, we get that <math>2x=z+3y</math> | Simplifying, we get that <math>2x=z+3y</math> |
Latest revision as of 23:13, 1 June 2021
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 heights 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.