Difference between revisions of "2005 AMC 10A Problems/Problem 25"
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Therefore, the desired ratio is <math>\frac{133\sin A}{392\sin A}=\frac{19}{56}\Longrightarrow \mathrm{(D)}</math> | Therefore, the desired ratio is <math>\frac{133\sin A}{392\sin A}=\frac{19}{56}\Longrightarrow \mathrm{(D)}</math> | ||
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+ | Note: <math>BC=39</math> was not used in this problem | ||
==See also== | ==See also== |
Revision as of 12:07, 8 November 2015
Problem
In we have , , and . Points and are on and respectively, with and . What is the ratio of the area of triangle to the area of the quadrilateral ?
Solution
Using this formula:
Since the area of is equal to the area of minus the area of ,
.
Therefore, the desired ratio is
Note: was not used in this problem
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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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