Difference between revisions of "2005 AMC 10A Problems/Problem 25"
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This means that <math>[BDC]=\frac{6}{19}\cdot3x=\frac{18x}{19}</math>. | This means that <math>[BDC]=\frac{6}{19}\cdot3x=\frac{18x}{19}</math>. | ||
Thus, <cmath>\frac{[ADE]}{[BCED]} = \frac{x}{\frac{18x}{19}+2x}= \boxed{\frac{19}{56}\Longrightarrow D}.</cmath> | Thus, <cmath>\frac{[ADE]}{[BCED]} = \frac{x}{\frac{18x}{19}+2x}= \boxed{\frac{19}{56}\Longrightarrow D}.</cmath> | ||
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+ | -Conantwiz2023 | ||
==Solution (trig)== | ==Solution (trig)== |
Revision as of 16:40, 10 January 2019
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 1(no trig)
We have that
But , so
Solution 2(no trig)
We can let . Since , . So, . This means that . Thus,
-Conantwiz2023
Solution (trig)
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 24 |
Followed by Last Problem | |
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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