Difference between revisions of "2005 AMC 10A Problems/Problem 25"
Dairyqueenxd (talk | contribs) (→Solution 1) |
Dairyqueenxd (talk | contribs) (→Solution 2 (no trig)) |
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We can let <math>[ADE]=x</math>. | We can let <math>[ADE]=x</math>. | ||
− | Since <math>EC=2 | + | Since <math>EC=2 \cdot EA</math>, <math>[DEC]=2x</math>. |
So, <math>[ADC]=3x</math>. | So, <math>[ADC]=3x</math>. | ||
− | This means | + | This means that <math>[BDC]=\frac{6}{19}\cdot3x=\frac{18x}{19}</math>. |
− | Thus, < | + | Thus, <math>\frac{[ADE]}{[BCED]} = \frac{x}{\frac{18x}{19}+2x}= \boxed{\textbf{(D) }\frac{19}{56}}.</math> |
-Conantwiz2023 | -Conantwiz2023 |
Revision as of 12:44, 14 December 2021
Contents
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
We have
But , so
Note: If it is hard to understand why , you can use the fact that the area of a triangle equals . If angle , we have that . - SuperJJ
Video Solution
CHECK OUT Video Solution: https://youtu.be/VXyOJWcpi00
Solution 2(no trig)
We can let .
Since , .
So, .
This means that .
Thus,
-Conantwiz2023
Solution 3(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
Solution 4
Let be on such that then we have Since we have Thus and Finally, after some calculations.
~ Nafer
~ LaTeX changes by tkfun
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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