Difference between revisions of "2005 AMC 10A Problems/Problem 25"
Dianefu math (talk | contribs) m (→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\cdotEA</math>, <math>[DEC]=2x</math>. |
So, <math>[ADC]=3x</math>. | So, <math>[ADC]=3x</math>. | ||
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>. |
Revision as of 22:03, 19 January 2020
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(no trig)
We have that
But , so
Solution 2(no trig)
We can let . Since $EC=2\cdotEA$ (Error compiling LaTeX. Unknown error_msg), . 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
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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