Difference between revisions of "2005 AMC 10A Problems/Problem 25"
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== Solution 4 == | == Solution 4 == | ||
− | + | Let <math>F</math> be on <math>AC</math> such that <math>DE\parallel BF</math> then we have | |
− | + | <cmath>\frac{<cmath>ABF</cmath>}{\[ADE]\}=(\frac{AD}{AE})^2=(\frac{19}{25})^2=\frac{361}{625}</cmath> | |
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==See also== | ==See also== |
Revision as of 22:42, 28 December 2019
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
,
.
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
\[\frac{<cmath>ABF\] (Error compiling LaTeX. Unknown error_msg)
}{\[ADE]\}=(\frac{AD}{AE})^2=(\frac{19}{25})^2=\frac{361}{625}</cmath>
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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