Difference between revisions of "2005 AMC 10A Problems/Problem 25"
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&= \frac{1}{[ABC]/[ADE] - 1} \\ | &= \frac{1}{[ABC]/[ADE] - 1} \\ | ||
&= \frac{1}{75/19 - 1} \\ | &= \frac{1}{75/19 - 1} \\ | ||
− | &= \boxed{\frac{19}{56}}. | + | &= \boxed{\frac{19}{56}\LongrightarrowD}. |
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
− | |||
==Solution (trig)== | ==Solution (trig)== |
Revision as of 18:40, 21 August 2018
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
\begin{align*} \frac{[ADE]}{[BCED]} &= \frac{[ADE]}{[ABC] - [ADE]} \\ &= \frac{1}{[ABC]/[ADE] - 1} \\ &= \frac{1}{75/19 - 1} \\ &= \boxed{\frac{19}{56}\LongrightarrowD}. \end{align*} (Error compiling LaTeX. Unknown error_msg)
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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