2005 AMC 10A Problems/Problem 25
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 (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}}.
\end{align*}
The answer is therefore
.
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) | ||
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