Difference between revisions of "2005 AMC 10A Problems/Problem 25"
Mathwiz0803 (talk | contribs) (→Solution (no trig)) |
Mathwiz0803 (talk | contribs) (→Solution (no trig)) |
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But <math>[BCED] = [ABC] - [ADE]</math>, so | But <math>[BCED] = [ABC] - [ADE]</math>, so | ||
+ | <cmath> | ||
\begin{align*} | \begin{align*} | ||
\frac{[ADE]}{[BCED]} &= \frac{[ADE]}{[ABC] - [ADE]} \ | \frac{[ADE]}{[BCED]} &= \frac{[ADE]}{[ABC] - [ADE]} \ | ||
Line 40: | Line 41: | ||
&= \boxed{\frac{19}{56}}. | &= \boxed{\frac{19}{56}}. | ||
\end{align*} | \end{align*} | ||
+ | </cmath> | ||
The answer is therefore <math>D</math>. | The answer is therefore <math>D</math>. | ||
Revision as of 19:55, 3 November 2016
Contents
[hide]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
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) | ||
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.