2004 AMC 12A Problems/Problem 8
- The following problem is from both the 2004 AMC 12A #8 and 2004 AMC 10A #9, so both problems redirect to this page.
Problem
In the overlapping triangles 
and 
sharing common side 
, 
and 
are right angles, 
, 
, 
, and 
and 
intersect at 
. What is the difference between the areas of 
and 
?
Solution
Solution 1
Since 
and 
, 
. By alternate interior angles and AA~, we find that 
, with side length ratio 
. Their heights also have the same ratio, and since the two heights add up to 
, we have that 
and 
. Subtracting the areas, 
.
Solution 2
Let ![$[X]$](http://latex.artofproblemsolving.com/f/6/3/f631f7a53d834de52b718ed6b496689e9658eadf.png)
represent the area of figure 
. Note that ![$[\triangle BEA]=[\triangle ABD]+[\triangle ADE]$](http://latex.artofproblemsolving.com/e/d/f/edf0f7f89e567669b122e8c013b4ca11e37dd631.png)
and ![$[\triangle BCA]=[\triangle ABD]+[\triangle BDC]$](http://latex.artofproblemsolving.com/0/8/6/086fd4bca85db5d287e525a878f3d9038f61c369.png)
.
![$[\triangle ADE]-[\triangle BDC]=[\triangle BEA]-[\triangle BCA]=\frac{1}{2} 8*4-\frac{1}{2} 6*4= 16-12=4$](http://latex.artofproblemsolving.com/b/d/a/bda9f11066b6d2492d624abb89375310e2605f0e.png)
.
See also
| 2004 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 7 |
Followed by Problem 9 |
| 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 12 Problems and Solutions | |
| 2004 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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. 
