2004 AMC 12A Problems/Problem 8
From AoPSWiki
(Redirected from 2004 AMC 10A Problems/Problem 9)
- 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 
?
Contents |
Solution
Solution 1
If we let ![[\ldots]](http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/e/9/1/e91765e61e85decb1d55bee8af8a32a6a3874818.gif)
denote area, ![[ABE] - [ABC] = [ADE] + [ABD] - [ABD] - [BDC] = [ADE] - [BDC]](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/4/8/e/48ea9c8a1eabae5e7e11e810aae7836594c1c680.gif)
. Using the given, ![[ABE] = \frac 12 \cdot 8 \cdot 4](http://alt2.artofproblemsolving.com/Forum/latexrender/pictures/9/4/d/94ddf815894baf388de114258cf05ee48a3e8df9.gif)
and ![[ABC] = \frac 12 \cdot 6 \cdot 4](http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/7/4/f/74f9aa0238a0db16362d314a210bad3c10230e8f.gif)
, and their difference is 
.
Solution 2
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, 
.
See also
| 2004 AMC 12A (Problems • 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 | ||
| 2004 AMC 10A (Problems • 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 | ||
