Difference between revisions of "2004 AMC 12A Problems/Problem 8"
(→See also) |
(→Solution 1) |
||
Line 9: | Line 9: | ||
__TOC__ | __TOC__ | ||
== Solution == | == Solution == | ||
− | + | == Solution 1 == | |
− | Since <math>AE \perp AB</math> and <math>BC \perp AB</math>, <math>AE \parallel BC</math>. By | + | |
+ | Since <math>AE \perp AB</math> and <math>BC \perp AB</math>, <math>AE \parallel BC</math>. By alternate interior angles and <math>AA\sim</math>, we find that <math>\triangle ADE \sim \triangle CDB</math>, with side length ratio <math>\frac{4}{3}</math>. Their heights also have the same ratio, and since the two heights add up to <math>4</math>, we have that <math>h_{ADE} = 4 \cdot \frac{4}{7} = \frac{16}{7}</math> and <math>h_{CDB} = 3 \cdot \frac 47 = \frac {12}7</math>. Subtracting the areas, <math>\frac{1}{2} \cdot 8 \cdot \frac {16}7 - \frac 12 \cdot 6 \cdot \frac{12}7 = 4</math> <math>\Rightarrow</math> <math>\boxed{\mathrm{(B)}\ 4}</math>. | ||
=== Solution 2 === | === Solution 2 === |
Revision as of 00:27, 21 July 2014
- 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 , 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 represent the area of figure . Note that and .
.
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.