Difference between revisions of "2004 AMC 12A Problems/Problem 8"
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Therefore, the difference is <math>4</math> | Therefore, the difference is <math>4</math> | ||
− | === Solution | + | === Solution 5 === |
− | We want to figure out <math> | + | We want to figure out <math>[\triangle ADE] - [\triangle BDC]</math>. |
Notice that <math>\triangle ABC</math> and <math>\triangle BAE</math> "intersect" and form <math>\triangle ADB</math>. | Notice that <math>\triangle ABC</math> and <math>\triangle BAE</math> "intersect" and form <math>\triangle ADB</math>. | ||
− | This means that <math> | + | This means that <math>[\triangle BAE] - [\triangle ABC)] = [\triangle ADE] - [\triangle BDC]</math> because <math>[\triangle ADB]</math> cancels out, which can be seen easily in the diagram. |
− | <math> | + | <math>[\triangle BAE] = 0.5 \cdot 4 \cdot 8 = 16</math> |
− | <math> | + | <math>[\triangle ABC] = 0.5 \cdot 4 \cdot 16 = 12</math> |
− | <math> | + | <math>[\triangle BDC] - [\triangle ADE] = 16 - 12 =\boxed{\mathrm{(B)}\ 4}</math> |
==Video Solution== | ==Video Solution== |
Latest revision as of 22:24, 28 September 2024
- The following problem is from both the 2004 AMC 12A #8 and 2004 AMC 10A #9, so both problems redirect to this page.
Contents
[hide]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 ?
Solutions
Solution 1
Looking, we see that the area of is 16 and the area of is 12. Set the area of to be x. We want to find - . So, that would be and . Therefore,
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Solution 2
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 3
Let represent the area of figure . Note that and .
Solution 4 (coordbash)
Put figure on a graph. goes from (0, 0) to (4, 6) and goes from (4, 0) to (0, 8). is on line . is on line . Finding intersection between these points,
.
This gives us the x-coordinate of D. So, is the height of , then area of is
Now, the height of is And the area of is
This gives us
Therefore, the difference is
Solution 5
We want to figure out . Notice that and "intersect" and form .
This means that because cancels out, which can be seen easily in the diagram.
Video Solution
Education, the Study of Everything
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.