Difference between revisions of "2004 AMC 12A Problems/Problem 8"
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<math>[\triangle ADE]-[\triangle BDC]=[\triangle BEA]-[\triangle BCA]=\frac{1}{2}\times8\times4-\frac{1}{2}\times6\times4= 16-12=4\Rightarrow\boxed{\mathrm{(B)}\ 4}</math> | <math>[\triangle ADE]-[\triangle BDC]=[\triangle BEA]-[\triangle BCA]=\frac{1}{2}\times8\times4-\frac{1}{2}\times6\times4= 16-12=4\Rightarrow\boxed{\mathrm{(B)}\ 4}</math> | ||
− | === Solution 3 === | + | === Solution 3 (coordbash)=== |
Put figure <math>ABCDE</math> on a graph. <math>\overline{AC}</math> goes from (0, 0) to (4, 6) and <math>\overline{BE}</math> goes from (4, 0) to (0, 8). <math>\overline{AC}</math> is on line <math>y = 1.5x</math>. <math>\overline{BE}</math> is on line <math>y = -2x + 8</math>. Finding intersection between these points, | Put figure <math>ABCDE</math> on a graph. <math>\overline{AC}</math> goes from (0, 0) to (4, 6) and <math>\overline{BE}</math> goes from (4, 0) to (0, 8). <math>\overline{AC}</math> is on line <math>y = 1.5x</math>. <math>\overline{BE}</math> is on line <math>y = -2x + 8</math>. Finding intersection between these points, | ||
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Therefore, the difference is <math>4</math> | Therefore, the difference is <math>4</math> | ||
+ | |||
+ | === Solution 4 === | ||
+ | We want to figure out <math>Area(\triangle ADE) - Area(\triangle BDC)</math>. | ||
+ | Notice that <math>\triangle ABC</math> and <math>\triangle BAE</math> "intersect" and form <math>\triangle ADB</math>. | ||
+ | |||
+ | This means that <math>Area(\triangle BAE) - Area(\triangle ABC) = Area(\triangle ADE) - Area(\triangle BDC)</math> because <math>Area(\triangle ADB)</math> cancels out, which can be seen easily in the diagram. | ||
+ | |||
+ | <math>Area(\triangle BAE) = 0.5 * 4 * 8 = 16</math> | ||
+ | |||
+ | <math>Area(\triangle ABC) = 0.5 * 4 * 16 = 12</math> | ||
+ | |||
+ | <math>Area(\triangle BDC) - Area(\triangle ADE) = 16 - 12 =\boxed{\mathrm{(B)}\ 4}</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/DlA71MBSviU | ||
+ | |||
+ | Education, the Study of Everything | ||
+ | |||
== See also == | == See also == |
Revision as of 07:25, 6 February 2023
- The following problem is from both the 2004 AMC 12A #8 and 2004 AMC 10A #9, so both problems redirect to this page.
Contents
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
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 .
Solution 3 (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 4
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.