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 == | ||
+ | |||
+ | Put figure <math>ABCDE</math> on 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, | ||
+ | |||
+ | <math>1.5x = -2x + 8</math>. | ||
+ | |||
+ | <math>3.5x = 8 </math> | ||
+ | |||
+ | <math> x = 8 \times \frac{2}{7}</math> | ||
+ | |||
+ | <math> = \frac{16}{7}</math> | ||
+ | |||
+ | This gives us the x-coordinate of D. | ||
+ | So, <math>\frac{16}{7}</math> is the height of <math>\triangle ADE</math>, then area of <math>\triangle ADE</math> is | ||
+ | <math>\frac{16}{7} \times 8 \times \frac{1}{2}</math> | ||
+ | <math> = \frac{64}{7}</math> | ||
+ | |||
+ | Now, the height of <math>\triangle BDC</math> is <math>4-\frac{16}{7} = \frac{12}{7}</math> | ||
+ | And the area of <math>\triangle BDC</math> is <math>6 \times \frac{12}{7} \times \frac{1}{2} = \frac{36}{7}</math> | ||
+ | |||
+ | This gives us <math>\frac{64}{7} - \frac{36}{7} = 4</math> | ||
+ | |||
+ | \dot{.\hspace{.095in}.}\hspace{.5in} The difference is <math>4</math> | ||
== See also == | == See also == |
Revision as of 15:42, 5 July 2018
- 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
[hide]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
Put figure on 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
\dot{.\hspace{.095in}.}\hspace{.5in} The difference is
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.