Difference between revisions of "2004 AMC 10B Problems/Problem 18"
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==Solution 3== | ==Solution 3== | ||
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+ | The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9), and let F be (12, 3). You can then use the shoelace theorem to find the area of DBF, which is 42. <math> \frac {42}{96} = \boxed{\frac 7{16}}</math> | ||
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+ | ==Solution 4== | ||
You can also place a point <math>X</math> on <math>CE</math> such that <math>CX</math> is <math>12</math>, creating trapezoid <math>CBFX</math>. Then, you can find the area of the trapezoid, subtract the area of the two right triangles <math>DFX</math> and <math>BCD</math>, divide by the area of <math>ABC</math>, and get the ratio of <math>7/16</math>. | You can also place a point <math>X</math> on <math>CE</math> such that <math>CX</math> is <math>12</math>, creating trapezoid <math>CBFX</math>. Then, you can find the area of the trapezoid, subtract the area of the two right triangles <math>DFX</math> and <math>BCD</math>, divide by the area of <math>ABC</math>, and get the ratio of <math>7/16</math>. |
Revision as of 21:30, 3 January 2020
Problem
In the right triangle , we have , , and . Points , , and are located on , , and , respectively, so that , , and . What is the ratio of the area of to that of ?
Solution 1
Let . Because is divided into four triangles, .
Because of triangle area, .
and , so .
, so .
Solution 2
First of all, note that , and therefore .
Draw the height from onto as in the picture below:
Now consider the area of . Clearly the triangles and are similar, as they have all angles equal. Their ratio is , hence . Now the area of can be computed as = .
Similarly we can find that as well.
Hence , and the answer is .
Solution 3
The area of triangle ACE is 96. To find the area of triangle DBF, let D be (4, 0), let B be (0, 9), and let F be (12, 3). You can then use the shoelace theorem to find the area of DBF, which is 42.
Solution 4
You can also place a point on such that is , creating trapezoid . Then, you can find the area of the trapezoid, subtract the area of the two right triangles and , divide by the area of , and get the ratio of .
See also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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.