Difference between revisions of "2004 AMC 10B Problems/Problem 18"
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Let <math>x = [DBF]</math>. Because <math>\triangle ACE</math> is divided into four triangles, <math>[ACE] = [BCD] + [ABF] + [DEF] + x</math>. | Let <math>x = [DBF]</math>. Because <math>\triangle ACE</math> is divided into four triangles, <math>[ACE] = [BCD] + [ABF] + [DEF] + x</math>. | ||
− | Because of <math>SAS</math> triangle area, \frac12 \cdot 12 \cdot 16 = \frac12 \cdot 9 \cdot 4 + \frac12 \cdot 3 \cdot 15 \cdot \sin(\angle A) + \frac12 \cdot 5 \cdot 12 \cdot \sin(\angle E) + x<math>. | + | Because of <math>SAS</math> triangle area, <math>\frac12 \cdot 12 \cdot 16 = \frac12 \cdot 9 \cdot 4 + \frac12 \cdot 3 \cdot 15 \cdot \sin(\angle A) + \frac12 \cdot 5 \cdot 12 \cdot \sin(\angle E) + x</math>. |
− | < | + | <math>\sin(\angle A) = \frac{16}{20}</math> and <math>\sin(\angle E) = \frac{12}{20}</math>, so <math>96 = 18 + 18 + 18 + x</math>. |
− | < | + | <math>x = 42</math>, so <math>\frac{[DBF]}{[ACE]} = \frac{42}{96} = \boxed{\textbf{(E)}\frac 7{16}}</math>. |
==Solution 2== | ==Solution 2== |
Revision as of 21:37, 24 January 2015
Contents
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 .
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 |
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