Difference between revisions of "2004 AMC 10B Problems/Problem 18"
(New page: == Problem == In the right triangle <math>\triangle ACE</math>, we have <math>AC=12</math>, <math>CE=16</math>, and <math>EA=20</math>. Points <math>B</math>, <math>D</math>, and <math>F<...) |
m (→Solution) |
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Line 52: | Line 52: | ||
label("$F$",F,NE); | label("$F$",F,NE); | ||
label("$3$",A--B,W); | label("$3$",A--B,W); | ||
− | label("$9$",0. | + | label("$9$",0.5*C + 0.5*B,4*W); |
label("$4$",C--D,S); | label("$4$",C--D,S); | ||
label("$12$",D--E,S); | label("$12$",D--E,S); |
Revision as of 15:12, 7 February 2009
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
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 |