Difference between revisions of "2003 AMC 10B Problems/Problem 20"
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<math>\triangle EFG \sim \triangle EAB</math> because <math>FG \parallel AB.</math> The ratio of <math>\triangle EFG</math> to <math>\triangle EAB</math> is <math>2:5</math> since <math>AB=5</math> and <math>FG=2</math> from subtraction. If we let <math>h</math> be the height of <math>\triangle EAB,</math> | <math>\triangle EFG \sim \triangle EAB</math> because <math>FG \parallel AB.</math> The ratio of <math>\triangle EFG</math> to <math>\triangle EAB</math> is <math>2:5</math> since <math>AB=5</math> and <math>FG=2</math> from subtraction. If we let <math>h</math> be the height of <math>\triangle EAB,</math> | ||
− | <cmath>\frac{2}{5} = \frac{h-3}{h} | + | <cmath>\frac{2}{5} = \frac{h-3}{h}</cmath> |
− | 2h = 5h-15 | + | <cmath>2h = 5h-15</cmath> |
− | 3h = 15 | + | <cmath>3h = 15</cmath> |
− | h = 5</cmath> | + | <cmath>h = 5</cmath> |
The height is <math>5</math> so the area of <math>\triangle EAB</math> is <math>\frac{1}{2}(5)(5) = \boxed{\textbf{(D)}\ \frac{25}{2}}</math>. | The height is <math>5</math> so the area of <math>\triangle EAB</math> is <math>\frac{1}{2}(5)(5) = \boxed{\textbf{(D)}\ \frac{25}{2}}</math>. |
Revision as of 00:00, 30 December 2015
- The following problem is from both the 2003 AMC 12B #14 and 2003 AMC 10B #20, so both problems redirect to this page.
Problem
In rectangle and . Points and are on so that and . Lines and intersect at . Find the area of .
Solution
because The ratio of to is since and from subtraction. If we let be the height of
The height is so the area of is .
See Also
2003 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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 |
2003 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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.