Difference between revisions of "2003 AMC 10B Problems/Problem 20"
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− | <math>\triangle EFG \sim \triangle EAB</math> because <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>\begin{align*}\frac{2}{5} &= \frac{h-3}{h}\\ | <cmath>\begin{align*}\frac{2}{5} &= \frac{h-3}{h}\\ |
Revision as of 23:15, 28 February 2012
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
\begin{align*}\frac{2}{5} &= \frac{h-3}{h}\\ 2h &= 5h-15\\ 3h &= 15\\ h &= 5 (Error compiling LaTeX. Unknown error_msg)
The height is so the area of is .
See Also
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 |