Difference between revisions of "2003 AMC 10B Problems/Problem 20"
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| − | <math>\triangle EFG \sim \triangle EAB</math> because <math>FB \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 | + | <math>\triangle EFG \sim \triangle EAB</math> because <math>FB \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}{h | + | <cmath>\begin{align*}\frac{2}{5} &= \frac{h-3}{h}\\ |
| − | 2h | + | 2h &= 5h-15\\ |
| − | 3h &= | + | 3h &= 15\\ |
| − | h &= | + | 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>. | |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2003|ab=B|num-b=19|num-a=21}} | {{AMC10 box|year=2003|ab=B|num-b=19|num-a=21}} | ||
Revision as of 19:47, 26 November 2011
Problem
In rectangle 
and 
. Points 
and 
are on 
so that 
and 
. Lines 
and 
intersect at 
. Find the area of 
.
Solution
![[asy] unitsize(8mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(5,0), C=(5,3), D=(0,3); pair F=(1,3), G=(3,3); pair E=(5/3,5); draw(A--B--C--D--cycle); draw(A--E); draw(B--E); pair[] ps={A,B,C,D,E,F,G}; dot(ps); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",E,N); label("$F$",F,SE); label("$G$",G,SW); label("$1$",midpoint(D--F),N); label("$2$",midpoint(G--C),N); label("$5$",midpoint(A--B),S); label("$3$",midpoint(A--D),W); [/asy]](http://latex.artofproblemsolving.com/1/a/c/1ac3fc6b18ce26a1ce9b9a1e0e9c2c73bed94a30.png)
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 | ||
