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 23:00, 29 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
![[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 
![\[\frac{2}{5} = \frac{h-3}{h}\]](http://latex.artofproblemsolving.com/1/4/0/140780e0cdc7959d758ae0976f1715fcc6ffbfa0.png)
![\[2h = 5h-15\]](http://latex.artofproblemsolving.com/d/1/8/d18ea1812352a00725cd0a4938c2adbc85eb9332.png)
![\[3h = 15\]](http://latex.artofproblemsolving.com/a/0/9/a094622098ec270ad387c750fdeb057742bb8196.png)
![\[h = 5\]](http://latex.artofproblemsolving.com/a/d/f/adf85c076b2be16b9a1042f4742a0572d100110d.png)
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. 
