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Difference between revisions of "2003 AMC 10B Problems/Problem 20"
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{{duplicate|[[2003 AMC 12B Problems|2003 AMC 12B #14]] and [[2003 AMC 10B Problems|2003 AMC 10B #20]]}}
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==Problem==
 
==Problem==
 
In rectangle <math>ABCD, AB=5</math> and <math>BC=3</math>. Points <math>F</math> and <math>G</math> are on <math>\overline{CD}</math> so that <math>DF=1</math> and <math>GC=2</math>. Lines <math>AF</math> and <math>BG</math> intersect at <math>E</math>. Find the area of <math>\triangle AEB</math>.
 
In rectangle <math>ABCD, AB=5</math> and <math>BC=3</math>. Points <math>F</math> and <math>G</math> are on <math>\overline{CD}</math> so that <math>DF=1</math> and <math>GC=2</math>. Lines <math>AF</math> and <math>BG</math> intersect at <math>E</math>. Find the area of <math>\triangle AEB</math>.
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==See Also==
 
==See Also==
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{{AMC12 box|year=2003|ab=B|num-b=13|num-a=15}}
 
{{AMC10 box|year=2003|ab=B|num-b=19|num-a=21}}
 
{{AMC10 box|year=2003|ab=B|num-b=19|num-a=21}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 01:00, 5 January 2014

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 $ABCD, AB=5$ and $BC=3$. Points $F$ and $G$ are on $\overline{CD}$ so that $DF=1$ and $GC=2$. Lines $AF$ and $BG$ intersect at $E$. Find the area of $\triangle AEB$.

$\textbf{(A) } 10 \qquad\textbf{(B) } \frac{21}{2} \qquad\textbf{(C) } 12 \qquad\textbf{(D) } \frac{25}{2} \qquad\textbf{(E) } 15$

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]

$\triangle EFG \sim \triangle EAB$ because $FG \parallel AB.$ The ratio of $\triangle EFG$ to $\triangle EAB$ is $2:5$ since $AB=5$ and $FG=2$ from subtraction. If we let $h$ be the height of $\triangle EAB,$ 

\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 $5$ so the area of $\triangle EAB$ is $\frac{1}{2}(5)(5) = \boxed{\textbf{(D)}\ \frac{25}{2}}$.

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

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