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Difference between revisions of "2003 AMC 10B Problems/Problem 20"

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<math>\textbf{(A) } 10 \qquad\textbf{(B) } \frac{21}{2} \qquad\textbf{(C) } 12 \qquad\textbf{(D) } \frac{25}{2} \qquad\textbf{(E) } 15</math>
 
<math>\textbf{(A) } 10 \qquad\textbf{(B) } \frac{21}{2} \qquad\textbf{(C) } 12 \qquad\textbf{(D) } \frac{25}{2} \qquad\textbf{(E) } 15</math>
  
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==Solution==
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==Solution 1==
  
 
<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>
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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>.
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==Solution 2==
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We can look at this diagram as if it were a coordinate plane with point <math>A</math> being <math>(0,0)</math>. This means that the equation of the line <math>AE</math> is <math>y=3x</math> and the equation of the line <math>EB</math> is <math>y=\frac{-3}{2}x+\frac{15}{2}</math>. From this we can set of the follow equation to find the <math>x</math> coordinate of point <math>e</math>:
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<cmath>3x=\frac{-3}{2}x+\frac{15}{2}</cmath>
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<cmath>6x=-3x+15</cmath>
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<cmath>9x=15</cmath>
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<cmath>x=\frac{5}{3}</cmath>
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We can plug this into one of our original equations to find that the <math>y</math> coordinate is <math>5</math>, meaning 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==

Revision as of 12:21, 3 January 2016

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 $\triangle AEB$.

[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]

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

Solution 1

$\triangle EFG \sim \triangle EAB$ 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 $\triangle EAB,$

\[\frac{2}{5} = \frac{h-3}{h}\] \[2h = 5h-15\] \[3h = 15\] \[h = 5\]

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

Solution 2

We can look at this diagram as if it were a coordinate plane with point $A$ being $(0,0)$. This means that the equation of the line $AE$ is $y=3x$ and the equation of the line $EB$ is $y=\frac{-3}{2}x+\frac{15}{2}$. From this we can set of the follow equation to find the $x$ coordinate of point $e$:

\[3x=\frac{-3}{2}x+\frac{15}{2}\] \[6x=-3x+15\] \[9x=15\] \[x=\frac{5}{3}\]

We can plug this into one of our original equations to find that the $y$ coordinate is $5$, meaning 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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