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

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 $\triangle EAB,$

$$\frac{2}{5} = \frac{h-3}{h}$$ $$2h = 5h-15$$ $$3h = 15$$ $$h = 5$$

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