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 $\triangle AEB$. Express your answer as a common fraction. [asy]
pair A,B,C,D,I,F,G;
A=(0,0);
B=(5,0);
C=(5,3);
D=(0,3);
F=(1,3);
G=(3,3);
I=(1.67,5);
draw(A--B--C--D--cycle,linewidth(0.7));
draw(A--B--I--cycle,linewidth(0.7));
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,NE);
label("$D$",D,NW);
label("$F$",F,SE);
label("$G$",G,SW);
label("$E$",I,N);
label("1",(0.5,3),N);
label("2",(4,3),N);
label("3",(0,1.5),W);
label("3",(5,1.5),E);
label("5",(2.5,0),S);
[/asy]
Let $H$ be the foot of the perpendicular from $E$ to $\overline{DC}$. Since $ CD= AB = 5$, $FG= 2$, and $ \triangle FEG$ is similar to $\triangle AEB$, we have \[\frac{EH}{EH+3} =\frac{2}{5},\quad \text{so} \quad 5EH =2EH + 6,\]and $EH = 2$. Hence \[[\triangle AEB] = \frac{1}{2}(2 + 3)\cdot 5 = \boxed{\frac{25}{2}}.\][asy]
pair A,B,C,D,I,F,G,H;
H=(1.66,3);
A=(0,0);
B=(5,0);
C=(5,3);
D=(0,3);
F=(1,3);
G=(3,3);
I=(1.67,5);
draw(A--B--C--D--cycle,linewidth(0.7));
draw(A--B--I--cycle,linewidth(0.7));
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,NE);
label("$D$",D,NW);
label("$F$",F,SE);
label("$G$",G,SW);
label("$E$",I,N);
label("1",(0.5,3),N);
label("2",(4,3),N);
label("3",(0,1.5),W);
label("3",(5,1.5),E);
label("5",(2.5,0),S);
draw(I--H,linewidth(0.7));
label("$H$",H,S);
[/asy]
\frac{25}{2}
Level 4
Geometry
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 $\triangle AEB$. Express your answer as a common fraction. [asy]
pair A,B,C,D,I,F,G;
A=(0,0);
B=(5,0);
C=(5,3);
D=(0,3);
F=(1,3);
G=(3,3);
I=(1.67,5);
draw(A--B--C--D--cycle,linewidth(0.7));
draw(A--B--I--cycle,linewidth(0.7));
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,NE);
label("$D$",D,NW);
label("$F$",F,SE);
label("$G$",G,SW);
label("$E$",I,N);
label("1",(0.5,3),N);
label("2",(4,3),N);
label("3",(0,1.5),W);
label("3",(5,1.5),E);
label("5",(2.5,0),S);
[/asy]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the area of triangle AEB, we first determine the coordinates of points A, B, C, D, F, and G based on the rectangle's dimensions. Let A be at (0, 0), B at (5, 0), C at (5, 3), and D at (0, 3). Point F is located at (1, 3) since DF = 1, and point G is at (3, 3) because GC = 2.

Next, we need to find the equations of lines AF and BG. The slope of line AF can be calculated as follows: the change in y over the change in x from A to F is (3 - 0) / (1 - 0) = 3. Thus, the equation of line AF, using point-slope form, is y = 3x.

For line BG, the slope is determined similarly: the change in y over the change in x from B to G is (3 - 0) / (3 - 5) = -3/2. The equation of line BG, again using point-slope form, is y = -3/2(x - 5), which simplifies to y = -3/2x + 15/2.

To find the intersection point E of lines AF and BG, we set the equations equal to each other: 3x = -3/2x + 15/2. Solving for x, we combine like terms: 3x + 3/2x = 15/2. This gives us (6/2 + 3/2)x = 15/2, leading to (9/2)x = 15/2. Thus, x = 15/9 = 5/3.

Substituting x = 5/3 back into the equation of line AF to find y, we have y = 3(5/3) = 5. Therefore, the coordinates of point E are (5/3, 5).

Now, we can calculate the area of triangle AEB using the formula for the area of a triangle given by vertices (x1, y1), (x2, y2), and (x3, y3): Area = 1/2 | x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2) |. Substituting the coordinates of points A(0, 0), B(5, 0), and E(5/3, 5), we have:

Area = 1/2 | 0(0 - 5) + 5(5 - 0) + (5/3)(0 - 0) | = 1/2 | 0 + 25 + 0 | = 1/2 * 25 = 25/2.

Thus, the final answer is 25/2.<|END_OF_LATENT|> The final answer is $\frac{25}{2}$.