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.

We are given a rectangle $ABCD$ with $AB = 5$ and $BC = 3$. Points $F$ and $G$ are located on side $\overline{CD}$ such that $DF = 1$ and $GC = 2$. Lines $AF$ and $BG$ intersect at point $E$. We are asked to find the area of triangle $\triangle AEB$.

### Step 1: Assigning Coordinates
Let's assign coordinates to the points:
- $A = (0, 0)$
- $B = (5, 0)$
- $C = (5, 3)$
- $D = (0, 3)$

Given $DF = 1$ and $GC = 2$, the coordinates of $F$ and $G$ are:
- $F = (1, 3)$ because $F$ is 1 unit to the right of $D$.
- $G = (3, 3)$ because $G$ is 2 units to the left of $C$.

### Step 2: Finding the Equations of Lines $AF$ and $BG$
**Line $AF$:**
- $A = (0, 0)$
- $F = (1, 3)$

The slope of $AF$ is:
\[
\text{slope of } AF = \frac{3 - 0}{1 - 0} = 3
\]
Thus, the equation of line $AF$ is:
\[
y = 3x
\]

**Line $BG$:**
- $B = (5, 0)$
- $G = (3, 3)$

The slope of $BG$ is:
\[
\text{slope of } BG = \frac{3 - 0}{3 - 5} = -\frac{3}{2}
\]
Thus, the equation of line $BG$ is:
\[
y = -\frac{3}{2}x + \frac{15}{2}
\]

### Step 3: Finding the Coordinates of $E$
We find the coordinates of $E$ by solving the system of equations for $AF$ and $BG$.

Setting $y = 3x$ equal to $y = -\frac{3}{2}x + \frac{15}{2}$:
\[
3x = -\frac{3}{2}x + \frac{15}{2}
\]
Multiply the entire equation by 2 to eliminate the fraction:
\[
6x = -3x + 15
\]
Add $3x$ to both sides:
\[
9x = 15
\]
Divide by 9:
\[
x = \frac{15}{9} = \frac{5}{3}
\]
Substituting $x = \frac{5}{3}$ into $y = 3x$:
\[
y = 3\left(\frac{5}{3}\right) = 5
\]
Thus, $E$ has coordinates $\left(\frac{5}{3}, 5\right)$.

### Step 4: Calculating the Area of $\triangle AEB$
We use the determinant formula to find the area of triangle $AEB$:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substitute the coordinates:
- $A = (0, 0)$
- $B = (5, 0)$
- $E = \left(\frac{5}{3}, 5\right)$

Thus:
\[
\text{Area} = \frac{1}{2} \left| 0(0 - 5) + 5\left(5 - 0\right) + \frac{5}{3}\left(0 - 0\right) \right|
\]
Simplifying:
\[
\text{Area} = \frac{1}{2} \left| 0 + 25 + 0 \right| = \frac{1}{2} \times 25 = \frac{25}{2}
\]

Therefore, the area of $\triangle AEB$ is $\boxed{\frac{25}{6}}$.

Learn how to find the area of triangle AEB within rectangle ABCD using coordinate geometry. This problem involves determining coordinates, calculating slopes of lines, and applying the formula for the area of a triangle defined by coordinates. Suitable for high school students.

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