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 $\boxed{\frac{25}{6}}$.