Trapezoid Side Length: Solving AB Where Area = 3AB and DC = 2AB

To solve this problem, we'll utilize the information given about trapezoid $ABCD$:

• $AB = AD$ which implies $x = x = AB$.
• $DC = 2 \times AB$ implies $DC = 2x$.
• The formula for the area of a trapezoid is given by: $\text{Area} = \frac{1}{2} \times (\text{Base}_1 + \text{Base}_2) \times \text{Height}$.
• The area of the trapezoid is stated as three times longer than side $AB$, giving us $\text{Area} = 3 \times x$.

The bases of trapezoid $ABCD$ are $AB = x$ and $DC = 2x$. Assume the height of trapezoid $ABCD$ is $h$.

Using the area formula, we have:
$\frac{1}{2} \times (x + 2x) \times h = 3x$

This simplifies to:
$\frac{3x}{2} \times h = 3x$

To find $h$, divide both sides by $\frac{3x}{2}$ this yields:
$h = 2$

Next, verify that when $h = 2$, the area calculation matches:
Substitute $h = 2$ back into the expression for area:
$\frac{1}{2} \times 3x \times 2 = 3x$, which holds true as $3x = 3x$.

Thus, the calculations confirm the length of side $AB$ is $2$.