To find the length of the base of the triangle, we can use the formula for the area of a triangle:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Let the height of the triangle be \( h \) cm. Then, the base of the triangle can be expressed as:
\[ \text{base} = h + 3 \text{ cm} \quad \text{(since the base is 3 cm longer than the height)} \]
According to the problem, the area of the triangle is 44 cm². We can set up the equation using the area formula:
\[ 44 = \frac{1}{2} \times (h + 3) \times h \]
Simplifying this, we multiply both sides by 2 to eliminate the fraction:
\[ 88 = (h + 3) \times h \]
Expanding the right side:
\[ 88 = h^2 + 3h \]
Rearranging the equation gives us a standard quadratic form:
\[ h^2 + 3h - 88 = 0 \]
Now, we can use the quadratic formula to find \( h \):
\[ h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 3 \), and \( c = -88 \):
Calculating the discriminant:
\[ b^2 - 4ac = 3^2 - 4 \times 1 \times (-88) = 9 + 352 = 361 \]
Now substituting this back into the quadratic formula:
\[ h = \frac{-3 \pm \sqrt{361}}{2 \times 1} = \frac{-3 \pm 19}{2} \]
Calculating the two possible values for \( h \):
- \( h = \frac{16}{2} = 8 \)
- \( h = \frac{-22}{2} = -11 \) (not a valid height since it can't be negative)
So, we have:
\[ h = 8 \text{ cm} \]
Now, substituting \( h \) back to find the base:
\[ \text{base} = h + 3 = 8 + 3 = 11 \text{ cm} \]
Thus, the length of the base is:
\[ \boxed{11 \text{ cm}} \]