**Solution:**

To find the base of the isosceles triangle, we need to use the formula for the area of a triangle.

The formula for the area of a triangle is given by:

**Area = (1/2) * base * height**

where the base is the length of the triangle's base and the height is the perpendicular distance from the base to the opposite vertex.

In the case of an isosceles triangle, the height is the line segment perpendicular to the base that bisects the opposite side.

Let's denote the base of the triangle as 'b' and the height as 'h'.

Given that the area of the triangle is 12 cm² and one of the equal sides is 5 cm, we can set up the following equation:

**12 = (1/2) * b * h**

We need one more equation to solve for both 'b' and 'h'. Since the triangle is isosceles, we know that the two equal sides are of length 5 cm each.

Using the Pythagorean theorem, we can find the height 'h' in terms of 'b':

**h² = 5² - (b/2)²**

**h² = 25 - (b²/4)**

Now we can substitute this expression for 'h' in the area equation:

**12 = (1/2) * b * √(25 - (b²/4))**

We can simplify this equation to solve for 'b' using algebraic methods.

Multiplying both sides by 2:

**24 = b * √(25 - (b²/4))**

Squaring both sides to eliminate the square root:

**576 = b² * (25 - (b²/4))**

Expanding and rearranging the terms:

**576 = 25b² - (b⁴/4)**

Multiplying both sides by 4 to remove the fraction:

**2304 = 100b² - b⁴**

Rearranging the terms:

**b⁴ - 100b² + 2304 = 0**

This is a quadratic equation in terms of 'b²'. We can solve it by factoring or by using the quadratic formula.

Once we find the values of 'b²', we can take their square roots to find the possible values of 'b'.

Finally, we can substitute these values of 'b' into the equation for 'h' to find the corresponding heights.

Therefore, by solving the quadratic equation and substituting the values of 'b' into the equation for 'h', we can find the possible values of the base and height of the isosceles triangle.