An isosceles triangle has sides A, B, and C, such that sides A and B have the same length. Side C has a length of #32 # and the triangle has an area of #16 #. What are the lengths of sides A and B?

Using the formula for the area of a triangle (A = \frac{1}{2} \times \text{base} \times \text{height}), where the base is any side of the triangle and the height is the perpendicular distance from the base to the opposite vertex, we can solve for the height of the triangle.

Let's denote the length of sides A and B as (x), which are the equal sides of the isosceles triangle.

Given that the area of the triangle is 16, we have:

[16 = \frac{1}{2} \times x \times h]

Where (h) is the height of the triangle.

Now, we need to find the height of the triangle. We can use the Pythagorean theorem to find the height:

[h^2 = 32^2 - \left(\frac{x}{2}\right)^2]

Now, we have two equations:

[16 = \frac{1}{2} \times x \times h] [h^2 = 32^2 - \left(\frac{x}{2}\right)^2]

We can substitute the second equation into the first to solve for (x), the length of sides A and B. After solving for (x), we can find the lengths of sides A and B.