Triangle A has an area of #4 # and two sides of lengths #8 # and #3 #. Triangle B is similar to triangle A and has a side with a length of #8 #. What are the maximum and minimum possible areas of triangle B?

Since the triangles are similar, sides are in same proportion.

The maximum and minimum possible areas of Triangle B can be found using the properties of similar triangles. Since Triangle B is similar to Triangle A, their corresponding sides are proportional.

Let the sides of Triangle B be $a$, $b$, and $c$, with $a = 8$ (the side corresponding to the side of length 8 in Triangle A).

Using the area formula for a triangle, $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$, and considering that the bases of Triangle A and Triangle B are proportional (as they correspond to sides of equal length), the ratio of the areas of Triangle A and Triangle B is the square of the ratio of their corresponding heights.

Given that the area of Triangle A is 4 and one of its sides is 8, we can find its height using the formula for the area of a triangle.

$4 = \frac{1}{2} \times 8 \times \text{height}$ $\text{height} = \frac{4}{8} = \frac{1}{2}$

Now, using this height, we can find the heights of similar triangles (Triangle B) using their corresponding sides.

For Triangle B, the corresponding side to the side of length 8 in Triangle A is also 8, so its height will also be $\frac{1}{2}$.

Now, we can calculate the minimum and maximum areas of Triangle B.

The minimum area occurs when one of the sides of Triangle B collapses to a line, resulting in a triangle with area $\frac{1}{2} \times 8 \times \frac{1}{2} = 2$.

The maximum area occurs when Triangle B is similar to Triangle A, resulting in a triangle with area $\frac{1}{2} \times 8 \times \frac{1}{2} = 2$.

Therefore, the minimum possible area of Triangle B is 2, and the maximum possible area of Triangle B is also 2.