Triangle A has an area of #6 # and two sides of lengths #6 # and #4 #. Triangle B is similar to triangle A and has a side with a length of #14 #. What are the maximum and minimum possible areas of triangle B?

To find the maximum and minimum possible areas of triangle B, we need to consider the relationship between the areas of similar triangles and the ratios of their corresponding sides. Since triangle B is similar to triangle A, their corresponding sides are proportional.

Let's denote the area of triangle B as ( A_B ).

Given that triangle A has an area of 6 and two sides of lengths 6 and 4, we can use the formula for the area of a triangle: ( A = \frac{1}{2} \times \text{base} \times \text{height} ).

For triangle A, let's consider the side lengths 6 and 4. The height can be found using these side lengths and the area given. So, the height of triangle A is ( h_A = \frac{2 \times A}{\text{base}} = \frac{2 \times 6}{6} = 2 ).

Now, for triangle B, we know one of its side lengths is 14, which corresponds to the side length of 6 in triangle A. Since triangle B is similar to triangle A, the ratio of corresponding sides is the same. Therefore, the corresponding side to the height of triangle B would be ( \frac{14}{6} ) times the height of triangle A.

So, the height of triangle B is ( h_B = \frac{14}{6} \times 2 = \frac{14}{3} ).

Now, using the formula for the area of a triangle, we find that the maximum possible area of triangle B occurs when the base is 14, and the height is ( \frac{14}{3} ). Therefore, the maximum area of triangle B is ( A_B = \frac{1}{2} \times 14 \times \frac{14}{3} = \frac{98}{3} ).

The minimum possible area of triangle B occurs when the base is 4 (corresponding to the side length of 4 in triangle A) and the height is ( \frac{14}{3} ). Therefore, the minimum area of triangle B is ( A_B = \frac{1}{2} \times 4 \times \frac{14}{3} = \frac{28}{3} ).