Triangle A has an area of #27 # and two sides of lengths #8 # and #6 #. Triangle B is similar to triangle A and has a side with a length of #6 #. What are the maximum and minimum possible areas of triangle B?

The maximum possible area of triangle B occurs when it is similar to triangle A and the side with length 6 in triangle B is the hypotenuse of triangle A. In this case, the other two sides of triangle B would be proportional to the sides of triangle A, with lengths of 8 and ( \sqrt{8^2 - 6^2} = \sqrt{64 - 36} = \sqrt{28} ).

The area of triangle B can be calculated using the formula for the area of a triangle:

[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} ]

In this case, the base is 8 and the height is ( \sqrt{28} ).

[ \text{Area} = \frac{1}{2} \times 8 \times \sqrt{28} = 4 \times \sqrt{28} ]

The minimum possible area of triangle B occurs when it is similar to triangle A and the side with length 6 in triangle B is one of the shorter sides of triangle A. In this case, the other two sides of triangle B would also be proportional to the sides of triangle A, with lengths of ( \frac{6}{8} \times 6 = \frac{9}{2} ) and ( \frac{6}{8} \times \sqrt{28} = \frac{3\sqrt{28}}{2} ).

The area of triangle B can be calculated using the same formula:

[ \text{Area} = \frac{1}{2} \times \frac{9}{2} \times \frac{3\sqrt{28}}{2} = \frac{27\sqrt{28}}{8} ]

So, the maximum possible area of triangle B is ( 4 \times \sqrt{28} ) and the minimum possible area is ( \frac{27\sqrt{28}}{8} ).