Triangle A has an area of #12 # and two sides of lengths #6 # and #9 #. Triangle B is similar to triangle A and has a side with a length of #15 #. What are the maximum and minimum possible areas of triangle B?

The maximum and minimum possible areas of Triangle B can be found by considering the ratios of the corresponding sides of similar triangles. Since Triangle B is similar to Triangle A, the ratio of corresponding sides of Triangle B to Triangle A is constant.

Let ( k ) be the ratio of corresponding sides of Triangle B to Triangle A.

Given that Triangle A has an area of 12 and sides of lengths 6 and 9, its area can be calculated using Heron's formula. Let ( s ) be the semi-perimeter of Triangle A.

[ s = \frac{6 + 9 + \sqrt{(6)^2 + (9)^2}}{2} = \frac{15 + \sqrt{117}}{2} ]

[ \text{Area of Triangle A} = \sqrt{s(s-6)(s-9)(s-\sqrt{117})} ]

[ 12 = \sqrt{s(s-6)(s-9)(s-\sqrt{117})} ]

Solve this equation to find the value of ( s ).

Once you find ( s ), use it to calculate the ratio ( k ) by dividing the corresponding side of Triangle B (15) by the corresponding side of Triangle A.

[ k = \frac{15}{s} ]

Then, find the maximum and minimum possible areas of Triangle B by applying the ratio ( k ) to the area of Triangle A.

[ \text{Maximum area of Triangle B} = k^2 \times \text{Area of Triangle A} ]

[ \text{Minimum area of Triangle B} = \left(\frac{k}{2}\right)^2 \times \text{Area of Triangle A} ]