# Triangle A has an area of #8 # and two sides of lengths #9 # and #12 #. Triangle B is similar to triangle A and has a side with a length of #25 #. What are the maximum and minimum possible areas of triangle B?

Max A =

Min A =

The largest similar triangle would have the given length of 25 as the shortest side, and the minimum area would have it as the longest side, corresponding to the 12 of the original.

We can use Heron’s Formula to solve for the Area with three sides. Ratios: 3.37:9 :12 = 12: 32: 42.7

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The maximum possible area of triangle B can be found when its side length is maximized. Since triangle B is similar to triangle A, the ratio of corresponding sides is constant. Therefore, if one side of triangle B is 25, then the corresponding side in triangle A would be (25/12) * 9 = 18.75. Now, to find the maximum area of triangle B, we use the formula for the area of a triangle, which is 0.5 * base * height. The base of triangle B is 25 and the corresponding height can be calculated using the base and the area of triangle A. So, the maximum area of triangle B = 0.5 * 25 * (8 / 18.75) ≈ 5.33 square units.

The minimum possible area of triangle B can be found when its side length is minimized. Again, using the ratio of corresponding sides, if one side of triangle B is 25, then the corresponding side in triangle A would be (25/12) * 9 = 18.75. To find the minimum area of triangle B, we use the formula for the area of a triangle. The base of triangle B is 25 and the corresponding height can be calculated using the base and the area of triangle A. So, the minimum area of triangle B = 0.5 * 25 * (8 / 18.75) ≈ 5.33 square units.

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