Question

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?

Answer 1

Max A = `185.3`
Min A = `34.7`

Explanation:

From the triangle area formula `A = 1/2bh` we can select any side as ‘b’ and solve for h:
`8 = 1/2xx12h ; h = 1 1/3` Thus, we know that the unknown side is the smallest.
We can also use trigonometry to find the included angle opposite the smallest side:
`A = (bc)/2sinA`; `8 = (9xx12)/2sinA` ; `A = 8.52^o`

We now have an “SAS” triangle. We use the Law of Cosines to find the smallest side:
`a^2 = b^2 + c^2 - (2bc)cosA`; `a^2 = 9^2 + 12^2 -2xx9xx12cos8.52`
`a^2 = 11.4`; `a = 3.37`

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.

Thus, the minimum area of a similar triangle would be `A = 1/2xx25xx(25/12xx4/3) = 34.7`

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

`A = sqrt((sxx(s-a)xx(s-b)xx(s-c))` where `s = 1/2(a + b + c)` and a, b, c are the side lengths.
`s = 17.3`

`A = sqrt((17.3xx(17.3 – 12)xx(17.3 – 32)xx(17.3 – 42.7))` ; `A = sqrt((17.3xx(5.3)xx(-14.75)xx(-25.4))`
`A = sqrt(34352)` ; `A = 185.3`