Question

Triangle A has an area of `24` and two sides of lengths `8` and `12`. Triangle B is similar to triangle A and has a side of length `5`. What are the maximum and minimum possible areas of triangle B?

Answer 1

`A_s = 4.16`
`A_L = 14.36`

Explanation:

Using either given side as a base, the third side can be calculated.
`A = 1/2 xx b xx h` ; `24 = 1/2 xx 8 xx h`

`h = 6` We use this as part of a right triangle to find the third side.
`C = 6.46`

NOW we know that the SMALLEST similar triangle "B" must have the largest side as 5, and the LARGEST similar triangle will have the smallest side with a length of 5. The corresponding ratios are:
`12:8:6.46` (original)
`5:3.33:2.69` (smallest)
`9.29:6.19:5` (largest)

Using Herron's Rule directly saves some further angle computation.
`s = (a + b+ c)/2`
`A = sqrt(((s(s-a)(s-b)(s-c))`

`5:3.33:2.69` (smallest)
`s = (5 + 3.33 + 2.69)/2 = 5.51`
`A = sqrt(((5.51(5.51-5)(5.51-3.33)(5.51-1.74))`
`A_s = 4.16`

`9.29:6.19:5` (largest)
`s = (9.29 + 6.19+ 5)/2 = 10.24`
`A = sqrt(((10.24(10.24-9.29)(10.24-6.19)(10.24-5))`
`A_L = 14.36`