Question

Triangle A has an area of `18` and two sides of lengths `9` and `6`. Triangle B is similar to triangle A and has a side of length `8`. What are the maximum and minimum possible areas of triangle B?

Answer 1

`32` and `128/9`

Explanation:

Let `alpha` be the angle between sides 9 and 6
`A=(9*6)/2*sinalpha`

`sinalpha=(18*2)/(9*6)=2/3 \ \ => cosalpha=sqrt5/3`

Let calculate third side `c` using law of cosines

`c^2= a^2+b^2-2abcosalpha=81+36-2*9*6*sqrt(5)/3=117-36sqrt(5)`

`c=sqrt(117-36sqrt(5))~=6.04`

So 9 is the longest side and 6 is the shortest, so the maximum area is obtained when 8 is proportional to 6 and the minimum when 8 is proportional to 9:

`A_(MAX)=(8/6)^2*18=16/9*18=32`

`A_(MIN)=(8/9)^2*18=64/81*18=128/9`