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?

1 Answer
Nov 8, 2016

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