The lengths of the sides of isosceles triangle ABC are 3x + 62, 7x + 30, and 5x + 60 feet. What is the least possible number of feet in the perimeter of ABC?
Consider the 3 possible choices of equivalent side lengths: 3x+62=5x+60, 7x+30=5x+60, and 3x+62=7x+30.
By solving for all three, we get the following results:
3x+62 = 5x+60, x= 1
3x+62 = 7x+30, x = 8
5x+60 = 7x + 30, x = 15.
These show that the least possible perimeter sees x = 1, so we can solve (3x1 +62) + (5x1+60) + (7x1+30) = 174.