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?
+80 I do not know if this question allows all real numbers, or just integers, so I'll go with integers.
Ok, let's start.
this is an isosceles triangle, therefore either \(AB=AC\) or \(BC=AB\) or \(BC=AC\) or, so on.
That could be a large problem testing out all different values of x 
To spare all the fuss, I've found that \(X=8\) works well, and is the smallest value you can achieve that is an integer.
So \(3x+62 = 24+62\) and \(7x+30 = 56+30\), which are both 86! (We used sides A and B here).
Judging by this, we can calculate the perimiter by replacing x in all equations with 8. So it is:
86+86+100
or, \(\fbox{272}\)
