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?
+128473 We have 3 equations that give us possible values for x
!st....assume that these two sides are equal
3x + 62 = 7x + 30
62 - 30 = 7x - 3x
32 = 4x
x = 8 and the perimeter = 3(8)+ 62 + 7(8)+ 30 + 5(8) + 60 = 272
Next assume that these two sides are equal
3x + 62 = 5x + 60
62 - 60 = 5x - 3x
2 = 2x
x = 1 The perimeter = 3(1) + 62 + 7(1) + 30 + 5(1) + 60 = 167
Finally, assume that
7x + 30 = 5x + 60
7 - 5x = 60 -30
2x = 30
x = 15.... it's obvious that this value of x will lead to the greatest perimeter
So....the perimeter is the least when 3x + 62 = 5x + 60 and the perimeter = 167
