SOLUTION: One leg of an isosceles triangle has a length of (x2 + 9)cm. The base of the isosceles triangle has a length of (5x + 12)cm. The perimeter of the triangle is 42 cm. Find the val

Algebra ->  Triangles -> SOLUTION: One leg of an isosceles triangle has a length of (x2 + 9)cm. The base of the isosceles triangle has a length of (5x + 12)cm. The perimeter of the triangle is 42 cm. Find the val      Log On


   

Question 247459: One leg of an isosceles triangle has a length of (x2 + 9)cm. The base of the isosceles triangle has a length of (5x + 12)cm. The perimeter of the triangle is 42 cm. Find the value(s) for x and the lengths of the sides.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
One leg of an isosceles triangle has a length of (x^2 + 9)cm.
The base of the isosceles triangle has a length of (5x + 12)cm.
The perimeter of the triangle is 42 cm.
Find the value(s) for x and the lengths of the sides.
--------------------------------
There are 2 equal legs and a base.
2(x^2+9) +5x+12 = 42
2x^2 + 5x + 18 - 42 = 0
2x^2 + 5x - 24 = 0
x = [-5 +- sqrt(25-4*2*-24)]/4
---
x = [-5 +- sqrt(217)]/4
---
Positive solution:
x = [-5+sqrt(217)]/4 = 2.4327..
---
leg1 = leg2 = x^2+9 = (2.4327)^2+9 = 14.9182
base = 5x+12 = 24.1635
================================================
Cheers,
Stan H.