SOLUTION: The area of a triangle is 250 in^2. Find the lengths of its legs if one leg is 5 inches longer than the other. i know area is 1/2bh. let x=shorter leg and x+5=longer leg so i

Algebra ->  Absolute-value -> SOLUTION: The area of a triangle is 250 in^2. Find the lengths of its legs if one leg is 5 inches longer than the other. i know area is 1/2bh. let x=shorter leg and x+5=longer leg so i      Log On


   

Question 37861: The area of a triangle is 250 in^2. Find the lengths of its legs if one leg is 5 inches longer than the other.
i know area is 1/2bh. let x=shorter leg and x+5=longer leg so i know area=1/2*x*x+5=250 i need to solve for x.
Found 2 solutions by fractalier, Kathy:
Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
You're on the right track...yes,
A = (1/2)bh and
250 = (1/2)(x)(x + 5)
now multiply both sides by two to clear fractions
500 = x^2 + 5x
x^2 + 5x - 500 = 0
Since this is factorable, we factor and solve
(x + 25)(x - 20) = 0 which leads us to
x = -25 and x = 20
but since distance is never negative, our short side is
x = 20 and the longer one is 25.

Answer by Kathy(4) About Me  (Show Source):
You can put this solution on YOUR website!
x2+2*5/2=250