SOLUTION: I need help solving a quadratic equation. The area of a field shaped like a right triangle is 600m2 ( squared). The legs of the field are fenced with steel at $10/m, while the h

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I need help solving a quadratic equation. The area of a field shaped like a right triangle is 600m2 ( squared). The legs of the field are fenced with steel at $10/m, while the h      Log On


   

Question 539715: I need help solving a quadratic equation.
The area of a field shaped like a right triangle is 600m2 ( squared). The legs of the field are fenced with steel at $10/m, while the hypotenuse is fenced with aluminum at $20/m. The perimeter of the field is 120m. The total cost of the fencing is $1700.
A. What is the length of each side of fence with steel?
B. what is the length of the side of the fence with aluminum?
Here's what I've tried:
xy/2= 600m2
x+y+z= 120
10x + 10y + 20z = $1700 or x+ y + 2z = $170
Not sure how to get rid of one of the variables.
Would greatly appreciate your assistance.
Daniel
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
xy/2= 600m2
xy=1200
x=1200/y
----
1200/y +y+z= 120
1200+y^2+yz=120y
y^2-120y+1200=-yz
multiply by 2
2y^2-240y+2400=-2yz
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x+ y + 2z = $170
1200/y +y+2z=170
1200+y^2+2yz=170y
y^2-170y+1200=-2yz
-----
Equate -2yz
y^2-170y+1200=2y^2-240y+2400
y^2-70y+1200=0
y^2-40y-30+1200=0
(y-40)(y-30)=0
y= 40 OR 30
The sides of triangle are 40,30,50
50 is the hypotenuse
Area = 1/2 * 30*40= 600 (check)
Now you know how to answer the questions
m.ananth@hotmail.ca