SOLUTION: I just finished a math exam and theres one question that I just couldn't figure out. It went like this: "A rectangular field has an 800m long fence around the perimeter. The area o

Algebra ->  Rectangles -> SOLUTION: I just finished a math exam and theres one question that I just couldn't figure out. It went like this: "A rectangular field has an 800m long fence around the perimeter. The area o      Log On


   

Question 463891: I just finished a math exam and theres one question that I just couldn't figure out. It went like this: "A rectangular field has an 800m long fence around the perimeter. The area of the field is 37500m^2. What are the diemensions of the field." I think im supposed to use Substitution so I wrote the formulas I know that relate to the question and I plugged in the information from the question into the equation so it was like this: "P(perimeter)=2*L(length)+2*width & A(area)=L*W. After this I dont know what to do. I can eliminate be the operations are different so the only option is to change an equation to isolate a variable to substitute it into the next equation but I don't know. No hurry gettin back to me the exams already done im just asking to ease my mind
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
We know the perimeter is 800 m and the area is 37500m^2, so

800 = 2L + 2W --> L + W = 400

LW = 37500

From the equation L + W = 400 we can isolate L and say L = 400 - W. Plug this into the expression for area:

(400-W)W = 37500

-W^2 + 400W - 37500 = 0

W^2 - 400W + 37500 = 0

By factoring or by the quadratic formula, we get W = 150 or W = 250. This implies the other dimension is 250 or 150 (order doesn't matter since we can choose a different L. The dimensions are basically 150x250).