Asked by Kestrel
a rectangular field with an area of 8000 m^2 is enclosed by 400 m of fencing , detemine the dimesions of the field to the nearest tenth of a metre
Answers
Answered by
Jai
first we represent the unknowns using variables:
let x = length
let y = width
recall that perimeter of rectangle is just,
P = 2x + 2y
since P is given (which is 400), we can get an expression of width in terms of length:
400 = 2x + 2y
400 - 2x = 2y
y = 200 - x
then to get the dimensions, we use the area given. recall that area of rectangle is just,
A = xy
substituting,
8000 = x(200 - x)
8000 = 200x - x^2
x^2 - 200x + 8000 = 0
then solve for x using quadratic equation. then also solve for y.
hope this helps~ :)
let x = length
let y = width
recall that perimeter of rectangle is just,
P = 2x + 2y
since P is given (which is 400), we can get an expression of width in terms of length:
400 = 2x + 2y
400 - 2x = 2y
y = 200 - x
then to get the dimensions, we use the area given. recall that area of rectangle is just,
A = xy
substituting,
8000 = x(200 - x)
8000 = 200x - x^2
x^2 - 200x + 8000 = 0
then solve for x using quadratic equation. then also solve for y.
hope this helps~ :)
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