A rectangular field with an area of 8000m² is enclosed by 400 m of fencing. Determine the dimensions of the field to the nearest tenth of a metre.
It has to be solved using a quadratic equation.
It has to be solved using a quadratic equation.
-
L+W=200
LW=8000
L+8000/L-200=0
L^2-200L+8000=0
From the quadratic formuls
L = 144.72 m
W = 55.28
Field is 144.7 by 55.2 metres
LW=8000
L+8000/L-200=0
L^2-200L+8000=0
From the quadratic formuls
L = 144.72 m
W = 55.28
Field is 144.7 by 55.2 metres
-
Area = l * w = 8000 m^2
Perimeter = 2l + 2w = 400 m
l = 200 - w
substitute: (200 - w) * w = 8000
solve the quadratic...
200w - w^2 = 8000
w^2 - 200w + 8000 = 0
w = [200 ± √(8000)]/2 = 100 ± 20√5 m
get out your calculator...make a contribution...go for it !
¶
Perimeter = 2l + 2w = 400 m
l = 200 - w
substitute: (200 - w) * w = 8000
solve the quadratic...
200w - w^2 = 8000
w^2 - 200w + 8000 = 0
w = [200 ± √(8000)]/2 = 100 ± 20√5 m
get out your calculator...make a contribution...go for it !
¶