Step-by-step explanation:
so, it is an ideal rectangular shape ? because that is what the problem description is "hiding".
if that assumption is true, then we have
l = length
w = width
perimeter = 32 = 2×l + 2×w
=> 16 = l + w
l = 16 - w
area = 60 = l×w
now we use the perimeter equation to substitute in the area equation
60 = (16 - w) × w = 16w - w²
so, we get the squared equation
-w² + 16w - 60 = 0
the solution for such an equation is
x (or w in our case) = (-b ± sqrt(b² - 4ac))/(2a)
a = -1
b = 16
c = -60
w = (-16 ± sqrt(256 - 4×-1×-60))/-2 =
= (-16 ± sqrt(256 - 240))/-2 =
= (-16 ± sqrt(16))/-2 = (-16 ± 4)/-2
w1 = (-16 + 4) / -2 = -12/-2 = 6 m
w2 = (-16 - 4) / -2 = -20/-2 = 10 m
l1 = 16 - 6 = 10 m
l2 = 16 - 10 = 6 m
so, one of length and width must be 6 meter and the other 10 meter long. it does not matter which, as a rectangle turned by 90 degrees is still the same rectangle.
