Perimeter (p) = 2×length (l) + 2×width (w)
p = 2l+2w
area (a) = l×w, so solve for one (I'll use l):
since p = 30, and a = 50, substitute the "a÷w" in for l in the perimeter equation:
Now plug in p and a values:
therefore width can be either 5 or 10 (but not both), so let's plug in:
l = a÷w = 50÷5 = 10
So if w = 5, then l = 10
D) 5 feet, and 10 feet
p = 2l+2w
area (a) = l×w, so solve for one (I'll use l):
since p = 30, and a = 50, substitute the "a÷w" in for l in the perimeter equation:
Now plug in p and a values:
therefore width can be either 5 or 10 (but not both), so let's plug in:
l = a÷w = 50÷5 = 10
So if w = 5, then l = 10
D) 5 feet, and 10 feet
