Arectangle has a perimeter of 30 feet and an area of 50 square feet. what are the dimensions of the rectangle? a) 2 feet, and 15 feet b) 2 feet, and 25 feet c) 4 feet, and 11 feet d) 5 feet, and 10 feet

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

Multiply 9 and 67. then multiply 3 with 67. lastly the answer that you got for 9x67 is to be subtracted by the answer you got for 3x67