The area of a rectangle is 72 square inches and the perimeter is 44 inches. Find its dimensions.
2 answers
One way to solve this is to find the factors of 72. Then see which of these will fit for the perimeter.
let the length and width be x and y respectively
then
2x + 2y = 44
x + y = 22
y = 22-x
area = xy = 72
x(22-x) = 72
22x - x^2 - 72 = 0
x^2 - 22x + 72 = 0
(x-4)(x-18) = 0
x = 4 or x = 18
if x = 4, then y = 22-4 = 18
if x = 18, then y = 22-18 = 4
the dimensions are 4 by 18 inches
then
2x + 2y = 44
x + y = 22
y = 22-x
area = xy = 72
x(22-x) = 72
22x - x^2 - 72 = 0
x^2 - 22x + 72 = 0
(x-4)(x-18) = 0
x = 4 or x = 18
if x = 4, then y = 22-4 = 18
if x = 18, then y = 22-18 = 4
the dimensions are 4 by 18 inches
1 answer