Question
The area of a rectangle is 72 square inches and the perimeter is 44 inches. Find its dimensions.
Answers
Ms. Sue
One way to solve this is to find the factors of 72. Then see which of these will fit for the perimeter.
Reiny
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