SOLUTION: The area of a rectangle is 384 square feet. If the perimeter is 80 feet, find the length and width of the rectangle.

Question 668609: The area of a rectangle is 384 square feet. If the perimeter is 80 feet, find the length and width of the rectangle.
Answer by swincher4391(1107) (Show Source): You can put this solution on YOUR website!
A = lw
P = 2l + 2w
384 = lw
80 = 2l + 2w

384/l = w
80 = 2l + 2*(384/l)
80l/l = (2l^2+ 2*384)/l
80l = 2l^2 + 768
2l^2-80l+768 = 0
2(l^2-40l + 384) = 0
l = 16, l =24
Case 1: l=16
If l =16 then w = 384/16 = 24
If l =24 then w = 384/24 = 16
So, we could have 16x24 or 24x16.

RELATED QUESTIONS
Find the length and width of a rectangle whose perimeter is 36 feet and whose area is 80... (answered by stanbon)

The length of a rectangle is 3 feet less than twice the width of the rectangle. If the... (answered by josgarithmetic)

find the length and width of a rectangle whose perimeter is 34 feet and whose area is 60... (answered by ReadingBoosters)

Find the length and width of a rectangle whose perimeter is 24 feet and whose area is 35... (answered by nanakwame)

Find the length and width of a rectangle whose perimeter is 32 feet and whose area is 55... (answered by josgarithmetic)

Find the length and width of a rectangle whose perimeter is 36 feet and whose area is 65... (answered by Alan3354,josgarithmetic)

Find the length and width of a rectangle whose perimeter is 30 feet and whose area is 50... (answered by Alan3354)

Find the length and width of a rectangle whose perimeter is 42 feet and whose area is 90... (answered by Alan3354)