Question

The area of a certain rectangle is 288 yd2. the perimeter is 68 yd. what are the dimensions of the rectangle?

Answer 1

P = 2(L + W)
P = 68
68 = 2(L + W)
68/2 = L + W
34 = L + W
34 - L = W

A = L * W
A = 288
W = 34 - L

288 = L(34 - L)
288 = 34L - L^2
L^2 - 34L + 288 = 0
(L - 16)(L - 18)

L - 16 = 0
L = 16

L - 18 = 0
L = 18

not exactly sure which (length or width) , but one is 16 yds and one is 18 yds