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

ANSWERS

2016-04-10 04:29:17

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

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