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Question

Find the dimensions of a rectangle whose area is 180 cm2 and whose perimeter is 54cm.

Answer 1

54 = 2(L + W)

27 = L + W

27 - L = W 

 

LW = 180

L(27 - L) = 180

27L - L² = 180

0 = L² - 27L + 180                FOIL

0 = (L - 12)(L - 15)

 

{12, 15} 

 

Length is the longer side. L = 15, W = 12

Answer 2

Area = (Length)(Width)
A = L*W
A = 180
Perimeter = 2L +2W
54 = 2L + 2W

A = L*W
Divide each side by W.
A / W = L*W/W
(A / W) = L
(180 / W) = L

P = 2L + 2W
P = 2(L + W)
P / 2 = [2(L+W)] / 2
P / 2 = L + W
(54 / 2) = L + W
27 = L + W
27 - W = L + W - W
27 - W = L

27 - W = (180 / W)
(27 - W)(W) = (W)(180 / W)
W(27 - W) = 180
27W - W2= 180
27W - W2 + W2 = 180 + W2
27W = W2 + 180
27W - 27W = W2 + 180 - 27W
0 = W2 - 27W + 180

From here, we need to figure out what two numbers can multiply to equal 180 and add to equal (-27). We then will put those numbers in the following form: (W + __)(W + __).

(-15)(-12) = 180
(-15) + (-12) = -15 - 12 = -27

(W - 12)(W - 15) = 0
Then, we can separate them out and solve.

(W - 12) = 0; (W - 15) = 0

W - 12 = 0
W - 12 + 12 = 0 + 12
W = 12

W - 15 = 0
W - 15 + 15 = 0 + 15
W = 15

So, the dimensions of the rectangle are: 15cm and 12cm.

You can double check this by plugging these numbers back into the Area and Perimeter equations:
A = LW
A = (15)(12)
A = 180
180 = 180

P = 2(L+W)
P = 2(15 + 12)
P = 2(27)
P = 54
54 = 54