The area of a rectangle is 90 square inches and the perimeter is 42 inches. What are its dimensions? Demonstrate your method

Question

The area of a rectangle is 90 square inches and the perimeter is 42 inches. What are its dimensions? Demonstrate your method.

Answer 1

area is length (l) * width (w)

lw = 90

perimeter = 2*length + 2*width

2l + 2w = 42

l + w = 21

l = 21 - w

substituting 21 - w for l in equation lw = 90:

(21 -w)w = 90

21w - w^2 = 90

-w^2 + 21w - 90 = 0

w^2 - 21w + 90 = 0

(w-15)(w-6) = 0

w = 15 or 6

since l + w = 21, l = 6 or 15

Answer 2

42 = 2(L+W)

90 = LW

L=90/W

42 = 2(90/W +W) = 180/W +2W

divide by 2

21 =90//W + W

multiply by W

21 W = 90 +W^2

W^2 -21W +90 = 0

complete the square or use the quadratic formula to solve for W

W^2 -21W + (21/2)^2 = -90 + (21/2)^2

(W-21/2)^2 = 441/4 - 360/4 = 81/4

W-21/2 = + or - 9/2

W = 21/2 + or - 9/2 = 15 or 6

L=90/15 or 90/6 = 6 or 15

the two sides are 6 and 15 inches