The width of a rectangle is one-third the length...?

Question

The width of a rectangle is one-third the length. The area of the rectangle is three hundred sixty-three square feet. What is the perimeter of the rectangle?

Answer 1

Length = l

Width = w = l/3

Area = lw = 363

Perimeter = 2l + 2w

A = lw (substitute the correct values)

363 = l(l/3) (multiply)

363 = l^2/3 (multiply both sides by 3)

1,089 = l^2 (square root both sides)

33 = l

And now find the width:

w = l/3

w = 33/3

w = 11

Finally, find the perimeter:

P = 2l + 2w

P = 2(33) + 2(11)

P = 66 + 22

P = 88 feet <===ANSWER

Answer 2

we can call the width of the rectangle w and the length 3w.

and area = width x length so:

363 = 3w^2

divide by 3

121 = w^2

square root

w = 11

therefore length = 33

so perimeter equals (2*11)+(2*33)

22+66 = 88

perimeter = 88feet

Answer 3

A = w (w/3) = 363

isolate w

w = 363/(w/3)

w = 1089/w

w^2 = 1089

w = 33

P = w + (w/3) + w + (w/3) = 8w/3 = 88

P=88

Answer 4

Find length (x):

363/x = 1/3x

1/3x² = 363

x² = 1,089

x = 33

Find perimeter:

= 2(33 + [33/3])

= 2(33 + 11)

= 2(44)

= 88

Answer: 88 feet is the perimeter

Proof area (363 feet):

= 33 * 11

= 363

Answer 5

A=LW

L=3W

3W*W=363

3W^2 = 363

W^2 = 121

W=11

P=2(W+L)

P=2(11+33) = 88

Answer 6

W=L/3

A=WL=L^2/3=363

L^2=121

L=11

P=2L+2W = 2L+2/3L = 8/3L=88/3 ft

Answer 7

A=LW=363

W=L/3

L*L/3=363

L^2=1098

L=33.13608

W=11.04536

P=2L+2W

P=2*33.13+2*11.05

P=66.26+22.1

P=88.36