Question

A rectangular garden has perimeter 66 ft and area 216 ft squared. Find the dimensions of the garden.

Answer 1

The perimeter is then

2L + 2W = 66

The area is

L*W = 216

Solve the first equation for L

2L = 66 - 2W

L = 66/2 - 2W/2 = 33 - W

Put this into the second equation

(33 - W)*W = 216

33W - W^2 = 216

rearranging

W^2 - 33W + 216 = 0

Using the quadratic formula

[-(-33) plus or minus sqrt((-33)*(-33) - 4 * 1 * 216)]/2

[33 plus or minus sqrt(1089 - 864)]/2

[33 plus or minus sqrt(225)]/2

[33 plus or minus 15]/2

(33 + 15)/2 = 48/3 = 16 = W

(33 - 15)/2 = 18/2 = 9 = W

Checking back into the original equations

L = 24 W = 9

Explanation:

Answer 2

Half the perimeter would be the sum of one length and one width.area = length x width.

X + y = 33 feet

Rewrite as x = 33-y

X x y = 216 feet^2

Replace x:

(33-y) x y = 216

Expand:

33y -y^2 = 216

Subtract 216 from both sides

33y -y^2 -216 = 0

Using quadratic formula solve for y

Y = 9 and 24

The dimensions would be 9 for by 24 ft