The area of a rectangular garden is 72 sq. feet. If the perimeter is 34 feet, what are

You need to find Length and Width such that
L×W = 72
and
2L+2W = 34

Rewrite the first equation to solve for L:
L = 72/W

Substitute 72/W for L in the second equation, then solve for W:
2(72/W) + 2W = 34
144/W + 2W = 34
144 + 2W² = 34W
2W² - 34W + 144 = 0
W² - 17W + 72 = 0
(W-9)(W-8) = 0
W = 8, 9

The dimensions are 8 feet wide and 9 feet long,
or 9 feet wide and 8 feet long.
 
The area of a rectangular garden is 72 square feet.

A = 72

The perimeter is 34 feet.

P = 34

What are the possible dimensions?

L = length
W = width

The perimeter is calculated by the formula

2L + 2L = P, so
2L + 2W = 34

The area is calculated by the formula

LW = A, so
LW = 72

All you have to do is solve

2L + 2W = 34
LW = 72

Let's reduce the first equation. Dividing it by 2,

L + W = 17
LW = 72

Two equations, two unknowns. Solve by substitution. From the first equation, since L + W = 17, W = 17 - L.
Plug in W = 17 - L into the second equation.

L(17 - L) = 72
17L - L^2 = 72
0 = L^2 - 17L + 72
0 = (L - 9)(L - 8)

Therefore

L = {9, 8}

When L = 9, W = 17 - 9 = 8.
When L = 8, W = 17 - 8 = 9

Therefore, the (only) possible dimension (since length and width can be interchangeable) would be

9 ft x 8 ft
 
There are tow possibilities:-

1) Width = 9 ft, height = 8ft.
So, area = 9*8 = 72
and perimeter = 2*(9+8) = 34

2) Width = 8 ft, height = 9ft.
So, area = 8*9 = 72
and perimeter = 2*(8+9) = 34

Milind
 
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