what is the shorter and longer dimensions?

Question

A security fence encloses a rectangular area on one side of a park in a city. Three sides of fencing are used, since the fourth side of the area is formed by a building. The enclosed area measures 2178 square feet. Exactly 132 feet of fencing is used to fence in three sides of this rectangle. What are the possible dimensions that could been used to construct this area? 

Answer 1

Hi Shane, there are several approaches to this problem. One way is to find all of the factors of 2178 and see which one fits the requirements for the given length of fencing. That answer is 33 time 66. The field is 33 feet wide and 66 feet long.

 

The algebraic method uses a modified perimeter equation plus the equation for the area of a rectangle.

 

Step 1:

 

Modified perimeter formula:

 

The formula for the perimeter of a rectangle is: P = 2L + 2W

 

We only have three sides, not four. The modified perimeter equation becomes: P = L + 2W. Note, we could use: P = 2L + W, and get the same answers.

 

I choose to use one L.

 

Solve the modified perimeter for W:

 

(P - L) / 2 = W

 

Substitute for P:

(132 - L) / 2 = W

 

132 / 2 - L / 2 = W

 

66 - L / 2 = W

 

The area of a rectangle:

 

A = L * W

 

Substitute for W:

 

A = L * (66 - L / 2)

 

We know A is 2178 square feet:

 

2178 = L * (66 - L / 2)

 

2178 = 66L - L² / 2

 

Multiply all term by 2 to clear the fraction:

 

4356 = 132L - L²

 

Move all terms to one side of the equal sign:

 

L² - 132L + 4356 = 0

 

Solve using the quadratic formula:

 

L = (132 ±√(132² - (4 * 4356)) / 2

 

L = (132 ±√0) / 2

 

L = 132 / 2

 

L = 66

 

Return to 66 - L / 2 = W, to solve for W:

 

66 - 66 / 2 = W

 

66 - 33 = W

 

W = 33

 

Answer:

 

The dimensions of the area enclosed is: 33 feet by 66 feet by 33 feet.

 

Questions?

 

Answer 2

P=2l+2w but there is only one length because of the building, so...

P=l+2w

132=l+2w

132-2w=l

A=lw

2178=(132-2w)(w)

2178=132w-2w²

2w²-132w+2178=0

divide both sides by 2

w²-66w+1089=0

factor

1089=3*3*11*11=(3*11)(3*11) and 33+33=66, the coefficient of the middle term

(w-33)(w-33)=0

w-33=0

w=33 feet

132-2*33=l

132-66=l

l=66 feet

you have a rectangle 33 feet by 66 feet by 33 feet by "the building"

check:

33+66+33=132

33*66=2178