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College Algebra - Area Word Problem

by Merl Jeant
(FL)











































a rectangular pool is 20 ft wide and 30ft long. want to make a walkway around edge of pool have enough material to cover 336 square ft. if the walkway is to be around the entire pool, how wide is the walkway?

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Jan 27, 2012
Area Word Problem
by: Staff


Question:

by Merl Jeant
(FL)


a rectangular pool is 20 ft wide and 30ft long. want to make a walkway around edge of pool have enough material to cover 336 square ft. if the walkway is to be around the entire pool, how wide is the walkway?


Answer:

Geometrically, this problem involves two rectangles of different sizes. They are NOT similar rectangles. The ratio of the length to the width of each rectangle is not the same.

The rectangle with the smallest size represents the pool. It is 20 ft wide and 30 ft long.

The large rectangle represents the pool and the walkway, together.

The walkway will have the same width all the way around the pool.


Conceptually, you can picture the small rectangle being placed on top of the larger rectangle and centered.



The area of the large rectangle (which represents the pool and walkway, together) MINUS the area of the small rectangle (which represents the pool) is the area of the walkway: 336 ft².

The area of the small rectangle (the pool) = 20 ft * 30 ft = 600 ft²

The area of the large rectangle (the pool + the walkway) = 600 ft² + 336 ft² = 936 ft²

W = width of the walkway


The length of the large rectangle = length of the small rectangle + W + W

The width of the large rectangle = width of the small rectangle + W + W

Length_large * Width_large = Area of the large rectangle = 936 ft²


(30 + W + W) * (20 + W + W) = 936


Solve for the width of the walkway, W


Expand the expression

(30 + 2W) * (20 + 2W) = 936

(2W + 30) * (2W + 20) = 936

4W² + 100W + 600 = 936


Divide each side of the equation by 4

(4W² + 100W + 600)/4 = 936/4

W² + 25W + 150 = 234


Subtract 234 from each side of the equation

W² + 25W + 150 - 234 = 234 - 234

W² + 25W - 84 = 0


Factor the quadratic expression

(W + 28)*(W - 3) = 0


Solve for W (there will be two solutions)

W + 28 = 0

W + 28 - 28 = 0 - 28

W + 0 = 0 - 28

W = - 28


W - 3 = 0

W - 3 + 3 = 0 + 3

W + 0 = 0 + 3

W = 3


W ∈ {- 28, 3}

However, W cannot have a negative value. The width of the walkway must have a positive value.

Therefore,

W = 3 ft


The length of the large rectangle = length of the small rectangle + W + W

Length_large = 30 + 2W

Length_large = 30 + 2*3

Length_large = 30 + 6

Length_large = 36 ft


The width of the large rectangle = length of the small rectangle + W + W

Width_large = 20 + 2W

Width_large = 20 + 2*3

Width_large = 20 + 6

Width_large = 26 ft


>>> The final answer is:

The walkway is 3 ft wide





Thanks for writing.
Staff

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