A particular rectangular swimming pool is twice as long as it is wide. There is a concrete walkway around the perimeter of the pool. The walkway is a constant feet wide. The walkway has an area of 100 square feet. What are the dimensions of the pool?

## What are the dimensions of the pool?

# 4 Answers

Hey Kami -- here's a "kitchen table" approach ... cut in half to form an "L" walkway of 50 sq.ft ... if ww width is 1ft, we have 2 squares of 1 sq.ft each plus the remaining 48 sq.ft ==> 48x1 ... so a 32x16 pool would work.

If ww width is 2ft, we have 2 squares of 4 sq.ft each plus 42 sq.ft ==> 21x2 ... a 14x7 pool works here, too. Best regards :)

Draw a rectangle and label one side L, and the other perpendicular side W. The parallel sides should be equal to L & W. We could care less about the walkway...it's extra information and the constant W means the pool isn't freaky and changing shapes...aka we are leaving theoretical physics out of this ;) ...if it is constant then so is the length

We will use two common equations: A=LW (Area= Length x Width) and P = 2L + 2W, where P is the Perimeter. (we won't actually need the perimeter equation in this problem, but it is always good to remember it.)

Recall, perimeter is the distance around the pool...so if you were to walk around it you would walk around the long part twice and the shorter part twice, hence the 2L + 2W.

We were given two items in this problem to supplement the equations we know...one is the area, A=100 and the other is that the length is twice as long as the width, L=2W. Let's use L=2W in the area equation since we have a numerical value for A and solve for W since we have an equation L in terms of W.

A=LW -> 100=(2W)W ->100=2W^{2 }-> 50=W^{2} -> √(50) = W

I don't have a calculator on me so I will leave it to you to take the square root of 50. Once you did this you will have a numerical answer for W, simply plug this number into L=2W and you will have the length. You can check this by multiplying the two together and getting 100 (or close to 100 if you use rounded values). Your L & W are in Feet!! (ft). If the problem ask you to find the perimeter (potential test question) then you would to the same process solving for L & W and then plug them into the perimeter equation.

I hope this helps!

v/r

Daniel

Let L = the length, and W = the width.

L = 2W ......(1)

(L+2)(W+2) - LW = 100 => 2W+2L+4 = 100 ......(2)

Plug (1) into (2), and solve for L,

3L = 96

L = 32 feet

W = L/2 = 16 feet

Answer: The length of the pool is 32 feet, and the width is 16 feet.

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Check:

34*18 - 32*16 = 100 ft^2. It works!

# Comments

Your answer only works if the width of the walkway is 1 foot. For any other constant width it will not work.

Let L be the length of the pool, W be the width of the pool, and C be the constant width of the walkway. Since the pool is twice as long as it is wide, L=2W. The area of the walkway is its width C times its length, 2L+2W+4C, which gives A=2LC+2WC+4C². With L=2W this becomes A=2(2W)C+2WC+4C², or A=6WC+4C². Now set this area equal to 100, 6WC+4C²=100, and solve for W: W=(50-2C²)/3, and the length is twice that, L=(100-4C²)/3.

## Comments

We are given the area of the walkway, not that of the pool, so you cannot set A=LW.

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