Question 492878: Sam has a 15 by 25 rectangular pool. He wants to pour a concrete deck of uniform width around the pool so the area of the pool and deck is 600 sq. ft. How wide is the deck?
Answer by lmeeks54(111)   (Show Source):
You can put this solution on YOUR website! This is actually a pretty straight forward problem...and easy to solve. But an unrealistic problem. We'll solve this one and then append some thoughts about what would make a better problem to solve.
Sam's pool is 15' x 25' = 375 sq ft. He wants a deck around the pool so that the pool + deck = 600 sq ft. All that is a bigger rectangle. Happily for us, he wants the deck width to be the same all the way around.
area 1, L * W w = A (pool)
area 2, (L + x) * (W + x) = new A (pool + deck)
Note, the requirement to have the deck be the same width around has us using the same "x" to add to both length L and width W to compute the new area.
So now, our equation looks like:
(x + 25) * (x + 15) = 600
On inspection, we can see that if we add x = 5 to both L & W, 30 x 20 = 600. Don't discount the value of inspection; but in this case, we want to go ahead and practice solving this quadratic equation.
Putting all the terms on the left side so the equation = 0, we step through:
x^2 + 40*x +375 = 600
subtract 600 from both sides to make the equality = 0, then:
x^2 + 40*x -225 = 0
This factors into:
(x-5) * (x+45) = 0
either of the equalities can be set to 0 to make the equality = 0, thus:
x = 5 or x = -45. Thinking in terms of our pool deck, x = 5' of extra dimension added to the length and width makes sense (and is the answer we got by inspection). However, x = -45' is a nonsensical answer in this instance, so we can ignore it.
Note, we are not quite done. The question is how wide is the deck? We solved, how much extra length and width do we need to make a 600 sq ft rectangle. We now need to divide the x = 5' in half so that for each of the length and width, 2.5' are added at each end and each side of the pool.
The answer to the question is: the deck is 2 1/2 feet wide all the way around the pool.
PS. I said we would think about a more realistic problem. Knowing how wide the deck is is one thing, but knowing how much concrete we need to buy to pour this deck is more interesting. We could easily determine the area of the deck because we know the area of the deck + pool and the pool by itself. The outer rectangle (the pool and deck) is 600 sq ft, the inner rectangle (the pool by itself) is 375 sq ft; therefore, the deck is 600 - 375 = 225 sq ft.
Then we can find out (or determine) how deep the concrete needs to be to construct the deck. Say the form to pour the concrete into is made from 2" x 4" boards, then the concrete is going to be 4" deep (or 1/3 of a foot). 225 sq ft x .333 ft = 75 cubic ft of cement. But we don't buy cement in cubic ft but in cubic yds. A cubic yd is 3' x 3' x 3' or 27 cu ft. Divide the 75 cu ft of cement by 27 and we get 2.78 cu yds of cement. Allow a little extra and now we have figured out how to estimate the quantity of cement we need to build the deck asked for in the first problem. Cheers...
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