February 10th, 2013 at 8:00:18 PM
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A rug merchant is hired to carpet a hallway formed by two circular concentric walls. Not knowing the radii of the circles he has his men carry in a 20' stick to measure the circumference. Unfortunately when he gets it in the door and closes the door it is completely wedged between the outside wall (at the tips) and the inside wall. Even though he is unable to use the stick, he still knows how much carpet he needs.

How much carpet does he need?

How much carpet does he need?

February 10th, 2013 at 8:31:57 PM
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To be clear two circular concetric walls is essentially a circular hallway.

I assume that the hallway is 20' thick. The amount of carpet required is

Let r be the radius of the outside circle.

The area of carpet is pi (r^2) - pi (r-20)^2 = pi(r^2 + 40r -400 - r^2 ) = pi(40r-400)

pi r^2 - pi (r-20)^2 = pi (-400 + 40r) where r is the radius of the outside wall.

I can't see how this can be solved without not knowing r. r > 20.

I assume that the hallway is 20' thick. The amount of carpet required is

Let r be the radius of the outside circle.

The area of carpet is pi (r^2) - pi (r-20)^2 = pi(r^2 + 40r -400 - r^2 ) = pi(40r-400)

pi r^2 - pi (r-20)^2 = pi (-400 + 40r) where r is the radius of the outside wall.

I can't see how this can be solved without not knowing r. r > 20.

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You want the truth! You can't handle the truth!

February 10th, 2013 at 8:45:11 PM
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314.1593 sq ft

I heart Crystal Math.

February 10th, 2013 at 8:46:36 PM
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Edit: recomputing....misunderstood the question the first time.

Quick answer: enough to cover the floor. ha ha.

Quick answer: enough to cover the floor. ha ha.

February 10th, 2013 at 8:55:02 PM
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Crystal, how do you make that assumption?

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You want the truth! You can't handle the truth!

February 10th, 2013 at 9:09:02 PM
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Quote:skrbornevryminrecomputing....misunderstood the question the first time.

Quick answer: enough to cover the floor. ha ha.

Ok. After having thought about this a little, I have determined that this cannot be solved with the information given, because we do not have the length of the hallway, only the width of it. At the very least it would be a circular room with the inner radius of 0' and the outer radius of 20 feet which is what I initially assumed. However, it could be a very long hallway such as that which would circle a basketball arena, for example, which was 20 feet wide.

Regardless of the area of the floor however, carpet comes off of a roll in rectangles (or squares) in most cases, so waste would be a factor as well. So until we get more info, I am sticking with my "quick answer" as stated above.

February 10th, 2013 at 9:19:03 PM
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Crystalmath is correct. The area of the outer circle minus the area of the inner circle will always be equal to the area of a circle with a radius of 10'

February 10th, 2013 at 9:21:50 PM
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Oh for heaven's sake. It's a carpet salesman. If he sent a stick that's 20 feet long he will tell you that you need 20 + 20 = 40 X 40 = 1600 square feet since carpet is sold in rectangular rolls, with a "liitle extra just in case you spill something later."

February 10th, 2013 at 9:57:11 PM
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Quote:AlanMendelsonOh for heaven's sake. It's a carpet salesman. If he sent a stick that's 20 feet long he will tell you that you need 20 + 20 = 40 X 40 = 1600 square feet since carpet is sold in rectangular rolls, with a "liitle extra just in case you spill something later."

Of course. But if these problems happened to a real salesman, he'd use a tape measure rather than a stick.

Donald Trump is a fucking criminal

February 10th, 2013 at 10:18:17 PM
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Quote:stargazerCrystalmath is correct. The area of the outer circle minus the area of the inner circle will always be equal to the area of a circle with a radius of 10'

I don't think so, because we do not know the diamater (or the radius) of the inner circle that the circular hallway goes around. (Imagine trying to carpet a circular racetrack around a circular field.) Since we don't know the length of the track, nor the diameter of the inner field, we cannot determine the area of the actual track itself. All we know is the width of the track, which isn't enough to compute the area of it.

I hope that I am not missing something...