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Area problem

  1. Jul 20, 2008 #1
    1. The problem statement, all variables and given/known data
    An architect wants to make an exposed gravel border of uniform width around a shed. The shed has dimensions of 10 by 11 feet. She only has enough gravel to cover 46 sq feet. How wide should the border be?


    2. Relevant equations
    Area= L x W
    46 sq ft is a limiting factor as well as the 110 sq feet of the shed.


    3. The attempt at a solution
    I basically did a trial and error because I didn't know how to set it up.
    I arbitrarily chose .5feet for the border.
    2((10ft+.5ft)x.5ft) for the two widths of the shed. Then I multiplied that by
    2((11ft+.5ft)x.5ft) for the two lengths of the shed.
    I got the answer of 23 ft so I seen that my original choice of .5 ft was two small
    and I doubled it to 1 ft.

    I didn't like doing it this way and would like someone to explain to me how to set up a way of solving this problem without trial and error please.
     
  2. jcsd
  3. Jul 20, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The very first thing you should do is draw a picture! Now specify your variables. Let L be the length and W be the width of the total area of shed and graveled border. You are told that the shed is 10 by 11 feet so the area of the shed is 110 square feet. Since there is enough gravel to cover 46 square feet, the area of shed and border must be 110+ 46= 156 square feet: LW= 156.

    Now, let's be a bit more precise. The question asks for the width of the border. I take that to mean that the border is the same width all around. Call that width x. Do you see, from your picture, that L= 11+ 2x and W= 10+ 2x? That's because you have a border on both sides of both length and width.

    Now LW= (11+ 2x)(10+ 2x)= 156. Can you multiply that out and solve the resulting equation?
     
  4. Jul 20, 2008 #3
    That is very sensible. When I multiply (11+2x)(10+2x)=156, I get 110+22x+20x+4x^2=156.
    Moving terms around I get, 4x^2+42x-46=0
    I tried using the quadratic formula but it didn't work out. And I'm not sure where I'm going wrong. If the answer is 1ft. I am way off.
     
  5. Jul 20, 2008 #4
    Hello, I tried it again and got it. Thanks for your help!
     
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