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Need lots of help!

  1. Jun 1, 2009 #1
    Need lots of help! :D

    1. The problem statement, all variables and given/known data

    A new rectangle with the word “STROM” is to be constructed inside the mat of an outside rectangle. The width of the mat surrounding the rectangular “STROM” logo will be the same. The area of the boundary around the word “STROM” will be the same as the logo itself. Determine the uniform width of the boundary.

    2. Relevant equations

    i made a picture of it including the measurements from the sheet:

    http://img188.imageshack.us/img188/4315/strompic.jpg [Broken]

    3. The attempt at a solution

    294.5 inches x 72 inches = 21204 inches

    i really dont know what to do, please show me step-by-step directions on how to solve this so I can understand and learn from it. Thanks!!!!:)
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 1, 2009 #2

    sylas

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    Re: Need lots of help! :D

    You'll learn even better when you show us the step by step solution! You've done the first step correctly. The units are "square inches". Now there's more information you were given. Can you find any other areas? What else has an area in this problem?
     
  4. Jun 1, 2009 #3
    Re: Need lots of help! :D

    i dont know what to do :(

    how am i to find the uniform width of the boundary??
     
  5. Jun 1, 2009 #4

    sylas

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    Re: Need lots of help! :D

    Step by step. Don't worry about the width yet. Can you find the area of anything else? There's the logo for example. What do you know about its area?
     
  6. Jun 1, 2009 #5
    Re: Need lots of help! :D

    well there weren't any measurements given to the logo, so how would i go about finding its area? as far as i know, the measurements from the mat rectangle have to do with something with its area, right? i cant seem to figure it out though. =\
     
  7. Jun 1, 2009 #6

    sylas

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    Re: Need lots of help! :D

    Try this sentence: "The area of the boundary around the word “STROM” will be the same as the logo itself."

    Can that help tell you the area of the logo itself? You've already obtained the total area.
     
  8. Jun 1, 2009 #7
    Re: Need lots of help! :D

    i still don't get it. :( i only got the total area which is 21204 inches. so you're saying its the same? so the logo's area is 21204 inches as well?

    ah damn i suck at this...lol.
     
  9. Jun 1, 2009 #8

    sylas

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    Re: Need lots of help! :D

    Hang in there; you're still ok. You got the first bit yourself, and now you're going to get the next.

    The total area is 21204. That area is a logo, and also a fixed width boundary.The logo is the same area as the boundary, and together they equal 21204.

    Whats the area of the logo? What's the area of the boundary?

    Cheers -- sylas
     
  10. Jun 1, 2009 #9
    Re: Need lots of help! :D

    10602? i divided the total area by 2.
     
  11. Jun 1, 2009 #10

    sylas

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    Re: Need lots of help! :D

    That's right.

    Now the next bit is going to take a bit of algebra. There are several ways of doing it but they all get the same answer.

    Use a variable "w" to represent the width of the boundary. The logo is a smaller rectangle inside the big one. What is its length and width? You won't be able to give them as numbers just yet, but you can get an expressing using "w".
     
  12. Jun 1, 2009 #11
    Re: Need lots of help! :D

    huh? I don't get it. :( so I have 10602 as the area, and now I need the legnth and width of it.

    10602 = L x W

    how do I get the length & width when i have the areA?
     
  13. Jun 1, 2009 #12

    sylas

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    Re: Need lots of help! :D

    One approach I use with this sort of problem is to give variables, for everything, and then write down as many equations as I can.

    For example. You have used L and W for the length and width of the logo. Now also use B for the width of the boundary.

    Can you write expression using L, W and B for the length and width of the whole mat?

    Length: 294.5 = ???[L,W,B]
    Width: 72 = ???[L,W,B]

    I'm also a big fan of drawing pictures. You can label your diagram with L, W and B, to show what lengths and widths they refer to in the problem.
     
  14. Jun 1, 2009 #13
    Re: Need lots of help! :D

    can you just give me the answer? i know thats probably the right way to learn in ur opinion, but hear me out. if you give me the answer then i can see how you got there by going backwards. i think that'd be better because i really can't figure this out and I'm not really understanding you :(
     
  15. Jun 1, 2009 #14

    sylas

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    Re: Need lots of help! :D

    Sorry. There are rules here about how we do this. I am probably not explaining as well as I could, and you can help me with that as well.

    Here's the diagram I suggested for you, labeled with L, W and B.
    strompic.jpg

    Can you give n expression in terms of the variables for the total length?

    294.5 = ???
     
  16. Jun 1, 2009 #15
    Re: Need lots of help! :D

    294.5 = n

    ???
     
  17. Jun 1, 2009 #16

    sylas

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    Re: Need lots of help! :D

    That's using a new variable for the total length, and you know its value. I often use variables for known values also.

    But what I am asking now is for a new expression, which captures something you know about the problem. Specifically, the total length "n" can be given in terms of your other variables L, W and B in the diagram.

    n = ???

    Replace the ??? with an expression involving the other variables.
     
  18. Jun 1, 2009 #17
    Re: Need lots of help! :D

    is it...

    294.5 = l + 2b

    72 = w + 2b
     
  19. Jun 1, 2009 #18

    sylas

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    Re: Need lots of help! :D

    That's right!

    You now have given three equations, with three unknowns. The other one is the equation you gave back in [post=2220405]msg #11[/post].

    You want to find "b". You can re-arrange the two equations here into the form

    L = ???
    W = ???

    where ??? are expressions using "b". You use these expressions for L and W in your earlier equation in [post=2220405]msg #11[/post], you will have one equation, with one variable, and you have to solve that for the answer.
     
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