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Word Problem

  1. Jul 12, 2005 #1
    Hello All

    I have this word problem that I am having problems with

    Here it is

    A circular piece of sheet metal has a diameter of 20 in. The edges are to be cut off to form a rectangle of area in^2. What are dimensions of the rectangle??

    Here is what I have figured out. Not sure if it of any use

    Formula of Circle is

    10^2 = y ^ 2 + x ^ 2

    Area of rectangle

    160 = LW

    Not sure how to go from here. Please help

    Thanks
     
  2. jcsd
  3. Jul 12, 2005 #2

    saltydog

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    So, as I see it you have a circle of radius 10 and want to draw a rectangle in it with a total area of 160 right? Look at the plot below (drawn to scale) and where I have theta. Can you figure out what the length and height has to be using some trig expressions?
     

    Attached Files:

  4. Jul 12, 2005 #3
    Don't think I have enough info to use trig expessions. All I know for that triangle is the hypotenuse is 10 in and is a right angle triangle.

    How can I determine the remaining info?
    if I use sin I need opposite side?
     
  5. Jul 12, 2005 #4
    IT’S ALL GEOMETRY…

    X^2 + Y^2 = 20^2 = 400 ----------------------DIAGONAL OF THE RECTANGLE

    XY = 160
    2XY = 320

    X^2 + 2XY + Y^2 = 400 + 320 = 720
    (X+Y) ^2= 720


    X+Y = 12 √5


    X^2 – 2XY + Y^2 = 400 – 320 = 80
    (X-Y)^ 2 = 80

    X-Y = 4 √5

    2 EQS., 2 UNKNOWNS

    2X = 4 √5 + 12 √5 = 16 √5


    X = 8 √5
    Y = 4√5


    :surprised
     
  6. Jul 12, 2005 #5
    Nice explanation plucker.
     
  7. Jul 12, 2005 #6
    thanks why did you multiply xy = 160 by 2 ??
     
  8. Jul 12, 2005 #7
    adding 2xy to x^2 + y^2 will yield to an expression x^2 + 2xy + y^2 which is the square of the sum of the sides of the rectangle...(x + y)

    subtracting 2xy to x^2 + y^2 will yield to an expression x^2 - 2xy + y^2 which is the square of the difference of the sides of the rectangle...(x - y)

    you will have 2 eqs and 2 unknows...
     
  9. Jul 13, 2005 #8
    how can you just add values to the equation?? How do you know when to do this?
     
  10. Jul 13, 2005 #9

    saltydog

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    Hey Powp. Sorry for not get getting back with you last night but looks like these guys helped you ok. Yea, I didnt' see the obvious algegra:

    [tex]x^2+y^2=400[/tex]

    [tex]xy=160[/tex]

    Solving for x or y in the second equation and substituting into the first leads to the same equation for both x and y:

    [tex]r^4-400r^2+160^2=0[/tex]

    Now, using the quadratic formula, just figure out which of the 4 roots make sense.

    I drew a pretty graph though. :wink:
     
    Last edited: Jul 13, 2005
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