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Balancing Areas of Two Rectangles

  1. Jul 2, 2012 #1
    1. The problem statement, all variables and given/known data

    Hi all hope this is posted in the right area, attached is an image showing what i have to figure out and I'm hoping theres a formula to do this instead of guess and check which I hope to avoid in future.

    Assuming all angles are 90° unless stated I wish to move the red line north-west a certain distance to create equal areas of the two shapes. As the line moves north-west the total area of both lots added together becomes larger.


    areaquestion.png


    2. Relevant equations

    Sadly i don't know any relevant equations that arent just basic knowlodge my hope is to get an equation where i substitute certain values for future problems.

    3. The attempt at a solution

    My attempts aren't even worth mentioning my crazy attempts at figuring it out result in disaster all i know is the answer must be be 1.xx


    Thanks in advance for any assistance with this problem I have.
     
  2. jcsd
  3. Jul 2, 2012 #2
    Only one variable here. Find all other constants.
    You can find the equation of equal area.
     
  4. Jul 2, 2012 #3

    Mentallic

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    Yep, it's definitely doable. For this question, the answer should be that the 22.46 length side should be shortened to 20.97, and thus the 15.41 side will be lengthened to 16.90.

    If you want to know how this is achieved (or are looking for a formula to simply plug your values in to spit out a result in future problems) then we'll start by substituting your numerical values for unknown constants and variables.

    Let the variables be the sides that need to be changed, thus

    [tex]x=15.41[/tex]
    [tex]y=22.46[/tex]

    The constants will be

    [tex]a=16.12[/tex]
    [tex]b=4[/tex]
    [tex]c=14.12[/tex]
    [tex]d=2.83[/tex]

    And let's consider that the angle in the corner doesn't change, so we don't need to make that a constant.

    The Area of A1 will be the South-East rectangle,

    [tex]A_1=x(a+b)[/tex]

    The Area of A2 is the North-West rectangle which includes the triangle bit. Now, the area of the rectangle is simply y*a, but to find the area of the triangle we need to use a bit of trigonometry. The hypotenuse is d, and the angle in this right triangle is 45o, thus the other sides are both,

    [tex]d\cos(45^o)=d\sin(45^o)=\frac{d}{\sqrt{2}}[/tex]

    And the area of a triangle is given by [tex]A=\frac{1}{2}bh[/tex] where in this case [tex]b=h=\frac{d}{\sqrt{2}}[/tex]

    So we now have,

    [tex]A_2=ya+\frac{d^2}{4}[/tex]

    Now we need both these areas to be equal, so we equate them,

    [tex]A_1=A_2[/tex]

    [tex]x(a+b)=ya+\frac{d^2}{4}[/tex]

    But what can we do with this? Well, we have one equation with 2 variables, so we can't really do much. We need another equation relating x and y.
    The only other thing we know about x and y is that their sum doesn't change, thus we can use

    [tex]x+y=k[/tex] in which your case, [tex]k=15.41+22.46=37.87[/tex]

    So now we have two equations with two unknowns, so we can find out the value of these unknowns.

    [tex]x+y=k[/tex]
    [tex]y=k-x[/tex]
    Substituting this into the first equation gives

    [tex]x(a+b)=(k-x)a+\frac{d^2}{4}[/tex]

    Now we solve for x,

    [tex]x(a+b)+(x-k)a=\frac{d^2}{4}[/tex]
    [tex]x(a+b+a)-ka=\frac{d^2}{4}[/tex]
    [tex]x(2a+b)=ka+\frac{d^2}{4}[/tex]
    [tex]x=\frac{ka+\frac{d^2}{4}}{2a+b}[/tex]

    And since [tex]y=k-x[/tex]
    We have
    [tex]y=k-\frac{ka+\frac{d^2}{4}}{2a+b}[/tex]

    You'll find that if you plug your values into these variables, you'll get the answers that I posted up the top.
     
  5. Jul 2, 2012 #4
    Thank you so much mentallic! thats an amazingly detailed reply for such a short time. Your working is sound and the answer is perfect.

    I figured this out just before you posted and i'll show you my working which is much less complex.

    I had written down (364-(16.12x)) = (310+(20.12x))

    I figured the small truncation was going to be obselete in this so i just used the above and then solved for x and made my formula A-a/B+b = x

    A = Largest Area
    a = Smallest Area
    B = Largest Side
    b = Smallest Side
    x = Amount line has to move to equal areas

    I actually can't even believe this worked and wouldn't even mind it being looked at and scrutinised.

    I'm still reading your post and doing my absoulute best to absorb every bit of information you posted. Once agaisnt thank you so much for a fast and informative response.
     
  6. Jul 2, 2012 #5

    Mentallic

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    No problem! If you have any questions or don't understand anything about it, just ask.

    That looks good :smile:
    And of course, the small triangle piece is only 2 units in area which is tiny compared to the rest of the land, but it will definitely make a small difference. It's upon you whether you think you need it or not though.
     
  7. Jul 2, 2012 #6
    Yea it does make a difference to the current area but if say the shape was just a standard rectangle with the same starting area of 364m² you would still end up moving the line the same distance.

    I finish decyphering your post and it was so comprehensive and helpful thank you again couldn't have asked for anything more! I now have to go read some other threads and hope I can give back in some way.
     
  8. Jul 2, 2012 #7

    Mentallic

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    Yep you're exactly right. It makes no difference that you've used the area in your formulae as opposed to how I've used the dimensions of the shapes, but the reason yours doesn't require knowing what the area of the triangle is (or any other irregular shapes at the ends) is because you're merely calculating how much the lines should move up or down, as opposed to how long the lines should be - that's the difference.

    No worries dude :smile: May I ask what this problem was for by the way?
     
  9. Jul 2, 2012 #8
    I am a draftsman for a surveying company and the client has asked for the lots to be created of equal size, so i ripped apart the cadastral information and quickly labelled it and took a screenshot.

    My logic of focusing on just the line is because this scenario happens a lot where the dividing line is moved slightly. I like understanding the logic behind both methods.

    EDIT: yea its not really homework i had no idea where to post my question though =[

    My boss said to just use guess and check and creep the line along, but I hate guess and check and knew you could figure an exact number, also future scenarios such as this will be simple now.
     
  10. Jul 2, 2012 #9

    Mentallic

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    Ahh cool :smile:

    As you should!


    Yeah I thought so. I was reluctant to post a full solution because it is against the rules on this forum, but I felt like I'd be forcing you to carry yourself through some maths that you likely haven't done in years.
    Except now I know you would've done just fine by the looks of it :smile:

    Guess and check? Well now you can tell him that you sat down for a few minutes, did some maths, and your first guess turned out perfect haha
     
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