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Geometry problem (rectangles)

  1. Mar 16, 2017 #1
    • Moved thread to homework forum
    There are three "concentric" rectangles, one inside the other, like in this figure:
    lqFobze.png

    The image is not perfect, but we know that the distance between the sides of the inner-most rectangle and the middle rectangle is always "X", while the distance between the middle rectangle and the outer rectangle is equal to 2X.

    The outer-most rectangle has sides of these measurements: 80; 100*√2
    The inner-most rectangle has an area of: 5600*√2 - 4200

    Calculate X.

    The way I tried to solve it was with an equation, by working on the area of the inner-most rectangle. Since its sides should be equal to the outer-rectangle sides minus a total of 6x, then we can say:

    (80-6x)*(100*√2 - 6x) = 5600*√2 - 4200

    I confirmed with wolfram alpha that this equation would give me the correct solution (5√2). It also gives another solution, which is geometrically incompatible (X would be too big to be a smaller part of the bigger triangle). So if it wasn't logically clear enough, I have the "mathematical evidence" that I haven't done anything wrong.

    However, the problem is that when I go and try to solve the equation, it becomes an awful mess of numbers which are incompatible with the typical level of difficulty of the exercises found on the textbook where this exercise comes from. The last passage before using the formula to solve the equation is:

    3*x2 - (40+50√2) x + (200√2 + 350) = 0

    I double-checked on wolfram alpha again, and it's all good to give the solution 5√2.

    However, I can't seem able to figure how to actually do the maths and calculate it by myself. The Δ I get is 10800 - 400√2, and I'm stuck there. I can extract the 400, but I'm still left with 20* √(27-√2) afterwards.

    Is there another way to set and solve the problem? If not, how am I supposed to solve the equation with basic high-school maths?
     
  2. jcsd
  3. Mar 16, 2017 #2

    phinds

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    this is a conceptually trivial, but arithmetically messy, quadratic equation. Do you know the formula for the answer to a quadratic equation? That is a high school math algebra derivation.
     
  4. Mar 16, 2017 #3

    mfb

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    I get a different expression to simplify.

    If you encounter double square roots like this, you can always assume that you can write them as ##a+b\sqrt 2##, then square both sides: ##(a+b \sqrt 2)^2 =a^2+2b^2 +2\sqrt 2 a b=2400+1600 \sqrt 2 ##. The square root terms have to match: ##2ab = 1600##, and the rational terms have to match: ##a^2+2b^2=2400##. Solving this leads to a=40, b=20.
     
  5. Mar 17, 2017 #4
    Let's try the Δ again...

    (40+50√2)2 - 12*(350+200√2) =
    = 1600+5000+4000√2 - 4200 - 2400√2 =
    = 2400 - 1600√2

    Three times I did this before, and three times I forgot to double the product between 40 and 50√2 AND that 4200 was negative. That's so lame lol

    Ok, so then I used the double-radicals formula, and I get Δ = (40+20√2)2

    Then I can continue solving the equation, and I do get the expected results: 5√2 and 5/3 * (8+7√2)

    What a mess lol

    Thanks for helping me solve the equation.

    Is there no other way to solve the problem though? :p This is really not the typical difficulty level of problems for us (it's usually a matter of setting up an equation, and then smooth-sailing maths from there to the end), so I wonder if there is maybe another, much simpler geometric approach.
     
  6. Mar 17, 2017 #5

    mfb

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    I don't see an easier approach. You can introduce a new variable for the side length, e. g. call the shorter side y, that gives a different quadratic equation but it leads to the same double square root.

    Instead of the central rectangle, you can also calculate the area that is cut away outside, but that will also lead to a quadratic equation, and I guess the same double square root again.
     
  7. Mar 17, 2017 #6
    Yea, I actually did that too before posting here, and it's exactly the same.

    Really weird to find a problem this "messy" in the middle of much easier ones, but I guess the authors just wanted to test our patience on this one.

    Thanks for the help.
     
  8. Mar 18, 2017 #7

    kuruman

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    Here is an easier approach. Move the rectangles so that their upper left corners coincide (see figure).
    Rectangles.png
    The difference in area between the larger and smaller rectangle is ##\Delta A = 80\times 100\sqrt{2}-(5600 \sqrt{2}-4200)##. But ##\Delta A## is also the gray shaded area, written as ##\Delta A = 80\times 6x+100 \sqrt{2}\times 6x - 36x^2##. The ##36x^2## is subtracted in order not to count the square at the lower corner twice.
    Finally solve the quadratic
    $$ 80\times 6x+100 \sqrt{2}\times 6x - 36x^2= 80*100 \sqrt{2}-(5600 \sqrt{2}-4200)$$
     
  9. Mar 18, 2017 #8

    mfb

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    That is nearly the same ugly quadratic equation we had before.
     
  10. Mar 18, 2017 #9

    kuruman

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    Indeed it is. Sorry.
     
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