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Math Help - Frustrating Geometry Problem

  1. #1
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    Frustrating Geometry Problem

    Ok... I have to do this problem attached.

    It's so frustrating. I have to find the length of squares 1, 2 and 3. Also, I have to show that square 3 is twice the area of the entire square. I'd like to share some of my work.

    I know each side is 1, so we know the diagonal is sqrt(2).

    I let the side of the square 1 be x. I let the side of square 3 be y-x. We know that x + y = 1. I then tried to find the hypotenuse of square 1, 2 (and consequently 3) as well as another one of 1.

    So 2*(hypot. of square 1) + hypot. of square 2 = sqrt(2)

    So, sqrt(2*(y-x)^2) + sqrt(2*sqrt(2x^2)) = sqrt(2)

    But when I solve this I get such a complicated answer and I'm so stuck.

    I also tried using proportions, like saying:

    1/sqrt(2) = x/(sqrt(2x^2)) ... but all my calculator gave me was x > 0 which wasn't helpful.

    Any help would be great
    Attached Thumbnails Attached Thumbnails Frustrating Geometry Problem-square.bmp  
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  2. #2
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    Quote Originally Posted by Startbucks View Post
    ... I have to find the length of squares 1, 2 and 3. Also, I have to show that square 3 is twice the area of the entire square. To me that's impossible: The included square can't be greater than the entire square. I'd like to share some of my work.

    I know each side is 1, so we know the diagonal is sqrt(2).

    I let the side of the square 1 be x. I let the side of square 3 be y-x. According to your sketch this is the side of square 2. We know that x + y = 1. Where from do you know this? I then tried to find the hypotenuse of square 1, 2 (and consequently 3) as well as another one of 1. Do you mean diagonal? Why?

    So 2*(hypot. of square 1) + hypot. of square 2 = sqrt(2)

    So, sqrt(2*(y-x)^2) + sqrt(2*sqrt(2x^2)) = sqrt(2)

    But when I solve this I get such a complicated answer and I'm so stuck.

    I also tried using proportions, like saying:

    1/sqrt(2) = x/(sqrt(2x^2)) ... but all my calculator gave me was x > 0 which wasn't helpful.

    Any help would be great
    I assume that:

    1. I have to show that square 3 is half the area of the entire square.

    2. We know that x + y = 1

    If so you know:

    y^2 = \frac12~\implies~y=\frac12 \cdot \sqrt{2} .........Length of side of square 3

    (1-x)^2=\frac12~\implies~x=1-\frac12 \cdot \sqrt{2}.........Length of side of square 1

    Consequently the side length of square 2 is:

    \frac12 \cdot \sqrt{2} - \left(1-\frac12 \cdot \sqrt{2}\right) =  \sqrt{2} - 1
    Attached Thumbnails Attached Thumbnails Frustrating Geometry Problem-threesquares.png  
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