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Math Help - Radical Equations Dimensions Problem

  1. #1
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    Radical Equations Dimensions Problem

    Been having some trouble with this problem, was wondering if anyone would be able to help me out.

    Given: Two rectangles have dimensions in meters expressed as radical equations.

    If the difference of the lengths of their diagonals is 2, find the dimensions of the rectangles.

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    Quote Originally Posted by Vesicant View Post
    Been having some trouble with this problem, was wondering if anyone would be able to help me out.

    Given: Two rectangles have dimensions in meters expressed as radical equations.

    If the difference of the lengths of their diagonals is 2, find the dimensions of the rectangles.

    Shown below:

    use Pythagoras to determine the length of each diagonal, then set up an equation ...

    short diagonal + 2 = long diagonal

    solve for x, and ATQ.
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    Quote Originally Posted by skeeter View Post
    use Pythagoras to determine the length of each diagonal, then set up an equation ...

    short diagonal + 2 = long diagonal

    solve for x, and ATQ.
    ah pythagoras? I was using the diagonal formula(d^2= width^2+height^2) and got D=5x^2+4x+1

    Care to tell me what answer I would get when using the pythagoram theorem so I know that I'm doing it right, I assumed I would have to find the diagonal first and then get the pythagorean equation for the two triangles.?
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    Quote Originally Posted by Vesicant View Post
    ah pythagoras? I was using the diagonal formula(d^2= width^2+height^2) and got D=5x^2+4x+1

    Care to tell me what answer I would get when using the pythagoram theorem so I know that I'm doing it right, I assumed I would have to find the diagonal first and then get the pythagorean equation for the two triangles.?
    set it up using Pythagoras and I'll let you know.
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    Set up with pythagoras the first would be

    sqrt x^2 + (sqrt2x+1)^2=c^2
    x^2+ (2x+1)(2x+1)=c^2
    X^2 +4x^2+2x+2x+1=c^2
    5x^2+4x+1=c^2

    The second:

    1^2+sqrt x^2=c^2
    1+x^2=c^2

    Hows that look?
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    Quote Originally Posted by Vesicant View Post
    Set up with pythagoras the first would be

    sqrt x^2 + (sqrt2x+1)^2=c^2
    x^2+ (2x+1)(2x+1)=c^2
    X^2 +4x^2+2x+2x+1=c^2
    5x^2+4x+1=c^2

    The second:

    1^2+sqrt x^2=c^2
    1+x^2=c^2

    Hows that look?
    first of all, (\sqrt{x})^2 = x and (\sqrt{2x+1})^2 = 2x+1

    make your corrections
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    Quote Originally Posted by skeeter View Post
    first of all, (\sqrt{x})^2 = x and (\sqrt{2x+1})^2 = 2x+1

    make your corrections
    sqrt x^2 + (sqrt2x+1)^2=c^2
    x+2x+1=c^2
    3x+1=c^2
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  8. #8
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    Quote Originally Posted by Vesicant View Post
    sqrt x^2 + (sqrt2x+1)^2=c^2
    x+2x+1=c^2
    3x+1=c^2
    ok ... now find c^2 for the other rectangle.
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    Quote Originally Posted by skeeter View Post
    ok ... now find c^2 for the other rectangle.
    1+x= c^2

    Now what?
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    Quote Originally Posted by Vesicant View Post
    1+x= c^2

    Now what?
    well, the first diagonal length is \sqrt{3x+1} , and the second is \sqrt{1+x}

    I believe the relative sizes of your rectangles in the sketch are backwards. the left one should be larger than the right ...

    you know the difference in the lengths between the two diagonals is 2.

    \sqrt{x+1} +  2 = \sqrt{3x+1}

    now solve for x and then determine the dimensions of each rectangle.
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    Quote Originally Posted by skeeter View Post
    well, the first diagonal length is \sqrt{3x+1} , and the second is \sqrt{1+x}

    I believe the relative sizes of your rectangles in the sketch are backwards. the left one should be larger than the right ...

    you know the difference in the lengths between the two diagonals is 2.

    \sqrt{x+1} +  2 = \sqrt{3x+1}

    now solve for x and then determine the dimensions of each rectangle.
    Alright so X=2

    Then I just input x into all of the original expressions I was given and then multiply width by height to get my dimensions?
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    Quote Originally Posted by Vesicant View Post
    Alright so X=2

    Then I just input x into all of the original expressions I was given and then multiply width by height to get my dimensions?
    x does not equal 2

    how did you get that?
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  13. #13
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    Quote Originally Posted by skeeter View Post
    x does not equal 2

    how did you get that?
    \sqrt{x+1} +  2 = \sqrt{3x+1}
    (\sqrt{x+1} +  2)^2 = (\sqrt{3x+1})^2
    x+1+4=3x+1
    4=2x
    4/2=2x/2
    X=2
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  14. #14
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    Quote Originally Posted by Vesicant View Post
    \sqrt{x+1} +  2 = \sqrt{3x+1}
    \textcolor{red}{(\sqrt{x+1} +  2)^2} = (\sqrt{3x+1})^2
    x+1+4=3x+1
    4=2x
    4/2=2x/2
    X=2
    left side ...

    (\sqrt{x+1} + 2)^2

    (x+1) + 4\sqrt{x+1} + 4
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  15. #15
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    Quote Originally Posted by skeeter View Post
    left side ...

    (\sqrt{x+1} + 2)^2

    (x+1) + 4\sqrt{x+1} + 4
    ah I see what I did wrong

    =3x+1

    (x+1)^2 + 4^2(\sqrt{x+1})^2 + 4^2=(3x+1)^2
    (x+1)(x+1)+16(x+1) + 16=(3x+1)(3x+1)
    x^2+2x+1+16x+32=9x^2+6x+1
    =-8x^2+4x-32
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