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Math Help - Geometry question I can't solve

  1. #1
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    Geometry question I can't solve

    The perimeter of an isosceles right triangle is 16+16√2 ((if you cant read that it is 16+16 * square root 2)). What is the length of the hypotenuse?

    I understand that it should be 2x + x√2 = 16+16√2, but not sure how to solve it.

    Any help would be great thanks!
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  2. #2
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    Hello, fwl200!

    The perimeter of an isosceles right triangle is: . 16+16\sqrt{2}
    What is the length of the hypotenuse?

    I understand that it should be: . 2x + x\sqrt{2} \:=\: 16+16\sqrt{2}
    but not sure how to solve it.

    The equal sides are x; the hypotenuse is x\sqrt{2}


    We have: . 2x + x\sqrt{2} \;=\;16 + 16\sqrt{2}

    Factor: . x(2 + \sqrt{2}) \;=\;16(1 + \sqrt{2})

    Hence: . x \;=\;\frac{16(1+\sqrt{2})}{2 + \sqrt{2}}


    Rationalize: . x \;=\;\frac{16(1+\sqrt{2})}{2+\sqrt{2}}\cdot\frac{2-\sqrt{2}}{2-\sqrt{2}} \;=\;\frac{16(2 - \sqrt{2} + 2\sqrt{2} - 2)}{4 - 2}

    . . . . . . x \;=\;\frac{16\sqrt{2}}{2} \;=\;8\sqrt{2}


    Therefore, the hypotenuse is: . x\sqrt{2} \;=\;(8\sqrt{2})(\sqrt{2}) \;=\;\boxed{16}

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  3. #3
    Senior Member nikhil's Avatar
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    Let length of equal sides be =x
    length of hypotenus be =y
    according to the question
    2x+y=16+16*2^(1/2)
    also
    y^2=2x^2 therfor
    y=2^(1/2) x
    substituting the value we get
    2x+2^(1/2) x=16+16*2^(1/2)
    therfor
    x(2+2^(1/2))= 16+16*2^(1/2)
    x=[16+16*2^(1/2)]/(2+2^(1/2))
    rationalise it by multiplying numerator and denominator by (2-2^(1/2)) by doing so we get
    x=8*2^(1/2)
    since
    y=x*2^(1/2)
    so y=16
    hence length of hypotenuse is 16 unit.
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  4. #4
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    Thanks!

    Thanks so much!
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