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Math Help - 45-45-90 Triangle

  1. #1
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    45-45-90 Triangle

    Hi

    Principle 5: The length of a leg opposite a 45 degree angle equals one-half the length of the hypotenuse times the square root of 2.

    a=\tfrac{1}{2}c\sqrt{2}

    why is this so? when I solve for 'a' in c^{2}=a^{2}+a^{2} I get \frac{c}{\sqrt{2}}=a

    Thanks.
    Last edited by norifurippu; June 1st 2008 at 11:16 AM.
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    A riddle wrapped in an enigma
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    Quote Originally Posted by norifurippu View Post
    Hi

    Principle 5: The length of a leg opposite a 45 degree angle equals one-half the length of the hypotenuse times the square root of 2.

    a=\tfrac{1}{2}c\sqrt{2}

    why is this so? when I solve for 'a' in c^{2}=a^{2}+a^{2} I get \frac{c}{\sqrt{2}}=a

    Thanks.
    When your final answer has a radical in the denominator, we must 'rationalize' the denominator. By that, I mean, remove the radical from the denominator. How do we do that? We simply mutiply both numerator and denominator by the radical. Same as multiplying by 1. Value doesn't change, just how the fraction looks. See:

    \frac{c}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}=\frac{c  \sqrt2}{2}=\frac{1}{2}c\sqrt2
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    Thanks!
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