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Math Help - Hypoteneuse of an isosceles right triangle

  1. #1
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    Hypoteneuse of an isosceles right triangle

    I am studying for the GMAT, and I have hit a snag. This is the question and answer.

    The perimeter of an isosceles right triangle is 16+16 sqr root(2). What is the length of the hypoteneuse?

    I know that the ratio of the sides of an isosceles right triangle is Side:Side:Side Sqr Root(2)

    However, I was under the impression that the hypoteneuse is always the side that is Side Sqr Root (2)

    But the answer to this question is that the hypoteneuse's length is actually 16. How can that be possible?
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  2. #2
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    Re: Hypoteneuse of an isosceles right triangle

    If the sides are 1 the hypothenuse is rad2. !^2 + 1^2 = (rad2)^2
    If sides are 16 the hyp is 16 rad 2
    If sides are 16/rad2 the hyp is 16


    bjh
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  3. #3
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    Re: Hypoteneuse of an isosceles right triangle

    Quote Originally Posted by bjhopper View Post
    If the sides are 1 the hypothenuse is rad2. !^2 + 1^2 = (rad2)^2
    If sides are 16 the hyp is 16 rad 2
    If sides are 16/rad2 the hyp is 16


    bjh
    Because the perimeter is 16+16^2, then wouldn't the sides have to both be 8 rad 2? The reason I ask is because I only see 2 terms in the perimeter equation, but I can't see how the length of the 2 sides are combined into the 16+16 rad 2 value.
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    Re: Hypoteneuse of an isosceles right triangle

    Quote Originally Posted by KingNathan View Post
    I am studying for the GMAT, and I have hit a snag. This is the question and answer.

    The perimeter of an isosceles right triangle is 16+16 sqr root(2). What is the length of the hypoteneuse?

    I know that the ratio of the sides of an isosceles right triangle is Side:Side:Side Sqr Root(2)

    However, I was under the impression that the hypoteneuse is always the side that is Side Sqr Root (2)

    But the answer to this question is that the hypoteneuse's length is actually 16. How can that be possible?
    Maybe a bit of a trick question..

    Label the perpendicular sides "x".
    Then use Pythagoras' theorem.

    x^2+x^2=2x^2

    so the hypotenuse length is

    \sqrt{2}x

    Then

    2x+\sqrt{2}x=16+16\sqrt{2}

    As the coefficients are inconsistent, we cannot solve for x right away.

    x(2+\sqrt{2})=16+16\sqrt{2}\Rightarrow\ x=\frac{16+16\sqrt{2}}{2+\sqrt{2}}

    Now you can use the surd conjugate to solve for x.

    Multiply by \frac{2-\sqrt{2}}{2-\sqrt{2}}

    You should obtain

    x=8\sqrt{2}
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    Re: Hypoteneuse of an isosceles right triangle

    Your ratio is right. We can divide throughout by \sqrt{2} to make life easier for us

    \dfrac{x+x}{\sqrt2} + x = 16+16\sqrt{2}

    Solve for x to give the length of the hypotenuse
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    Re: Hypoteneuse of an isosceles right triangle

    Quote Originally Posted by e^(i*pi) View Post
    Your ratio is right. We can divide throughout by \sqrt{2} to make life easier for us

    \dfrac{x+x}{\sqrt2} + x = 16+16\sqrt{2}

    Solve for x to give the length of the hypotenuse
    Thanks to everyone for helping out. This is my challenge. I am studying for the GMAT. In order to survive on this test, I need to be able to do these types of questions quickly. For every answer I can figure out in 15 seconds, gives me 2 extra minutes to work on a very difficult question (which is worth more points under the grading scale). It is multiple choice. The idea of the Side:Side:Side Rad2 ratio should be open and shut for me. I see a perimeter and the goal is to quickly seperate the sides and the hypoteneuse. With only 2 terms in the perimeter formula, that makes it more difficult. For example, I see 16+16 rad 2 and I assume that the 16 is the sum of the 2 sides ( 8 and 8), but that doesn't make sense because then the hypoteneuse would have to be 8 rad 2. This leads me to believe that its the other way around, and the sides are actually 8 rad 2, and 8 rad 2. So my question now is how would I quickly figure out the hypoteneuse (Presumably at this point, 8 rad 2 * rad 2) to be the number 16? Is there a formula for that?
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    Re: Hypoteneuse of an isosceles right triangle

    Quote Originally Posted by KingNathan View Post
    Thanks to everyone for helping out. This is my challenge. I am studying for the GMAT. In order to survive on this test, I need to be able to do these types of questions quickly. For every answer I can figure out in 15 seconds, gives me 2 extra minutes to work on a very difficult question (which is worth more points under the grading scale). It is multiple choice. The idea of the Side:Side:Side Rad2 ratio should be open and shut for me. I see a perimeter and the goal is to quickly seperate the sides and the hypoteneuse. With only 2 terms in the perimeter formula, that makes it more difficult. For example, I see 16+16 rad 2 and I assume that the 16 is the sum of the 2 sides ( 8 and 8), but that doesn't make sense because then the hypoteneuse would have to be 8 rad 2. This leads me to believe that its the other way around, and the sides are actually 8 rad 2, and 8 rad 2. So my question now is how would I quickly figure out the hypoteneuse (Presumably at this point, 8 rad 2 * rad 2) to be the number 16? Is there a formula for that?
    Exactly!

    16\sqrt{2}=2x

    since 2x\ne\ 16
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    Re: Hypoteneuse of an isosceles right triangle

    Quote Originally Posted by Archie Meade View Post
    Exactly!

    16\sqrt{2}=2x

    since 2x\ne\ 16
    Perfect. Thanks again for your assistance.
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    Re: Hypoteneuse of an isosceles right triangle

    Hi KingNathan,
    The key is to divide the perimeter into three parts

    16 + 16rad2 = perimeter = 16 + 8rad2 + 8rad2. 8rad2 *rad2=16


    bjh
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    Re: Hypoteneuse of an isosceles right triangle

    I think, to answer quickly,
    you just need to make the equations consistent.

    2x+x\sqrt{2}=16+16\sqrt{2}

    x\ne\ 16,\;\;\;2x\ne\ 16

    2x+x\sqrt{2}=16\sqrt{2}+16

    \Rightarrow\ 2x+x\sqrt{2}=2(8\sqrt{2})+(8\sqrt{2})\sqrt{2}
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