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Math Help - isosceles theorem?

  1. #1
    Junior Member
    Joined
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    isosceles theorem?

    I have two isosceles triangles that have the same base length, x.
    For one of the isosceles triangles, the sides opposite the base angles have length sqrt(11).
    For the other isosceles triangle, the sides opposite the base angles have length sqrt(17).
    There is another triangle that has one side length x, one side length sqrt(11), and one side sqrt(17).

    what is the value of x?

    I do not see how to apply any law of sines/cosines without knowing an angle. I do not see how to apply any congruence theorems.

    I suspect i am missing some theorem about isosceles triangles, perhaps?

    Given two sides of a triangle, i ought to be able to find the third, no?

    Thanks for your help.
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  2. #2
    MHF Contributor
    Joined
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    Quote Originally Posted by n0083 View Post
    I have two isosceles triangles that have the same base length, x.
    For one of the isosceles triangles, the sides opposite the base angles have length sqrt(11).
    For the other isosceles triangle, the sides opposite the base angles have length sqrt(17).
    There is another triangle that has one side length x, one side length sqrt(11), and one side sqrt(17).

    what is the value of x?

    I do not see how to apply any law of sines/cosines without knowing an angle. I do not see how to apply any congruence theorems.

    I suspect i am missing some theorem about isosceles triangles, perhaps?

    Given two sides of a triangle, i ought to be able to find the third, no?

    Thanks for your help.
    This is a tricky question, I found out.
    I think it cannot be solved the usual ways.

    The x will stretch as it wishes but there are limits to its stretch.

    The minimum stretch, or squeeze, would be when x tends to be zero.
    If x = 0, there would be no isosceles triangles left, and there would be no separate, the 3rd, triangle. Hence x cannot be zero.
    So that the 3rd triangle would exist, the x should tend to become [sqrt(17) -sqrt(11)]

    The maximum stretch would be when x tends to be 2sqrt(11).
    Then the isosceles triangle with two sqrt(17) sides would be visible, while the isosceles triangle with two sqrt(11) sides would tend to be flat or a straight line only.
    The 3rd triangle would exist because any one side would be less than the sum of the other two sides.

    Therefore,
    [sqsrt(17) -sqrt(11)] < x < 2sqrt(11) -------------answer.
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  3. #3
    Junior Member
    Joined
    Oct 2008
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    thanks for the excellent response.

    I consider this problem solved.

    thanks again.
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