Geometry

Find area of a triangle with side lengths 15 15 and 8

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  1. simplest way:
    Heron's Formula
    A = √(s(s-a)(s-b)(s-c)), where s = (1/2)the perimeter.

    s = (1/2)(15 + 15 + 8) = 19
    s-a = 19-15 = 4
    s-b = 19-15 = 4
    s-c = 19-8 - 11
    area = √(19(4)(4)(11) = √3344
    = appr 57.83

    2nd way: works in this case because it is isosceles.
    Sketch the triangle, let the angle between the two equal sides be 2θ
    Draw a perpendicular from that angle to the base.
    sinθ = 4/15
    θ = 15.466..
    2θ = 30.932..

    area = (1/2)(15)(15)sin30.932..
    = 57.83 , same as above

    There are other ways, but these two work nicely. The 2nd way of course only worked so fine, because we had an isosceles triangle

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    posted by Reiny
  2. Thank you!

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  3. An easier way would be to find the perpendicular and use pythag. Thm

    c^2 = a^2 + b^2
    15^2 = 4^2 + b^2
    b = 14.45

    A = 1/2 bh
    A= 1/2 8 x 14.45

    you will get the same answer as above.

    Depending on the level you are working at this might be what your teacher is looking for.

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    posted by John

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