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Math Help - Simple Triangle Question

  1. #1
    iwz
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    Simple Triangle Question

    The triangle PQR isn't a square triangle.
    PQ = 9.9cm
    QR = 22.3cm
    RP = 18.1cm

    (Going around the triangle anti clockwise)

         P


    Q             R

    Calculate the area of the triangle.

    Please help me and state how you did it
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  2. #2
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    Cosine rule....

    a^2 = b^2 + c^2 - 2bcCosA

    2bcCosA = b^2 + c^2 - a^2

    CosA = (b^2 + c^2 - a^2) / 2bc

    CosA = (9.9^2 + 22.3^2 - 18.1^2) / 441.51

    A = arccos((9.9^2 + 22.3^2 - 18.1^2) / 441.51) = 0.919389657 radians

    Area = 0.5bcSinA = 87.8 cm^2
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by iwz View Post
    The triangle PQR isn't a square triangle.
    PQ = 9.9cm
    QR = 22.3cm
    RP = 18.1cm

    (Going around the triangle anti clockwise)

         P


    Q             R

    Calculate the area of the triangle.

    Please help me and state how you did it
    This problem can also be solved using Heron's formula

    A=\sqrt{s(s-a)(s-b)(s-c)}, \mbox{ where } s=\frac{a+b+c}{2}

    and a,b,c are the side lengths of the triangle.

    s=\frac{9.9+22.3+18.1}{2}=25.15
    so the Area is
    A=\sqrt{(25.15)(25.15-9.9)(25.15-22.3)(25.15-18.1)} \approx 87.79cm^2
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  4. #4
    iwz
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    Quote Originally Posted by sean.1986 View Post
    Cosine rule....

    a^2 = b^2 + c^2 - 2bcCosA

    2bcCosA = b^2 + c^2 - a^2

    CosA = (b^2 + c^2 - a^2) / 2bc

    CosA = (9.9^2 + 22.3^2 - 18.1^2) / 441.51

    A = arccos((9.9^2 + 22.3^2 - 18.1^2) / 441.51) = 0.919389657 radians

    Area = 0.5bcSinA = 87.8 cm^2
    Thanks, that's what i was looking for
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