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Math Help - area of a triangle

  1. #1
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    area of a triangle

    Find the area of the triangle with vertices , , and .
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by wvlilgurl View Post
    Find the area of the triangle with vertices , , and .
    are you aware of Heron's formula?

    Of course, there are other ways to approach this, like the (1/2)base*height formula
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  3. #3
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    Quote Originally Posted by wvlilgurl View Post
    Find the area of the triangle with vertices , , and .
    Let A(-1, -4) , B(3, -3), and C(-4, 4) be the coordinates of triangle. Please see attached diagram.

    Using distance formula find

     BC = a = \sqrt{(3+4)^2+(-3-4)^2}= \sqrt{98}

     CA = b = \sqrt{(-4+1)^2+(4+4)^2}= \sqrt{73}

     AB = c = \sqrt{(-1-3)^2+(-4+3)^2}= \sqrt{17}

    Now, use cos law to find angle C

     \cos C = \frac{a^2+b^2-c^2}{2ab}

     \cos C = \frac{98+73-17}{2\sqrt{98}\sqrt{73}}

     \cos C = 0.910366

    C = 24.44 degrees

    Area of triangle ABC = \frac{1}{2}ab \;\sin C

    = \frac{1}{2}\sqrt{98}\sqrt{73} \;\sin 24.44

    = 17.5 square units.
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  4. #4
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    The formula for arrea of a triangle is

    Area of triangle ABC = \frac{1}{2}ab \;\sin C

    Area of triangle ABC = \frac{1}{2}bc \;\sin A

    Area of triangle ABC = \frac{1}{2}ca \;\sin B
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