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Math Help - Prove that the area of the quadrilateral

  1. #1
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    Prove that the area of the quadrilateral

    A quadrilateral ABCD. The points BCD form an equilateral triangle. If AB=u; AD=v and Angle BAD= theta, prove that the area of the quadrilateral is

    \frac{1}{4}(\sqrt{3}u^2+\sqrt{3}v^2+2uv\sin \theta -2\sqrt{3}uv\cos \theta )

    thanks as always!
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  2. #2
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    Quote Originally Posted by BabyMilo View Post
    A quadrilateral ABCD. The points BCD form an equilateral triangle. If AB=u; AD=v and Angle BAD= theta, prove that the area of the quadrilateral is

    \frac{1}{4}(\sqrt{3}u^2+\sqrt{3}v^2+2uv\sin \theta -2\sqrt{3}uv\cos \theta )

    thanks as always!
    let BD = BC = CD = s

    using the cosine law ...

    s^2 = u^2+v^2-2uv\cos{\theta}

    area of triangle BCD ...

    A_1 = \frac{1}{2}s^2 \sin(60^\circ) = \frac{\sqrt{3}}{4}(u^2+v^2-2uv\cos{\theta})

    area of triangle ABD ...

    A_2 = \frac{1}{2}uv\sin{\theta}

    area of quad ABCD = A_1 + A_2

    can you finish from here?
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