Prove that the area of the quadrilateral

**BabyMilo** · January 19th 2010, 01:41 PM

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!

**skeeter** · January 19th 2010, 02:32 PM

Originally Posted by BabyMilo

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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