In a quadrilateral $$ABCD$$, $$AB = 3$$, $$BC = 4$$, $$CD = 5$$, $$\angle ABC = \angle BCD = 120^\circ$$. Find the area of the quadrilateral.

Correct Answer: 20.35

Solution

Take $$B(0, 0)$$ and $$C(4, 0)$$. Then $$A = \left(-\frac{3}{2}, \frac{3\sqrt{3}}{2}\right)$$ and $$D = \left(\frac{13}{2}, \frac{5\sqrt{3}}{2}\right)$$. By the shoelace formula the area is $$\frac{1}{2}\left(10\sqrt{3} + \frac{27\sqrt{3}}{2}\right) = \frac{47\sqrt{3}}{4} \approx 20.35$$ sq. units.

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