Answer:
Step-by-step explanation:
Consider ABCD is a parallelogram and AB=CD=6 and BC=DA=12, Let AE be the height of the parallelogram, then from ΔAED, we have
\(\frac{AE}{AD}=sin60^{\circ}\)
⇒\(\frac{AE}{12}=\frac{\sqrt{3}}{2}\)
⇒\(AE=6\sqrt{3}\)
Then, the area of parallelogram=\(Base{\times}height\)
=\(CD{\times}AE\)
=\(6{\times}6\sqrt{3}\)
=\(36\sqrt{3}sq units\)
Therefore, the area of parallelogram is \(36\sqrt{3}sq units\).
