A trapezium is such that three of its sides have lengths as 6 cm, then the length (in cm) of the fourth side such that the area of trapezium is maximum, is

From the figure above, Area of the trapezium
$$\begin{gathered} =6\times 6\sin \theta +2\times \frac{1}{2}\times 6\cos \theta \times 6\sin \theta \\ A=36\sin \theta +36\sin \theta \cos \theta \\ A=36\sin \theta +18\sin 2\theta \\ \frac{dA}{d\theta }=0\Rightarrow 36\cos \theta +36\cos 2\theta =0 \\ 2{\cos }^2\theta +\cos \theta -1=0 \\ \cos \theta =\frac{1}{2}or\cos \theta =-1 \\ for\theta =\frac{\pi }{3},\frac{d^{2}A}{d{\theta }^2}<0 \end{gathered}$$ 
So, area is maximum when  $\theta =\frac{\pi }{3}$
Length of fourth side 
$=\left(6\cos \frac{\pi }{3}\right)\times 2+6=12cm$