Three sides of a trapezium are each equal to $k$ cm.Prove that the greatest possible area of the trapezium is $\frac{3\sqrt3 k^2}{4}$ sq cm. Three sides of a trapezium are each equal to $k$ cm.Prove that the greatest possible area of the trapezium is $\frac{3\sqrt3 k^2}{4}$ sq cm.
I let two non-parallel sides and one of the parallel sides as $k$(shorter one).I know that area of the trapezium is $\frac{1}{2}\times $sum of parallel sides $\times$ height.But in this question,neither height is given nor longest side is given.How should i formulate the equation of the area? 


The diagram shows the sides $k$ and included angles $\theta$ and $\pi-2\theta $ of the trapezium. Then its area $A$ can be expressed in terms of $k$ and $\theta$ now.
$$
\begin{aligned}
A &=k^{2} \sin \theta+\frac{k^{2}}{2} \sin (\pi-2 \theta) \\
&=\frac{k^{2}}{2}(2 \sin \theta+\sin 2 \theta)
\end{aligned}
$$
$$
\begin{aligned}
&\quad \frac{d A}{d \theta}= k^{2}(\cos \theta+\cos 2 \theta)=0 \\\Leftrightarrow & \quad \cos \theta+\cos 2 \theta=0 \\
\Leftrightarrow & \quad  2 \cos ^{2} \theta+\cos \theta-1=0 \\
\Leftrightarrow &\quad (2 \cos \theta-1) (\cos \theta+1)=0\\ 
\Leftrightarrow &\quad \cos \theta=\frac{1}{2} \text { or }-1 \textrm{ (rejected) }
\end{aligned}
$$
Hence
$$
\begin{aligned}
A_{\max } &=\frac{k^{2}}{2}\left(2 \cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right) =\frac{3 \sqrt{3}}{4} k^{2} \quad  cm^2
\end{aligned}
$$
