<s>[INST] 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? 
 [/INST] Notice, let the unknown side be $x$ which is parallel to one of three equal sides each of length $k$ then the normal distance between the parallel sides can be determined as follows (using pytagorean theorem in a right triangle)
$$\sqrt{k^2-\left(\frac{x-k}{2}\right)^2}$$
the area of the trapezium is given as 
$$A=\frac{1}{2}(x+k)\sqrt{k^2-\left(\frac{x-k}{2}\right)^2}\tag 1$$
differentiating (1) w.r.t. $x$, we get
$$\frac{dA}{dx}=\frac{1}{2}(x+k)\frac{-(x-k)}{4\sqrt{k^2-\left(\frac{x-k}{2}\right)^2}}+\frac{1}{2}\sqrt{k^2-\left(\frac{x-k}{2}\right)^2}$$
$$=\frac{1}{8}\frac{x^2-k^2+4k^2-4\left(\frac{x-k}{2}\right)^2}{\sqrt{k^2-\left(\frac{x-k}{2}\right)^2}}=\frac{1}{4}\frac{(-x^2+kx+2k^2)}{\sqrt{k^2-\left(\frac{x-k}{2}\right)^2}}$$
Again differentiating w.r.t. $x$, we get 
$$\frac{d^2A}{dx^2}=\frac{x^3-3kx^2-7k^2x+5k^3}{16\left(k^2-\left(\frac{x-k}{2}\right)^2\right)^{3/2}}$$
Now, for maximum or minimum, setting $\frac{dA}{dx}=0$, we get 
$$\frac{1}{4}\frac{(-x^2+kx+2k^2)}{\sqrt{k^2-\left(\frac{x-k}{2}\right)^2}}=0$$
$$x^2-kx-2k^2=0$$
$$x=\frac{-(-k)\pm\sqrt{(-k)^2-4(1)(-2k^2)}}{2(1)}=\frac{k\pm 3k}{2}$$
But, side $k>0$ hence, we accept $$x=\frac{k+3k}{2}=2k$$
It can be checked that $\frac{d^2A}{dx^2}=-\frac{13}{6\sqrt 3}<0$ at $x=2k$. Hence, the area is maximum at $x=2k$ 
Hence, substituting $x=2k$, the greatest possible area is $$A_{\text{max}}=\frac{1}{2}(2k+k)\sqrt{k^2-\left(\frac{2k-k}{2}\right)^2}$$
$$=\frac{3k}{2}\sqrt{k^2-\frac{k^2}{4}}=\frac{3k}{2}\sqrt{\frac{3k^2}{4}}$$
$$\color{red}{A_{\text{max}}=\frac{3\sqrt 3\ k^2}{4}\ \text{sq. cm.}}$$
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