<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] 
Since we have an isosceles trapezium (non-parallel sides of equal length), we can "clip off" a triangular section and arrange the portions into a rectangle of dimensions $ \ k + x \ $ by $ \ h \ \ , $ with $ \ k \ $ being the hypotenuse of a right triangle with legs $ \ x \ $ and $ \ h \ = \ \sqrt{k^2 - x^2} \ \ . $  The area of the trapezium is then given by
$$ A \ \ = \ \ (k + x)·\sqrt{k^2 - x^2} \ \ = \ \ \sqrt{(k + x)^2·(k^2 - x^2)} \ \ = \ \ \sqrt{(k + x)^3·(k  - x )} \ \ . $$
Without the use of calculus, we can take $ \ x \ = \ r·k \ $ to express the area as $ \  \sqrt{(k + rk)^3·(k  - rk )}   $ $ = \ k^2·\sqrt{(1 + r )^3·(1  - r )} \ = \ k^2·\sqrt{2·(1 + r )^3 \ - \ (1  + r )^4} \ \ . $  Making the further substitution $ \ s \ = \ 1 + r \ \ , $ the area is then $ \ k^2·\sqrt{2 s^3  -  s^4} \ = \ k^2·\sqrt{ s^3 · (2  -  s)} \ \ . $  Applying the AM-GM inequality, we obtain
$$ \sqrt[4]{\left(\frac{s}{3} \right)^3·(2 - s )}
 \ \ \le \ \   \frac{\frac{s}{3} \ + \  \frac{s}{3} \ + \ \frac{s}{3} \ + \ (2 \ - \ s)}{4} \ \ = \ \ \frac{1}{2}  $$ $$ \Rightarrow \ \ A \ = \ k^2 \ · \ \sqrt{3^3} \ · \ \left[ \ \sqrt[4]{\left(\frac{s}{3} \right)^3·(2 - s )} \ \right]^2  \ \ \le \ \  k^2 \ · \ 3\sqrt{3} \ · \left( \frac{1}{2} \right)^2 \ \ ,  $$
with equality occurring for $ \ s \ = \ \frac32 \ \Rightarrow \ r \ = \ \frac12 \ \Rightarrow \ x \ = \ \frac{1}{2}·k \ \ . $
With the use of differentiation, we have
$$ \frac{dA}{dx} \ \ = \ \ \frac{3}{2}·(k + x)^{1/2}·(k - x)^{1/2} \  - \ \frac{1}{2}·(k + x)^{3/2}·(k - x)^{-1/2} \ \ = \ \ 0 $$
$$ \Rightarrow \ \ 3·(k - x) \ \ = \ \ k + x \ \ \Rightarrow \ \ x \ = \ \frac{k}{2} \ \ . $$
[We then have $ \frac{d^2A}{dx^2} \   = \   \frac{2x^2-2kx-k^2}{(k - x)^{3/2}·(k + x)^{ 1/2}} \ \ < \ \ 0 \ \   $ for $ \ x \ = \ \frac{k}{2} \ \ . \ ] $
So the maximum area of the trapezium is $ \ \sqrt{\left(\frac{3k}{2} \right)^3·\left(\frac{ k}{2} \right) } \ = \ \frac{3\sqrt3}{4}·k^2 \ \ . $
(The factor of $ \ \frac{ \sqrt3}{4}·k^2 \ $ is perhaps a bit of a tip-off that a $ \ 60º \ $ angle is involved (as seen in Lai's solution), as this is the area of an equilateral triangle of side $ \ k \ \ .   $  This actually suggests another approach:  as the equilateral triangle maximizes the area of an isosceles triangle, Lai's parallelogram can be subdivided into three equilateral triangles with the indicated maximal total area.
Yet another way -- which would need to be made rigorous -- is to surmount the isosceles trapezium with an isosceles triangle.  The area is maximized when the large "total" triangle is equilateral, which occurs for $ \ y \ = \ z \ = \ k \ \ $ in the diagram below.  This maximal area is $ \ \frac{\sqrt{3}}{4}·(2k)^2 \ \ , $ of which the maximal-area isosceles trapezium has $ \ \frac34 \ $ of that area.)

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