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? 
 A: You can parametrize the trapezium via its height $h$. Then the longer side is
$$ 2 \sqrt{k^2-h^2} + k $$ and its area is
$$ h (\sqrt{k^2-h^2} + k)\ .$$
The derivative for $h$ is
$$k - \frac{h^2}{\sqrt{-h^2 + k^2}} + \sqrt{-h^2 + k^2}\ ,$$
which is zero for $h=0$ or $h=\pm \frac{\sqrt{3}}{2}k$. $h=\frac{\sqrt{3}}{2}k$ gives the proposed optimum, which you can be sure about after also checking the border case $h=k$ (which gives an area of $k^2$).
A: 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 $$A_{\text{max}}=\frac{1}{2}(2k+k)\sqrt{k^2-\left(\frac{2k-k}{2}\right)^2}$$
$$=\frac{3k}{2}\sqrt{k^2-\frac{k^2}{4}}=\frac{3k}{2}\sqrt{\frac{3k^2}{4}}$$
$$\color{red}{A_{\text{max}}=\frac{3\sqrt 3\ k^2}{4}\ \text{sq. cm.}}$$
A: 
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
$$
\begin{aligned}
A_{\max } &=\frac{k^{2}}{2}\left(2 \cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right) =\frac{3 \sqrt{3}}{4} k^{2} \quad  cm^2
\end{aligned}
$$
A: 
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.)

