The lengths of the sides of a triangle are $10+x^2$, $10+x^2$ and $20-2x^2$. If for $x=k$, the area of the triangle is maximum, then $3{\mathrm{k}}^2$ is equal to :

 $\begin{array}{l}\mathrm{a}=20-2{\mathrm{x}}^2, \mathrm{b}=10+{\mathrm{x}}^2,\mathrm{c}=10+{\mathrm{x}}^2\\ \\ =\frac{a+b+c}{2}\\ =20\\ \\ \Delta =\sqrt{s(s-a)(s-b)(s-c)}\\ \\ =\sqrt{20\left(2x^2\right)\left(10-x^2\right)\left(10-x^2\right)}\\ \\ =2\sqrt{10}\sqrt{x^2{\left(10-x^2\right)}^2}\\ \\ =2\sqrt{10}\left|x\left(10-x^2\right)\right|\\ \\ \begin{array}{l}=2\sqrt{10}\left|10x-x^3\right|\\ \\ S=10x-x^3\\ \\ \frac{ds}{dx}=10-3x^2\\ \\ \frac{ds}{dx}=0\Rightarrow x^2=\frac{10}{3}\\ \\ 3{\mathrm{x}}^2=10\end{array}\end{array}$