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 $$3k^2$$ is equal to

The sides of the triangle are

$$10+x^2,\qquad 10+x^2,\qquad 20-2x^2$$

Hence, the triangle is isosceles.

The semi-perimeter is

$$s=\frac{(10+x^2)+(10+x^2)+(20-2x^2)}{2}=20$$

Using Heron's formula,

$$\Delta^2=s(s-a)(s-b)(s-c)$$

$$=20(10-x^2)(10-x^2)(2x^2)$$

$$=40x^2(10-x^2)^2$$

Therefore,

$$\Delta=2\sqrt{10}\,x(10-x^2)$$

To maximize area, maximize

$$f(x)=10x-x^3$$

Now,

$$f'(x)=10-3x^2$$

For maximum area,

$$10-3x^2=0$$

$$3x^2=10$$

Since maximum occurs at

$$x=k$$

we get

$$3k^2=10$$

Hence,

$$\boxed{10}$$