The area of an isoceles right triangle is...

Hello LUANDATO.

Let's take a look at you question.

We have an isoceles right triangle with area: $$A=\ 18x^2+6x+\frac{1}{2}=\frac{1}{2}\cdot\left(36x^2+12x+1\right)=\frac{1}{2}\cdot\left(6x+1\right)\cdot\left(6x+1\right).$$ It implies that the length of two cathetus are: 6x+1.

So, the lenght of the third side (hypotenuse) is: $$hip=\sqrt{\left(6x+1\right)^2+\left(6x+1\right)^2}=\sqrt{2\left(6x+1\right)^2}=\left(6x+1\right)\sqrt{2}=\left(6x+1\right)\cdot2^{\frac{1}{2}}.$$

Finally, the perimeter of the triangle is $$\left(6x+1\right)+\left(6x+1\right)+\left(6x+1\right)\cdot2^{\frac{1}{2}}=2\left(6x+1\right)+2^{\frac{1}{2}}\left(6x+1\right)=\left(2+2^{\frac{1}{2}}\right)\left(6x+1\right).$$

So, the correct answer is C.

I hope this explanation may help you.

I'm available if you'd like a follow up.

Regards.