Calculate surface of a triangle It's quite a simple question solvable with many methods but the $2$ of them I picked don't agree.
First of I have chosen the triangle with corners: $(3,0,0) \quad (0,3,0) \quad (0,0,6)$
Thus, this one here:

First Method is by evaluating the surface Integral: $A = \int_{S}1\,\mathrm{dS} = \int_{(u,v)}1\,|n|\,\mathrm{du\,dv}$
As parametrization I came up with: $\psi(u,v) = \left(\begin{array}{cc}3\,u\\ 3\,v \\ 6-6\,u-6\,v\end{array}\right) \quad u \in[0,1] \quad v \in [0,1] \quad u+v<1$
Hence $n = \partial_u\psi \times \partial_v\psi = \left(\begin{array}{cc}18\\ 18 \\ 9\end{array}\right)\quad$ $\Rightarrow \quad  \displaystyle{A = \int_0^1 \int_0^{1-u}} \sqrt{2\cdot18^2+9^2}\,\mathrm{dv\,du} = 13,5$
Next Method is by using straight geometry: $A=  \frac{1}{2}\,g\,h$ (base times hight)
It should be: $A = \frac{1}{2}\,\sqrt{3^2+3^2}\,h = \frac{1}{2}\,\sqrt{18}\,\sqrt{6^2+\left(\sqrt{3^2-\frac{1}{2}\sqrt{18}}\right)^2} \approx 13,9$
Now again, where does this difference come from? Probably one could check by calculation $\textbf{det}$

You must have:
$$A = \frac{1}{2}\,\sqrt{3^2+3^2}\,h = \frac{1}{2}\,\sqrt{18}\,\sqrt{6^2+\left(\sqrt{3^2-\left(\frac{1}{2}\sqrt{18}\right)^\color{red}{2}}\right)^2} =13.5.$$
Or you can note that the triangles at the base are isosceles:

So $AC=BC=\frac{\sqrt{18}}{2}$ is what you need:
$$A = \frac{1}{2}\,\sqrt{3^2+3^2}\,h = \frac{1}{2}\,\sqrt{18}\,\sqrt{6^2+\left(\frac{\sqrt{18}}{2}\right)^2} =13.5.$$
Alternatively, you can use Heron's formula:
$$a=3\sqrt{2},b=c=3\sqrt5,\\
S=\sqrt{p(p-a)(p-b)(p-c)}=\\
\sqrt{\frac{3\sqrt{2}+6\sqrt5}{2}\cdot \frac{6\sqrt{5}-3\sqrt2}{2}\cdot \frac{3\sqrt{2}}{2}\cdot \frac{3\sqrt{2}}{2}}=\\
\sqrt{\frac{162\cdot 18}{16}}=\frac{27}{2}=13.5.$$
