What´s the value of the area of the triangle below? For reference: The sides of an acute-angled triangle measure
$3\sqrt2$, $\sqrt{26}$ and $\sqrt{20}$.
Calculate the area of ​​the triangle (Answer:$9$)
My progress...
Is there any way other than Heron's formula since the accounts would be laborious or algebraic manipulation for the resolution?
$p=\frac{\sqrt{18}+\sqrt{20}+\sqrt{26}}{2}\implies S_{ABC} =\sqrt{p(p-\sqrt{20})(p-\sqrt{26})(p-\sqrt{18})}$

Use cosine law,
$c^2=a^2+b^2-2ab\cos{\gamma}$
$(\sqrt{26})^2=(3\sqrt2)^2+(\sqrt{20})^2-2\cdot3\sqrt2\sqrt{20}cos\gamma$
$\implies cos(\gamma) =\frac{1}{\sqrt{10}}\Rightarrow sin(\gamma)=\frac{3}{\sqrt{10}}$
$A=\frac12\cdot 3\sqrt2\cdot \sqrt{20}\cdot sin(\gamma)$
$\implies...A=9$
