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?
$p=\frac{\sqrt{18}+\sqrt{20}+\sqrt{26}}{2}\implies S_{ABC} =\sqrt{p(p-\sqrt{20})(p-\sqrt{26})(p-\sqrt{18})}$


If $a = \sqrt{18}, b = \sqrt{20}, c = \sqrt{26}$, first note that $a^2 + b^2 \gt c^2$ where $c$ is the longest side. That means it is an acute-angled triangle and the orthocenter is inside the triangle. So if we drop a perp from vertex $A$, then using Pythagoras,
$AD^2 = 20 - x^2 = 26 - (\sqrt{18} - x)^2$
$ \implies x = \sqrt 2$ and we get $AD = \sqrt{18}$
So, $A = \frac 12 a \cdot AD = 9$
By the way, if you notice that $26 - 8 = 20 - 2$ and $\sqrt{8} + \sqrt{2} = a$, you can get to altitude $AD = \sqrt{18} ~$ a bit faster.
