<s>[INST] Calculate the area of the triangular region ABC How to calculate the area of the triangle from the figure

 [/INST] Let $R>3$ be the circumradius of $ABC$ and $a,b,c$ its side lengths. We have
$$ a=2\sqrt{R^2-(R-2)^2},\qquad b=2\sqrt{R^2-(R-3)^2},\qquad c=2\sqrt{R^2-(R-1)^2} $$
hence
$$ a^2=16R-16,\qquad b^2=24R-36,\qquad c^2 = 8R-4 $$
and we also have
$$ R^2 = \frac{a^2 b^2 c^2}{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)} $$
so $R$ is the root of a cubic polynomial, and 
$$ R = \color{red}{2+2\cos\frac{\pi}{9}}=4\cos^2\frac{\pi}{18}\approx 3.87938524.$$
It follows that the area of the given triangle equals
$$ \Delta=4\sqrt{6+10 \cos\left(\frac{\pi }{9}\right)+4 \cos\left(\frac{2 \pi }{9}\right)} \approx \color{red}{17.1865547} $$
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