Calculate the area of the triangular region ABC How to calculate the area of the triangle from the figure

 A: 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
$$ \Delta=4\sqrt{6+10 \cos\left(\frac{\pi }{9}\right)+4 \cos\left(\frac{2 \pi }{9}\right)} \approx \color{red}{17.1865547} $$
A: Let $O$ be the centre of the circle and $r$ its radius.
Then $AB=2\sqrt{r^2-(r-1)^2}=2\sqrt{2r-1}$,  $BC=2\sqrt{r^2-(r-2)^2}=2\sqrt{4r-4}$, $AC=2\sqrt{r^2-(r-3)^2}=2\sqrt{6r-9}$.
Now depending on what edge we take as base, the area of $ABC$ can be expressed in three forms:
$$\sqrt{2r-1}\cdot (2r-1)=\sqrt{4r-4}\cdot (2r-2)=\sqrt{6r-9}\cdot (2r-3).$$
By squaring, we get
$$8r^3 -12r^2+6r-1 = 16r^3-48r^2+48r-16=24r^3-108r^2+162r-81$$
and from this by subtraction
$$ 8r^3-36r^2+42r-15=0,\quad 8r^3-60r^2+114r-65=0$$
and by another subtraction and division by $2$
$$ 12r^2-36r+25=0.$$
From this, 
$$ 0=3\cdot (8r^3-36r^2+42r-15)-(2r+3)\cdot (12r^2-36r+25)=76r-78$$
so that $r=\frac{39}{38}$ -- but unfortunately, this does not fulfil the original equation (so either I made a mistake or there is no solution; but in the former case I assume you can find the correct solution from this)
