Fix any point $R$. Let $\phi_1$, $\phi_2$ and $\phi_3$ be angles, relative to $R$, then $0 < \phi_1 < \phi_2 < \phi_3 < 2 \pi$, i.e the triple $\{\phi_1, \phi_2, \phi_3\}$ is the first, second and the third order statistics of the uniformly distributed angle on the interval $(0, 2\pi)$.
The area of the triangle, is then
$$
A = 4 \sin \left( \frac{\phi_2-\phi_1}{2} \right) \sin \left( \frac{\phi_3-\phi_2}{2} \right) \sin \left( \frac{\phi_3-\phi_1}{2} \right)
$$
and the perimeter
$$
p = 2 \left( \sin \left( \frac{\phi_2-\phi_1}{2} \right) + \sin \left( \frac{\phi_3-\phi_2}{2} \right) + \sin \left( \frac{\phi_3-\phi_1}{2} \right) \right)
$$
Given that $\frac{\phi_3 -\phi_1}{2} = \frac{\phi_3 -\phi_2}{2} + \frac{\phi_2 -\phi_1}{2}$, let $\alpha = \frac{\phi_2-\phi_1}{2}$ and $\beta = \frac{\phi_3-\phi_2}{2}$, then
$$
A(\alpha, \beta) = 4 \sin \alpha \sin \beta \sin (\alpha+\beta) \qquad
p(\alpha, \beta) = 2 \left( \sin \alpha + \sin \beta + \sin (\alpha+\beta) \right)
$$
The distribution of $\alpha$ and $\beta$ are not hard to find. Indeed, distribution of $\phi_1$, $\phi_2$ and $\phi_3$ ($[\chi]$ stands for Iverson bracket):
$$
\mathrm{d} F(\phi_1, \phi_2, \phi_3) = \frac{3!}{(2 \pi)^3} \Big[ 0 < \phi_1 < \phi_2 < \phi_3 < 2 \pi \Big]
\mathrm{d} \phi_1 \mathrm{d} \phi_2 \mathrm{d} \phi_3
$$
Changing variables to $\alpha = \frac{\phi_2-\phi_1}{2}$, $\beta=\frac{\phi_3-\phi_2}{2}$ and $\gamma = \frac{1}{3} \left( \phi_1 + \phi_2 + \phi_3 \right)$ we get, noting that the Jacobian equals $4$,
$$
\mathrm{d} F(\alpha, \beta, \gamma) = \frac{3!}{(2 \pi)^3} \Big[ 0 < \alpha < \pi, 0 < \beta < \pi-\alpha, \frac{4 \alpha + 2 \beta}{3} <\gamma< 2\pi - \frac{2 \alpha + 4 \beta}{3} \Big] \cdot 4 \cdot \mathrm{d} \alpha \, \mathrm{d} \beta \, \mathrm{d} \gamma
$$
Integrating over $\gamma$ we find the joint pdf for $(\alpha, \beta)$:
$$
\mathrm{d} F\left(\alpha,\beta\right) = \frac{3!}{\pi^3} \left( \pi - \alpha - \beta \right) \Big[ 0 < \alpha < \pi, 0< \beta < \pi - \alpha \Big] \, \, \mathrm{d} \alpha \, \mathrm{d} \beta
$$
This means that $\left\{ \frac{\alpha}{\pi}, \frac{\beta}{\pi} \right\}$ follows Dirichlet distribution with parameters $\{1,1,2\}$.
Using the interpretation in terms of the Dirichlet distribution, it is not hard to determine the mean and the variance of the area and the perimeter:
This answer to a slightly different problem
gives a useful diagram showing how to compute the distance between the points
of tangency of two circles and a line, given that the circles are externally tangent
(as yours are).
From this we see that if we label the three points of tangency $A,$ $B,$ and $C$
(in sequence from the leftmost such point to the rightmost in your diagram), then
considering just the two circles of radius $r_1$ and $r_3$, which touch
the line at $A$ and $B$,
$$ |AB| = 2\sqrt{r_1 r_3}. $$
For the other two pairs of circles we get
$ |BC| = 2\sqrt{r_2 r_3} $ and
$ |AC| = 2\sqrt{r_1 r_2}.$
We can also see that $ |AC| = |AB| + |BC|;$ substituting the formulas we just found
for those three lengths (or perhaps even better still, labeling the three distances
$ 2\sqrt{r_1 r_3},$ $ 2\sqrt{r_2 r_3},$ and $2\sqrt{r_1 r_2}$ on your diagram),
$$ 2\sqrt{r_1 r_3} + 2\sqrt{r_2 r_3} = 2\sqrt{r_1 r_2}.$$
I can't think of a visual representation of the last step, but algebraically,
you can divide all three terms by $2\sqrt{r_1 r_2 r_3}$ to get
$$ \frac1{\sqrt{r_2}} + \frac1{\sqrt{r_1}} = \frac1{\sqrt{r_3}},$$
which is your simplified formula. You can then get the other formula by further
algebraic manipulation.
Best Answer
Not a problem:), and this is how you can arrive at the result!
We know that $\Delta = \frac{abc}{4R}$
Now I will assume that $c$ is the hypotenuse. Draw a circum-circle around your triangle you can easily observe by Thales theorem that $c$ is the diameter of the circle . Hence $c=2R$, Therefore $\Delta = \frac{ab}{2}$ or Area is triangle = $\frac{1}{2}$ * base * height.