Sue Liu of Madras College, St Andrew's sent a solution to this problem which depends on the use of Heron's Formula for the area of a triangle. Here is Sue's method.

The incircle divides the sides of the triangle into lengths $c\text{'}=a+b$, $b\text{'}=b+x$ and $a\text{'}=x+a$ as shown in the diagram. The semi-perimeter of the triangle is given by $s=x+a+b$ and from Heron's formula the area of the triangle is

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{}\\ \multicolumn{1}{c}{A}& =& \sqrt{(s(s-a\text{'})(s-b\text{'})(s-c\text{'}))}\\ \multicolumn{1}{c}{}& =& \sqrt{((a+b+x)\mathrm{abx})}\end{array}}$

Also, the triangle is divided into three smaller triangles and the total area is given by

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{}\\ \multicolumn{1}{c}{A}& =& \frac{1}{2}(a+b)r+\frac{1}{2}(b+x)r+\frac{1}{2}(x+a)r\\ \multicolumn{1}{c}{}& =& (a+b+x)r.\end{array}}$

Equating the two answers

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{}\\ \multicolumn{1}{c}{\sqrt{\mathrm{abx}(a+b+x)}}& =& (a+b+x)r\\ \multicolumn{1}{c}{\mathrm{abx}(a+b+x)}& =& (a+b+x){}^2{r}^2\\ \multicolumn{1}{c}{\mathrm{abx}}& =& (a+b+x)r^2\\ \multicolumn{1}{c}{x}& =& \frac{(a+b)r^2}{(\mathrm{ab}-r^2).}\end{array}}$

Hence

${\displaystyle \begin{array}{c}\multicolumn{1}{c}{}\\ \multicolumn{1}{c}{A}& =& (a+b+x)r\\ \multicolumn{1}{c}{}& =& \frac{r((a+b)(\mathrm{ab}-r^2)+(a+b)r^2)}{\mathrm{ab}-r^2}\\ \multicolumn{1}{c}{}& =& \frac{\mathrm{abr}(a+b)}{\mathrm{ab}-r^2.}\end{array}}$

An alternative method, not using Heron's formula, is based on finding x in terms of $a$ and $b$ using the tangents of the angles at the centre of the circle.