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 ' = a + b, b ' = b + x and a ' = 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

A = √   (s(s − a ' )(s − b ' )(s − c ' ))   = √   ((a + b + x)abx)

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

A =  1 2 (a + b)r +  1 2 (b + x)r +  1 2 (x + a)r = (a + b + x)r .

Equating the two answers

√   abx(a + b + x)   = (a + b + x)r abx(a + b + x) = (a + b + x)2 r2 abx = (a + b + x)r2 x =  (a + b)r2 (ab − r2).

Hence

A = (a + b + x)r =  r((a + b)(ab − r2) + (a + b)r2) ab − r2 =  abr(a + b) ab − r2.

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.