Eduardo from the British School, Manila,
Tomas from Malmesbury School and Philip used the tangent formula
and Alex and also 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. Well done all of
you.

FIRST
METHOD
Using the tangents of the angles at the centre of the circle
and the formula
$\tan X= \tan(A + B) =\frac{\tan A + \tan B} {\tan A \tan B 
1} $ where $A + B + X = 180^o$,
$\frac{x}{r}= \frac{\frac{a}{r} +
\frac{b}{r}}{\frac{ab}{r^2}1} = \frac{r(a + b)}{ab  r^2}.$
The area of the triangle is
$(a + b + x)r = (a + b)r +
\frac{r^3(a + b)}{ab  r^2}= \frac{(a + b)abr}{ab  r^2}$ as
required.

SECOND METHOD
This method uses Heron's formula. The lines from the centre of the
circle to the edges each meet the tangents to the circle forming
right angles as shown. The triangle has been rotated/reflected so
that the side with length (a+b) is the base. (However it is rotated
it is the same shape of unknown angles and lengths). As the
tangents to a circle from an external point are equal in length
(easily proved using congruent triangles) the other lengths of the
sides of the triangle can be found. 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 semiperimeter of the triangle is given by $s = x + a + b$ and
from Heron's formula the area of the triangle is
$A = \sqrt{(s(s  a')(s  b')(s 
c'))}$
$= \sqrt{((a + b + x)abx)} $
Also, the triangle is divided into three smaller triangles and the
total area is given by
$ A = \frac{1}{2}(a + b)r +
\frac{1}{2}(b + x)r + \frac{1}{2}(x + a)r $
$=(a + b + x)r $.
Equating the two answers
$ \sqrt{abx(a + b + x)} = (a + b +
x)r $
$ abx(a + b + x) = (a + b + x)^2
r^2 $
$ abx = (a + b + x)r^2 $
$x = \frac{(a + b)r^2}{ (ab 
r^2).} $
Hence
$ A = (a + b + x)r$
$ = \frac{r((a + b)(ab  r^2) + (a
+ b)r^2)}{ ab  r^2}$
$ = \frac{abr(a + b)}{ab  r^2.}
$
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.