Or search by topic
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.
|