Find the area of ​the triangular region ABC below

Question

For reference: Calculate the area of ​​the triangular region $ABC$; if $R = 36$m and r = $9$m (T ➔ tangency point).(Answer:$324m^2$)

My progress: By property: $AC =2\sqrt{R.r}=2\sqrt{36.9}=36$

By property: $\frac{1}{\sqrt x}=\frac{1}{\sqrt R}+ \frac{1}{\sqrt r}\implies\frac{1}{\sqrt x} =\frac{1}{\sqrt{36}}+\frac{1}{\sqrt{9}}\\ \therefore x = 4\\ SACOO': = \frac{(36+9).36}{2}=810\\ S_{ABC} = \frac{36.BT}{2} = 18BT$

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Answer 1

Realize that $AT$ and $TC$ are also tangent segments, you can compute their lengths same way you found $AC$.

Next, extend $OO'$ to meet $AC$ in $D$. Now triangles $BTD$ and $OAD$ can be shown to be similar with ratio $1/2$. Hint : Join $O'C$ and find some $3:4:5$ triangles.

This will yield $BT=OA/2=18$.