#### Answer

$A = \left[ 5~\frac{sin~A~sin~B}{sin~(A+B)}\right]~R^2$

#### Work Step by Step

The angle $C = 180^{\circ}-(A+B)$
$sin~C = sin~(180^{\circ}-(A+B))$
$sin~C = sin~(180^{\circ})~cos~(A+B)-sin~(A+B)~cos~(180^{\circ})$
$sin~C = (0)~cos~(A+B)-sin~(A+B)~(-1)$
$sin~C = sin~(A+B)$
We can use the law of sines to write an expression for $r$:
$\frac{r}{sin~A} = \frac{R}{sin~C}$
$r = \frac{R~sin~A}{sin~C}$
$r = \frac{R~sin~A}{sin~(A+B)}$
Let $x$ be the length of the side AC. We can use the law of sines to write an expression for $x$:
$\frac{x}{sin~B} = \frac{R}{sin~C}$
$x = \frac{R~sin~B}{sin~C}$
$x = \frac{R~sin~B}{sin~(A+B)}$
We can find the area of the triangle ABC:
$Area = \frac{1}{2}r~x~sin~C$
$Area = \frac{1}{2}~(\frac{R~sin~A}{sin~(A+B)})~(\frac{R~sin~B}{sin~(A+B)})~sin~(A+B)$
$Area = \frac{1}{2}~\frac{R^2~sin~A~sin~B}{sin~(A+B)}$
$Area = \frac{1}{2}~\frac{sin~A~sin~B}{sin~(A+B)}~R^2$
We can find the total area $A$ of ten such triangles:
$A = 10 \times \frac{1}{2}~\frac{sin~A~sin~B}{sin~(A+B)}~R^2$
$A = \left[ 5~\frac{sin~A~sin~B}{sin~(A+B)}\right]~R^2$