From the figure below
$\angle DAB = 30^\circ$, $\angle EBA = 30^\circ$, $\angle ABD = 60^\circ$
$r^2 + 10^2 = 20^2$
$r^2 = 300$
$r = 17.32 \, \text{ cm}$
$BF = AD = r = 17.32 \, \text{ cm}$
$EF^2 + 10^2 = r^2$
$EF^2 + 100 = 300$
$EF = 14.14 \, \text{ cm}$
$BE = BF - EF = 17.32 - 14.14$
$BE = 3.18 \, \text{ cm}$
Sine law for triangle ABE
$\dfrac{r}{\sin \angle EBA} = \dfrac{BE}{\sin \angle EAB}$
$\dfrac{17.32}{\sin 30^\circ} = \dfrac{3.18}{\sin \angle EAB}$
$\sin \angle EAB = \dfrac{3.18 \sin 30^\circ}{17.32}$
$\angle EAB = 5.27^\circ$
$\angle DAE = \angle DAB - \angle EAB = 30^\circ - 5.27^\circ$
$\angle DAE = 24.73^\circ$
$A_{BED} = A_{ABD} - A_{ABE} - A_{AED}$
$A_{BED} = \frac{1}{2}(10)(20)\sin 60^\circ - \frac{1}{2}(20)(3.18) \sin 30^\circ - \dfrac{\pi(17.32^2)(24.73^\circ)}{360^\circ}$
$A_{BED} = 86.60 - 15.9 - 64.74$
$A_{BED} = 5.96 \, \text{ cm}^2$
Required Area
$A_{required} = A_{ABC} - 6A_{BED}$
$A_{required} = \frac{1}{2}(20^2) \sin 60^\circ - 6(5.96)$
$A_{required} = 137.44 \, \text{ cm}^2$ answer
