From triangle BAD
$(R - 10)^2 + 30^2 = R^2$
$(R^2 - 20R + 100) + 900 = R^2$
$20R = 1000$
$R = 50 \, \text{ cm}$
$\sin (\frac{1}{2}\theta) = \dfrac{30}{R} = \dfrac{30}{50}$
$\theta = 73.74^\circ$
Area of segment CDE
$A_{CDE} = A_{BDCE} - A_{BDE}$
$A_{CDE} = \dfrac{\pi R^2 \theta_{deg}}{360^\circ} - \frac{1}{2}R^2 \sin \theta$
$A_{CDE} = \dfrac{\pi (50^2)(73.74^\circ)}{360^\circ} - \frac{1}{2}(50^2) \sin 73.74^\circ$
$A_{CDE} = 408.76 \, \text{ cm}^2$
Required area = Area of cirlce of radius R - Area of semi-circle of radius 30 cm - Area of segment CDE
$A = \pi R^2 - \frac{1}{2}\pi (30^2) - A_{CDE}$
$A = \pi (50^2) - \frac{1}{2}\pi (30^2) - 408.76$
$A = 6031.50 \, \text{ cm}^2$ answer
