Radius of four circles
$r = 30 ~ \text{cm}$
From triangle ECO
$\cos \angle EOC = \dfrac{CE}{OC}$
$\cos 30^\circ = \dfrac{30}{OC}$
$OC = 20\sqrt{3} ~ \text{cm}$
Cosine law for triangle OCD
$OD^2 = CD^2 + OC^2 - 2(CD)(OC) \cos (\frac{1}{2}\theta)$
$30^2 = 30^2 + (20\sqrt{3})^2 - 2(30)(20\sqrt{3}) \cos (\frac{1}{2}\theta)$
$1200 - 1200\sqrt{3} \cos (\frac{1}{2}\theta) = 0$
$\cos (\frac{1}{2}\theta) = 1 / \sqrt{3}$
$\theta = 109.47^\circ$
Area of the circular segment of central angle θ
$A_{segment} = \frac{1}{2}r^2 (\theta_{rad} - \sin \theta_{deg})$
$A_{segment} = \frac{1}{2}(30^2) \left[ 109.47^\circ \left( \dfrac{\pi}{180^\circ} \right) - \sin 109.47^\circ \right]$
$A_{segment} = 435.52 ~ \text{cm}^2$
Required area
$A = A_{circle} - 6A_{segment}$
$A = \pi (30^2) - 6(435.52)$
$A = 214.32 ~ \text{cm}^2$ answer
