From triangle ABC

$\cos \alpha = \dfrac{10}{20}$

$\alpha = 60^\circ$

At corner A

$\beta = 90^\circ - \alpha = 90^\circ - 60^\circ$

$\beta = 30^\circ$

Area of rectangle ABDE

$A_{ABDE} = 20(10)$

$A_{ABDE} = 200 \, \text{ cm}^2$

Area of triangle ABC

$A_{ABC} = \frac{1}{2}(10)(20) \sin \alpha = 100 \sin 60^\circ$

$A_{ABC} = 50\sqrt{3} \, \text{ cm}^2$

Area of sector ACE

$A_{ACE} = \dfrac{\pi r^2 \beta_{deg}}{360^\circ} = \dfrac{\pi (20^2)(30^\circ)}{360^\circ}$

$A_{ACE} = \frac{100}{3}\pi \, \text{ cm}^2$

Area of CED

$A_{CED} = A_{ABDE} - A_{ABC} - A_{ACE}$

$A_{CED} = 200 - 50\sqrt{3} - \frac{100}{3}\pi$

$A_{CED} = 8.68 \, \text{ cm}^2$

A similar problem that involves the same area as A_{CED} was solved using integration. Look for the solution of area ABC in the following link: Solution by integration.

Required Area

$A_{required} = A_{square} - 8A_{CED}$

$A_{required} = 20^2 - 8(8.68)$

$A_{required} = 330.56 \, \text{ cm}^2$ *answer*