Area of quarter circle AECDB

$A_{AECDB} = \frac{1}{4}\pi (20^2) = 100\pi \, \text{ cm}^2$

$A_{AECDB} = 314.16 \, \text{ cm}^2$

Area of sector AFCGB

$A_{AFCGB} = \frac{1}{6}\pi (20^2) = \frac{200}{3}\pi \, \text{ cm}^2$

$A_{AFCGB} = 209.44 \, \text{ cm}^2$

Area of segment AFCE

$A_{AFCE} = \frac{1}{6}\pi (20^2) - \frac{1}{2}(20^2)\sin 60^\circ$

$A_{AFCE} = \frac{200}{3}\pi - 100\sqrt{3} \, \text{ cm}^2$

$A_{AFCE} = 36.23 \, \text{ cm}^2$

Area of BGCD

$A_{BGCD} = A_{AECDB} - A_{AFCGB} - A_{AFCE}$

$A_{BGCD} = 314.16 - 209.44 - 36.23$

$A_{BGCD} = 68.49 \, \text{ cm}^2$

Required Area

$A_{required} = A_{square} - 4A_{BGCD} = 20^2 - 4(68.49)$

$A_{required} = 126.04 \, \text{ cm}^2$ *answer*

## Comments

## you can solve this without

you can solve this without any use of trigonometry, just triangles (pythagorean theorem) and the area of a circle.

## You may take a picture of

You may take a picture of your solution and upload it here. Tnx.