Area enclosed by pairs of overlapping quarter circles

Area CEB is equal to area of the square OAEF minus area of the quarter circle OABCF minus four times the area of CED.
 

 

$A_{CED} = 8.68 \, \text{ cm}^2$       → See how it was found
 

Area of CEB
$A_{CEB} = A_{square} - A_{OABCF} - 4A_{CED}$

$A_{CEB} = 20^2 - \frac{1}{4} \pi (20^2) - 4(8.68)$

$A_{CEB} = 51.12 \, \text{ cm}^2$
 

Required Area
$A_{required} = 4A_{CEB} = 4(51.12)$

$A_{required} = 204.48 \, \text{ cm}^2$           answer