Solution:

Connect AB, BC, CD, DA, AC and BD as shown in the diagram. Then AC = BD = 20cm and the area of ABCD is $\frac{1}{2}\times AC\times BD=200$ cm². The area of each of the intersecting circles is $\pi \times {{10}^{2}}=314c{{m}^{2}}$ . Let S cm² be the area of each of the 7 identical parts of the shaded region. By symmetry, 2S cm² is the difference of the area of the circle and the square ABCD. Hence 2S = 314 – 200 = 114 so that S = 57. It follows that the desired area is 7S = 399 cm².