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Since AE = BE = AB, \(\Delta ABE\) is an equilateral triangle
\(S_{ABE}=\dfrac{6^2\sqrt{3}}{4}=9\sqrt{3}\) (cm2)
The area of the sector ABE is : \(\dfrac{6^2\pi.60}{360}=6\pi\) (cm2)
The area of the shaded region is :
\(\dfrac{6^2\pi.90}{360}.2-\left[9\sqrt{3}+2\left(6\pi-9\sqrt{3}\right)\right]=6\pi+9\sqrt{3}\) (cm2)
The answer is : 6 + 9 + 3 = 18