(1) Let two circles intersect at point A and point B, connecting O 1A, O2A, O 1B, O2B. ..

Then s sound =S drill +4S bow.

∫S rhombus =2S△AO 1O2, △O 1O2A is a regular triangle with side length r 。

∴S△AO 1O2=3r24，S bow =60πr2360-3r24=πr26？ 3r24。

∴S Yin =2×3r24+4(πr26? 3r24)=23πr2-32r2。

(2) The shaded areas in Figure 2 are:

S Yin =S△O 1O2O3+3S bow

∫△o 1o 2 o 3 is a regular triangle with side length r,

∴S△O 1O2O3=3r24.

∴S bow = 60 π R2360-3R24.

s Yin = 3r 24+3(60πr 2360-3r 24)=πR22？ 3r22。

(3) extend the intersection of O2O 1 and ⊙O 1 at point A and ⊙O 1 and ⊙O4 at point B. 。

According to (1), so1bo4 =12 (23 π R2-3R22).

∫so 1AB = S sector AO 1O