以前回答过：
假设 E 离 A、B 较近,F 离 B、C 较近,G 离 C、D 较近,H 离 A、D 较近.
则：
连接 AF、AG,则有 ∠FAD = ∠GAB = 60°.
∴ ∠FAG = ∠FAD + ∠GAB - ∠DAB
= 60° + 60°+ 90°
= 30°
{①
∴弧FG = 2aπ * 30°/ 360° = aπ/6
同理 弧EF = 弧FG = 弧GH = 弧HE = aπ/6
∴相交部分周长为 4aπ/6 = 2aπ/3
}
{②
∴扇形AFG面积 S1 = πaa * 30°/ 360°= πaa/12
设正方形中心点为 O ,连接 OA、OF,延长 OF,交 AD 于 P
则有：FP ⊥ AD,AP = a/2
又∵ AF = a
∴根据勾股定理,FP = (a/2)*√3
∴OF = PF - OP = (a/2)*√3 - a/2 = a/2 * (- 1 + √3)
∴S△AOF = OF * AP / 2 = aa/8 * (- 1 + √3)
同理 S△AOG = S△AOF
设 OF、OG、弧FG 围成图形的面积为 S2,则：
S2 = S1 - S△AOF - S△AOG
= S1 - 2 * S△AOF
= πaa/12 - 2 * aa/8 * (- 1 + √3)
= aa/4 * (π/3 + 1 - √3)
由题意知,四条弧相交部分面积 = 4 * S2 = aa * (π/3 + 1 - √3)
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