如下圖所示，注意到圓弧部分的麵積可相互抵消(S1=S4, S2=S3)，因此此題比初看要好處理些。
S_blue - S_yellow
= ( S_AFC + S_GDEH + S1 + S3) - (S_FCDG + S_HEBK + S2 + S4)
= S_AFC + S_GDEH - S_FCDG - S_HEBK + (S1-S4) + (S3 - S2)
= S_AFC + S_GDEH - S_FCDG - S_HEBK
= S_AFC + (S_AHE - S_AGD) - (S_AGD - S_AFC) - (S_ABK - S_AHE)
= S_AFC + S_AHE - S_AGD - S_AGD + S_AFC - S_ABK + S_AHE
= 2*S_AFC + 2*S_AHE - 2*S_AGD - S_ABK
##### let t := S_ABK
= 2t*1/16 + 2t*9/16 - 2t*1/4 - t
= t * 1/16 * (2 + 18 - 8 - 16)
= t/16 * (-4)
= -1/4 * t
= -1/4 * 1/2 * AB*DK
= -1/4 * AD^2
= -1/4 * 2^2
= -1
