解：?$(1) $?由題圖可得，
?$S_{1}=a^2-b^2，$??$S_{2}=2b^2-ab。$?
?$ (2) a - b = 8，$??$ab = 13，$?則
?$ \begin {aligned}S_{1}+S_{2}&=a^2-b^2+2b^2-ab\\&=a^2+b^2-ab\\&=(a - b)^2+ab\\&=8^2+13\\&=64 + 13\\&=77\end {aligned}$?
?$ $?所以?$S_{1}+S_{2}$?的值為?$77。$?
?$ (3) $?由題圖可得?$S_{3}=a^2+b^2-\frac {1}{2}b(a + b)-\frac {1}{2}a^2$?
?$=\frac {1}{2}(a^2+b^2-ab)$?
?$ $?因為?$S_{1}+S_{2}=a^2+b^2-ab = 34，$?
所以?$S_{3}=\frac {1}{2}×34 = 17。$?
