解：?$(1) ②$?兩個(gè)正方形的面積之和
?$\begin {aligned}S &= x^2+(a - x)^2\\&=x^2+a^2-2ax+x^2\\&=2x^2-2ax + a^2\end {aligned}$?
?$ (2) $?因?yàn)樗倪呅?$APCD、$?四邊形?$PBEF $?均為正方形，
?$AP = x，$??$BP = a - x，$
?所以?$CF=PF - PC=a - x - x=a - 2x。$?
?$ $?陰影部分的面積?$S = S_{正方形APCD}+S_{正方形PBEF}+S_{\triangle FCD}-S_{\triangle ABD}-S_{\triangle EFB}$?
?$ \begin {aligned}&=x^2+(a - x)^2+\frac {1}{2}x·(a - 2x)-\frac {1}{2}x·a-\frac {1}{2}(a - x)^2\\&=x^2+\frac {1}{2}(a - x)^2+\frac {1}{2}x·(a - 2x)-\frac {1}{2}x·a\\&=x^2+\frac {1}{2}(a^2-2ax+x^2)+\frac {1}{2}(ax - 2x^2)-\frac {1}{2}ax\\&=x^2+\frac {1}{2}a^2-ax+\frac {1}{2}x^2+\frac {1}{2}ax - x^2-\frac {1}{2}ax\\&=\frac {1}{2}x^2+\frac {1}{2}a^2-ax\end {aligned}$?
