$ 解：(3)S_{陰影}=\frac {1}{2}S_{正方形ABCD}-S_{△AHE}-S_{梯形HEBC}+\frac {1}{4}S_{正方形EFGH}$
$=\frac {1}{2}a2-\frac {1}{2}×\frac {1}{2}a×b-\frac {1}{2}(a+b)×\frac {1}{2}a+\frac {1}{4}b2$
$=-\frac {ab}{2}+\frac {a2}{4}+\frac {b2}{4}$
$=\frac {1}{4}(a-b)2$
$∵a+b=12，ab=7$
$∴(a-b)2=(a+b)2-4ab$
$=144-28$
$=116$
$則\frac {1}{4}(a-b)2=29$
$∴陰影部分面積之和是29$
