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