已知正三棱锥S-ABC的高SO=h，斜高SM=n，求经过SO的中点且平行于底面的截面△A1B1C1的面积．
设底面正三角形的边长为a，
在RT△SOM中SO=h，SM=n，
∴OM=
，n2-l2
又MO=
a，即a=3 6 6 3
，n2-l2
∴s△ABC=
a2=33 4
(n2-l2)，3
∴截面面积为3 4
(n2-l2)．3
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