【关键词】相似、光影
【答案】解：（1）由题意可知：∠BAC?∠EDF?90?，?BCA??EFD．
∴△ABC∽△DEF． ∴
ABDE?ACDF80DE60900即，?··········································································· 2分 ．∴DE=1200（cm）．
所以，学校旗杆的高度是12m． ····································································· 3分 （2）解法一： 与①类似得：
ABGN?ACGH80GN60156即，? ．∴GN=208． ····································································································· 4分 在Rt△NGH中，根据勾股定理得：
NH2?156?208?260.
222∴NH=260． ····································································································· 5分 设?O的半径为rcm，连结OM，
∵NH切?O于M，∴OM?NH．··································································· 6分 则∠OMN??HGN?90?，又∠ONM?∠HNG． ∴△OMN∽△HGN．∴
OMHG?ONHN． ···························································· 7分
又ON?OK?KN?OK?(GN?GK)?r?8． ∴
r156?r?8260，解得：r=12．
所以，景灯灯罩的半径是12cm． ····································································· 9分
B 80cm A 200cm E KNMOC D 900cm 图2
F
60cm G 156cm 图3
H
图1
解法二： 与①类似得：
ABGN?ACGH，即
80GN?60156．
∴GN=208． ····································································································· 4分
设?O的半径为rcm，连结OM， ∵NH切?O于M，∴OM?NH．··································································· 5分 则∠OMN??HGN?90?，又∠ONM?∠HNG， ∴△OMN∽△HGN． ∴
OMHG?MNGN4即，r156?MN208 ········································································ 6分 ．∴MN?又ON?OK?KN?OK?(GN?GK)?r?8． ······················· 7分 r，3在Rt△OMN中，根据勾股定理得：
2?4?2即r2?9r?36?0． r??r???r?8?，3??2解得：r1?12，r2??3（不合题意，舍去）
所以，景灯灯罩的半径是12cm．