?BP?BF?PG, QBE//PG, ??ECF∽?GCP
?EFCEPG?CG 设BP?BF?PG?y,
?15?yy?2025 ?y?253 则BP?253 在Rt?PBC中,PC?2510BC3,cos?PCB?25310PC?2510?103③若BP?9,
解法一:连接GF,(如图3)
Q?GEF??BAE?90o, QBF//PG,BF?PG
∴四边形BPGF是平行四边形
QBP?BF,
?平行四边形BPGF是菱形
?BP//GF, ??GFE??ABE, ??GEF∽?EAB
?EFABGF?BE ?BEgEF?ABgGF?12?9?108
解法二:如图2,
Q?FEC??PBC?90o,
5
?EFC??PFB??BPF, ??EFC∽?BPC
?EFCEBP?CB 又Q?BEC??A?90o, 由AD//BC得?AEB??EBC,
??AEB∽?EBC
?ABCEBE?CB ?AEEFBE?BP ?BEgEF?AEgBP?12?9?108
解法三:(如图4)过点F作FH?BC,垂足为HS?BPFBFBFS?四边形PFEGEF?PG?BE图4
BFS?BFCEFBE??BCEFS?? ?BEC12?BC12?9EFBE?12 ?BEgEF?12?9?108
(2018齐齐哈尔)
6
7
类型2 分割与剪接 (2018海南)
(2018嘉兴)将一张正方形纸片按如图步骤①,②沿虚线对折两次,然后沿③中平行于底边的虚线剪去一个角,展开铺平后的图形是()
(2018菏泽)
8
(2018枣庄)
(2018嘉兴)
9
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