将边长为a的正方形OABC绕顶点O按顺时针方向旋转角α(0°
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(1)∵∠EA1O=∠MC1O=Rt∠,OA1=OC1,∠1=∠2

∴△OC1M≌△OA1E

(2)作OH⊥MN于H

∵OE=OM,∠EON=∠M0N,ON=ON,

∴ONM≌△ONE

∴∠3=∠4,又∵∠OA1N=∠OHN=Rt∠,ON=ON

∴OHN≌△OA1N,

∴OH=OA1=a

(3)∵∠OHM=∠OC1M=Rt∠,OH=OC1,OM=OM

∴OHM≌△OC1M,∴HM=C1M

p=NM+B1M+B1N

=NH+MH+B1M+B1N

=A1B1+B1C1

=2a