题目内容
如图,把矩形
沿
对折后使两部分重合,若
,则
=( )
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A.110° | B.115° | C.120° | D.130° |
B
分析:根据折叠的性质及∠1=50°可求出∠2的度数,再由平行线的性质即可解答.
解答:
解:∵四边形EFGH是四边形EFBA折叠而成,
∴∠2=∠3,
∵∠2+∠3+∠1=180°,∠1=50°,
∴∠2=∠3=
(180°-50°)=
×130°=65°,
又∵AD∥BC,
∴∠AEF+∠EFB=180°,
∴∠AEF=180°-65°=115°
解答:
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∴∠2=∠3,
∵∠2+∠3+∠1=180°,∠1=50°,
∴∠2=∠3=
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又∵AD∥BC,
∴∠AEF+∠EFB=180°,
∴∠AEF=180°-65°=115°
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