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,垂直的定义,人教版 七年级,下,第,5,章 相交线与平行线,5.1.2,D,1,2,3,4,5,A,B,6,7,60,或,120 ,答 案 呈 现,温馨,提示,:,点击,进入,讲评,习题链接,如图,若,CD,EF,,,1,2,,则,AB,EF,,请说明理由,(,补全解题过程,),解:因为,CD,EF,,,所以,1,_(,垂直的定义,),所以,2,1,_.,所以,AB,_,EF,(,垂直的定义,),90,1,90,已知在同一平面内:两条直线相交成直角;,两条直线互相垂直;,一条直线是另一条直线的垂线,那么下列因果关系:;中,正确的有,(,),A,0,个,B,1,个,C,2,个,D,3,个,2,D,3,A,【,教材,P,8,习题,T,5,变式,】【2021,北京,】,如图,点,O,在直线,AB,上,,OC,OD,.,若,AOC,120,,则,BOD,的大小为,(,),A,30,B,40,C,50,D,60,【2020,乐山,】,如图,,E,是直线,CA,上一点,,FEA,40,,射线,EB,平分,CEF,,,GE,EF,,则,GEB,(,),A,10,B,20,C,30,D,40,B,4,在直线,AB,上任取一点,O,,过点,O,作射线,OC,,,OD,,使,OC,OD,,当,AOC,30,时,,BOD,的度数是,_,_,_,60,或,120 ,5,【,点拨,】,本题易因只考虑,OC,,,OD,在直线,AB,同侧的情况,而忽略了,OC,,,OD,在直线,AB,两侧的情况,以致漏解而致错,6,(2),判断,OD,与,AB,的位置关系,并说明理由,解:,OD,AB,.,理由:由,(1),知,AOC,COD,45,,,所以,AOD,AOC,COD,90,,,所以,OD,AB,(,垂直的定义,),将一副三角尺的两个直角顶点重合在一起,按如图位置放置,(1),如图,若,BOC,50,,求,AOD,的度数;,7,解,:因为,AOB,90,,,BOC,50,,,所以,AOC,AOB,BOC,90,50,40.,又因为,COD,90,,所以,AOD,AOC,COD,40,90,130.,(2),如图,若,BOC,60,,求,AOD,的度数;,解:,因为,AOB,90,,,COD,90,,,BOC,60,,,所以,AOD,360,AOB,BOC,COD,360,90,60,90,120.,(3),如图,猜想,AOD,与,BOC,的关系,并说明理由,解:,AOD,与,BOC,互补,理由,如下,:,因为,AOB,90,,,COD,90,,,所以,90,BOC,90,AOD,360,,,所以,AOD,BOC,180,,,即,AOD,与,BOC,互补,
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