资源描述
,垂线,沪科版,七,年级,下,第,10,章 相交,线、平行线与平移,10.1.2,目标一垂直的,定义,D,1,2,3,4,5,B,C,6,7,8,B,答 案 呈 现,温馨,提示,:,点击,进入,讲评,习题链接,9,B,60,或,120,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,B,【,2021,毫州月考】,如图,已知,a,b,,垂足为,O,,,直,线,c,经过点,O,,则,1,与,2,的关系一定成立的是,(,),A,相等,B,互余,C,互补,D,互为对顶角,【,点拨,】,如图,,2,3(,对顶角相等,),,又因为,a,b,,所以,1,3,90,,所以,1,2,90,,所以,1,与,2,互余故选,B.,C,4,如,图,,OA,OC,,,OB,OD,,那么,(,),A,1,2,B,2,3,C,1,3,D,1,2,3,B,5,【,中考,孝感】,如图,直线,AB,,,CD,相交于点,O,,,OE,CD,,垂足为,O,.,若,BOE,40,,则,AOC,的度数为,(,),A,40,B,50,C,60,D,140,6,B,【,中考,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,40,7,60,或,120,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,_,【,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,以致漏解而出错,8,(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,),9,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;,解,:因为,AOB,90,,,BOC,50,,,所以,AOC,AOB,BOC,90,50,40.,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,(4),如图,,若,BOC,:,AOD,7,:,29,,求,BOC,和,AOD,的度数,解:,由,(3),知,BOC,AOD,180.,因为,BOC,:,AOD,7,:,29,,所以设,BOC,7,x,,则,AOD,29,x,,,所以,7,x,29,x,180,,解和,x,5.,所以,BOC,35,,,AOD,145.,
展开阅读全文