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时间:2018-04-07
《八年级数学平移旋转在几何证明中的应用(图形的平移与旋转)拔高练习》由会员上传分享,免费在线阅读,更多相关内容在学术论文-天天文库。
1、八年级数学平移旋转在几何证明中的应用(图形的平移与旋转)拔高练习试卷简介:本试卷共5道选择题,全面考察学生对旋转和平移这一部分知识的掌握学习建议:北师版八年级上册第三章图形的平移与旋转,平移和旋转的定义和性质要熟练掌握。一、单选题(共5道,每道20分)1.(呼和浩特)把∠A是直角的△ABC绕A点沿顺时针方向旋转85°,点B转到点E得△AEF,则以下列结论错误的是( )A.∠BAE=85° B.AC=AF C.EF=BC D.∠EAF=85° 2.如图,在△ABC中,∠CAB=70°.在同一平面内,将△ABC绕点A旋转到△AB′C′的位置,使得
2、CC′∥AB,则∠B′AB=_________ A.70° B.35° C.45° D.40° 3.如图,△ABC中,AC=5,中线AD=7,△EDC是由△ADB绕D点旋转所得到的,则AB边的取值范围是() A.13、D为底的梯形. A.120° B.90° C.60° D.30° 5.已知:如图,在正方形ABCD外取一点E,连接AE、BE、DE.过点A作AE的垂线交DE于点P.若AE=AP=1,PB=.下列结论: ①△APD≌△AEB; ②点B到直线AE的距离为2; ③EB⊥ED; ④S△APD+S△APB=1+; ⑤S正方形ABCD=4+. 其中正确结论的序号是() A.①③④ B.①②⑤ C.③④⑤ D.①③⑤ 内部资料仅供参考第3页共3页9JWKffwvG#tYM*Jg&6a*CZ7H$dq8KqqfHVZFedswSyXTy#&QA9wkxFyeQ4、^!djs#XuyUP2kNXp6X4NGpP$vSTT#UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmYWpazadNu##KN&MuWFA5ux^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmYWpazadNu##KN&MuWFA5uxY7JnD6YWRrWwc^vR9CpbK!zn%Mz849Gx^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh55、pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmUE9aQ@Gn8xp$R#͑Gx^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmYWpazadNu##KN&MuWFA5uxY7JnD6YWRrWwc^vR9CpbK!zn%Mz849Gx^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmYWpazad6、Nu##KN&MuWFA5ux^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmYWpazadNu##KN&MuWFA5uxY7JnD6YWRrWwc^vR9CpbK!zn%Mz849Gx^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z8vG#tYM*Jg&6a*CZ7H$dq8KqqfHVZFedswSyXTy#&QA9wkxFyeQ^!djs#XuyUP7、2kNXpRWXmA&UE9aQ@Gn8xp$R#͑Gx^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmYWpazadNu##KN&MuWFA5uxY7JnD6YWRrWwc^vR9CpbK!zn%Mz849Gx^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmYWpaza8、dNu##KN&MuWFA5ux^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&k
3、D为底的梯形. A.120° B.90° C.60° D.30° 5.已知:如图,在正方形ABCD外取一点E,连接AE、BE、DE.过点A作AE的垂线交DE于点P.若AE=AP=1,PB=.下列结论: ①△APD≌△AEB; ②点B到直线AE的距离为2; ③EB⊥ED; ④S△APD+S△APB=1+; ⑤S正方形ABCD=4+. 其中正确结论的序号是() A.①③④ B.①②⑤ C.③④⑤ D.①③⑤ 内部资料仅供参考第3页共3页9JWKffwvG#tYM*Jg&6a*CZ7H$dq8KqqfHVZFedswSyXTy#&QA9wkxFyeQ
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