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九年级数学解答题一般
题目
如图①,现有三张形状大小完全相同的三角形纸片叠合到一起,其中AB=ACAB=AC,B=C=α\angle B=\angle C=\alpha.老师让同学们以"三角形的旋转"为主题,通过小组合作探究,提出问题一展示一集体谈论,解决问题.
(1)(1)"希望"小组提出问题:
将图11中的ABC\triangle ABC以点CC为旋转中心,顺时针旋转角度α\alpha,得到DEC\triangle DEC,再将ABC\triangle ABC以点AA为旋转中心,逆时针旋转角度α\alpha,得到AFG\triangle AFG,连接DGDG,得到图②,请判断四边形AEDGAEDG的形状,并说明理由;
(2)(2)"善学"小组提出问题:
将图①中的ABC\triangle ABC以点CC为旋转中心,顺时针旋转9090^{\circ},得到DEC\triangle DEC,再将ABC\triangle ABC以点AA为旋转中心,逆时针旋转9090^{\circ},得到AFG\triangle AFG,连接AEAE,DFDF,DGDG,得到图③请判断四边形ACDGACDG的形状,并说明理由;
老师根据上面小组的探究提出:
(3)(3)α=75\alpha =75^{\circ},则图③中,EDF=______.\angle EDF=\_\_\_\_\_\_.
知识点:旋转对称图形、相似图形、图形的相似章节:第24章 相似三角形 / 第1节 相似形 / 24.1 放缩与相似形

答案与解析

答案

(1)四边形AEDGAEDG是平行四边形,理由如下:
\because旋转,
AC=CD=AG\therefore AC=CD=AGAB=DEAB=DEGAC=α\angle GAC=\alphaDEC=B=α\angle DEC=\angle B=\alpha
DEC=GAC\therefore \angle DEC=\angle GAC
AG\therefore AGDEDE
AB=AC\because AB=AC
AG=DE\therefore AG=DE
\therefore四边形AEDGAEDG是平行四边形;
(2)(2)四边形ACDGACDG是正方形,理由如下:
\because旋转,
AC=CD=AG\therefore AC=CD=AGAB=DEAB=DEGAC=90=ACD\angle GAC=90^{\circ}=\angle ACD
AG\therefore AGCDCD
\therefore四边形ACDGACDG是平行四边形,
GAC=90\because \angle GAC=90^{\circ}
\therefore四边形ACDGACDG是矩形,
AC=CD=AG\because AC=CD=AG
\therefore四边形ACDGACDG是正方形;
(3)(3)连接GEGE

B=ACB=α=75\because \angle B=\angle ACB=\alpha =75^{\circ}
BAC=30\therefore \angle BAC=30^{\circ}
\because旋转,
CDE=GAF=30\therefore \angle CDE=\angle GAF=30^{\circ}AB=DE=AC=CDAB=DE=AC=CD
\because四边形ACDGACDG是正方形,
GD=CD=AC=AG\therefore GD=CD=AC=AGGDC=AGD=90\angle GDC=\angle AGD=90^{\circ}
GDE=60\therefore \angle GDE=60^{\circ}DG=DEDG=DE
GDE\therefore \triangle GDE是等边三角形,
GE=GD=AG\therefore GE=GD=AGGDE=60\angle GDE=60^{\circ}
AGE=30\therefore \angle AGE=30^{\circ}
GAE=GEA=75\therefore \angle GAE=\angle GEA=75^{\circ}
FAE=45\therefore \angle FAE=45^{\circ}
\because四边形AEDFAEDF是平行四边形,
EAF=EDF=45\therefore \angle EAF=\angle EDF=45^{\circ}
故答案为:4545^{\circ}.

解析

(1)四边形AEDGAEDG是平行四边形,理由如下:
\because旋转,
AC=CD=AG\therefore AC=CD=AGAB=DEAB=DEGAC=α\angle GAC=\alphaDEC=B=α\angle DEC=\angle B=\alpha
DEC=GAC\therefore \angle DEC=\angle GAC
AG\therefore AGDEDE
AB=AC\because AB=AC
AG=DE\therefore AG=DE
\therefore四边形AEDGAEDG是平行四边形;
(2)(2)四边形ACDGACDG是正方形,理由如下:
\because旋转,
AC=CD=AG\therefore AC=CD=AGAB=DEAB=DEGAC=90=ACD\angle GAC=90^{\circ}=\angle ACD
AG\therefore AGCDCD
\therefore四边形ACDGACDG是平行四边形,
GAC=90\because \angle GAC=90^{\circ}
\therefore四边形ACDGACDG是矩形,
AC=CD=AG\because AC=CD=AG
\therefore四边形ACDGACDG是正方形;
(3)(3)连接GEGE

B=ACB=α=75\because \angle B=\angle ACB=\alpha =75^{\circ}
BAC=30\therefore \angle BAC=30^{\circ}
\because旋转,
CDE=GAF=30\therefore \angle CDE=\angle GAF=30^{\circ}AB=DE=AC=CDAB=DE=AC=CD
\because四边形ACDGACDG是正方形,
GD=CD=AC=AG\therefore GD=CD=AC=AGGDC=AGD=90\angle GDC=\angle AGD=90^{\circ}
GDE=60\therefore \angle GDE=60^{\circ}DG=DEDG=DE
GDE\therefore \triangle GDE是等边三角形,
GE=GD=AG\therefore GE=GD=AGGDE=60\angle GDE=60^{\circ}
AGE=30\therefore \angle AGE=30^{\circ}
GAE=GEA=75\therefore \angle GAE=\angle GEA=75^{\circ}
FAE=45\therefore \angle FAE=45^{\circ}
\because四边形AEDFAEDF是平行四边形,
EAF=EDF=45\therefore \angle EAF=\angle EDF=45^{\circ}
故答案为:4545^{\circ}.

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