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八年级数学填空题一般
题目
在学习全等三角形的知识时,数学兴趣小组发现这样一个模型:它是由两个共顶点且顶角相等的等腰三角形构成.在相对位置变化时,始终存在一对全等三角形.通过查询资料,他们得知这种模型称为"手拉手模型",兴趣小组进行了如下操作:
(1)(1)观察猜想
如图11,在ABC\triangle ABC中,分别以ABAB,ACAC为边向外作等腰直角ABD\triangle ABD和等腰直角ACE\triangle ACE,BAD=CAE=90\angle BAD=\angle CAE=90^{\circ},连接BEBE,CDCD,则BEBECDCD的数量关系为______,位置关系为______;
(2)(2)类比探究
如图22,在ABC\triangle ABC中,分别以ABAB,ACAC为边作等腰直角ABD\triangle ABD和等腰直角ACE.BAD=CAE=90\triangle ACE.\angle BAD=\angle CAE=90^{\circ},点DD,EE,CC在同一直线上,AMAMACE\triangle ACECECE边上的高,猜想DCDC,BCBC,AMAM之间的数量关系并说明理由;
(3)(3)解决问题
运用(1)(2)\left(1\right)\left(2\right)中所积累的经验和知识,完成下题:如图33,要测量池塘两岸相对的两点DD,CC的距离,已经测得ACB=45\angle ACB=45^{\circ},DAB=90\angle DAB=90^{\circ},AB=ADAB=AD,AC=152AC=15\sqrt{2}米,BC=40BC=40米,CDCD的长为______米.
知识点:相似形综合题章节:第6章 图形的相似 / 6.7 用相似三角形解决问题

答案与解析

答案

(1)ABD\left(1\right)\because \triangle ABDACE\triangle ACE都是等腰直角三角形,
AB=AD\therefore AB=ADAC=AEAC=AE
BAD=CAE=90\because \angle BAD=\angle CAE=90^{\circ}
BAD+CAB=CAE+CAB\therefore \angle BAD+\angle CAB=\angle CAE+\angle CAB,即BAE=CAD\angle BAE=\angle CAD
CAD\triangle CADEAB\triangle EAB中,
{AB=ADBAE=CADAE=AC\left\{\begin{array}{l}{AB=AD}\\{∠BAE=∠CAD}\\{AE=AC}\end{array}\right.
CAD\therefore \triangle CADEAB(SAS)\triangle EAB\left(SAS\right)
CD=BE\therefore CD=BEACD=AEB\angle ACD=\angle AEB
BEBECDCD交于点OOACACBEBE交于点FF
AFE=OFC\because \angle AFE=\angle OFC
COF=CAE=90\therefore \angle COF=\angle CAE=90^{\circ}
BECD\therefore BE\bot CD.

故答案为:BE=CDBE=CDBECDBE\bot CD.
(2)DC=BC+2AM(2)DC=BC+2AM,理由如下,
ABD\because \triangle ABDACE\triangle ACE都是等腰直角三角形,
AB=AD\therefore AB=ADAC=AEAC=AE
BAD=CAE=90\because \angle BAD=\angle CAE=90^{\circ}
BADEAB=CAEEAB\therefore \angle BAD-\angle EAB=\angle CAE-\angle EAB,即DAE=BAC\angle DAE=\angle BAC
ADE\triangle ADEABC\triangle ABC中,
{AE=ACDAE=BACAD=AB\left\{\begin{array}{l}{AE=AC}\\{∠DAE=∠BAC}\\{AD=AB}\end{array}\right.
ADE\therefore \triangle ADEABC(SAS)\triangle ABC\left(SAS\right)
DE=BC\therefore DE=BC
AC=AE\because AC=AEAMCEAM\bot CE
EC=2EM\therefore EC=2EM
ACE\because \triangle ACE为等腰直角三角形,AMCEAM\bot CE
AEM=EAM=45\therefore \angle AEM=\angle EAM=45^{\circ}
EM=AM\therefore EM=AM
EC=2AM\therefore EC=2AM
DC=DE+EC=BC+2AM\therefore DC=DE+EC=BC+2AM.
(3)(3)如图,作AMACAM\bot AC,使AM=ACAM=AC,连接BMBMCMCM,则ACM\triangle ACM为等腰直角三角形.
按照第二问思路同理可证:BAM\triangle BAMDAC(SAS)\triangle DAC\left(SAS\right)
BM=CD\therefore BM=CD
ACM\because \triangle ACM是等腰直角三角形,
ACM=45\therefore \angle ACM=45^{\circ}
ACB=45\because \angle ACB=45^{\circ}
BCM=90\therefore \angle BCM=90^{\circ}
AC=152=AM\because AC=15\sqrt{2}=AM
CM=AM2+AC2=30\therefore CM=\sqrt{AM^{2}+AC^{2}}=30
RtBCMRt\triangle BCM中,BC=40BC=40
BM=BC2+CM2=50\therefore BM=\sqrt{BC^{2}+CM^{2}}=50米,
CD=50\therefore CD=50米,
故答案为:5050.

解析

(1)ABD\left(1\right)\because \triangle ABDACE\triangle ACE都是等腰直角三角形,
AB=AD\therefore AB=ADAC=AEAC=AE
BAD=CAE=90\because \angle BAD=\angle CAE=90^{\circ}
BAD+CAB=CAE+CAB\therefore \angle BAD+\angle CAB=\angle CAE+\angle CAB,即BAE=CAD\angle BAE=\angle CAD
CAD\triangle CADEAB\triangle EAB中,
{AB=ADBAE=CADAE=AC\left\{\begin{array}{l}{AB=AD}\\{∠BAE=∠CAD}\\{AE=AC}\end{array}\right.
CAD\therefore \triangle CADEAB(SAS)\triangle EAB\left(SAS\right)
CD=BE\therefore CD=BEACD=AEB\angle ACD=\angle AEB
BEBECDCD交于点OOACACBEBE交于点FF
AFE=OFC\because \angle AFE=\angle OFC
COF=CAE=90\therefore \angle COF=\angle CAE=90^{\circ}
BECD\therefore BE\bot CD.

故答案为:BE=CDBE=CDBECDBE\bot CD.
(2)DC=BC+2AM(2)DC=BC+2AM,理由如下,
ABD\because \triangle ABDACE\triangle ACE都是等腰直角三角形,
AB=AD\therefore AB=ADAC=AEAC=AE
BAD=CAE=90\because \angle BAD=\angle CAE=90^{\circ}
BADEAB=CAEEAB\therefore \angle BAD-\angle EAB=\angle CAE-\angle EAB,即DAE=BAC\angle DAE=\angle BAC
ADE\triangle ADEABC\triangle ABC中,
{AE=ACDAE=BACAD=AB\left\{\begin{array}{l}{AE=AC}\\{∠DAE=∠BAC}\\{AD=AB}\end{array}\right.
ADE\therefore \triangle ADEABC(SAS)\triangle ABC\left(SAS\right)
DE=BC\therefore DE=BC
AC=AE\because AC=AEAMCEAM\bot CE
EC=2EM\therefore EC=2EM
ACE\because \triangle ACE为等腰直角三角形,AMCEAM\bot CE
AEM=EAM=45\therefore \angle AEM=\angle EAM=45^{\circ}
EM=AM\therefore EM=AM
EC=2AM\therefore EC=2AM
DC=DE+EC=BC+2AM\therefore DC=DE+EC=BC+2AM.
(3)(3)如图,作AMACAM\bot AC,使AM=ACAM=AC,连接BMBMCMCM,则ACM\triangle ACM为等腰直角三角形.
按照第二问思路同理可证:BAM\triangle BAMDAC(SAS)\triangle DAC\left(SAS\right)
BM=CD\therefore BM=CD
ACM\because \triangle ACM是等腰直角三角形,
ACM=45\therefore \angle ACM=45^{\circ}
ACB=45\because \angle ACB=45^{\circ}
BCM=90\therefore \angle BCM=90^{\circ}
AC=152=AM\because AC=15\sqrt{2}=AM
CM=AM2+AC2=30\therefore CM=\sqrt{AM^{2}+AC^{2}}=30
RtBCMRt\triangle BCM中,BC=40BC=40
BM=BC2+CM2=50\therefore BM=\sqrt{BC^{2}+CM^{2}}=50米,
CD=50\therefore CD=50米,
故答案为:5050.

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