(宜宾专版)2019年中考数学总复习第一编教材知识梳理篇第4章图形的初步认识与三角形第15讲等腰三角形与直角三角形(精练)试题.doc

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1、1第十五讲 等腰三角形与直角三角形(时间:45分钟)一、选择题1.如图,已知等腰三角形ABC,ABAC,若以点B为圆心,BC长为半径画弧,交腰AC于点E,则下列结论一定正确的是( C )A.AEEC B.AEBEC.EBCBAC D.EBCABE,(第1题图) ,(第2题图)2.某城市几条道路的位置关系如图所示,已知ABCD,AE与AB的夹角为48,若CF与EF的长度相等,则C的度数为( D )A.48 B.40 C.30 D.243.如图,ABD是以BD为斜边的等腰直角三角形,BCD中,DBC90,BCD60,DC的中点为E,AD与BE的延长线交于点F,则AFB的度数为( B )A.30 B

2、.15 C.45 D.25,(第3题图) ,(第5题图)4.已知等腰三角形的两条边长分别为2和3,则它的周长为( D )A.7 B.8 C.5 D.7或85.如图,在ABC中,ACB90,CDAB,垂足为D,点E是AB的中点,CDDEa,则AB的长为( B )A.2a B.2 a C.3a D.24336.一次函数yx1的图象交x轴于点A,交y轴于点B.点C在x轴上,且使得ABC是等腰三角形,符合题意的点C有( C )A.2个 B.3个 C.4个 D.5个二、填空题7.如图,在ABC中,B_25_.8.如图,已知 直线l 1l 2,将等边三角形如图放置,若40,则等于_20_.,(第7题图)

3、,(第8题图) ,(第9题图)29.如图,已知ABC中,ABAC,AD平分BAC,E是AC的中点,若AB6,则DE的长为_3_.10.如图,已知 RtABE中,A90,B60,BE10,D是线段AE上的一动点,过D作CD交BE于C,并使得CDE30,则CD长度的取值范围是_0CD5_.,(第10题图) ,(第11题图)11.如图,在ABC中,ABAC6,AB的垂直平分线交AB于点E,交BC于点D,连结AD,若AD4,则DC_5_.12.在边长为4的等边三角形ABC中,D为BC边上的任意一点,过点D分别作DEAB,DFAC,垂足分别为E、F,则DEDF_2 _.3三、解答题13.如图, 在ABC

4、中,ADBC,EF垂直平分AC,交AC于点F,交BC于点E,且BDDE.(1)若C40,求BAD的度数;(2)若AC5,DC4,求ABC的周长.解:(1)EF垂直平分AC,AECE,CEAC40.ADBC,BDDE,ABAE,BBEA2C 80,BAD908010;(2)由(1)知AEECAB.BDDE,ABBDDEAEDECEDC,C ABC ABBCAC2DCAC24513.14.在等边ABC中,点D、E分别在边BC、AC上,若CD2,过点D作DEAB,过点E作EFDE,交BC的延长线于点F,求EF的长. 解:ABC是等边三角形,BACB60.DEAB,EDCB60,EDC是等边三角形,D

5、EDC2.在 RtDEF中,DEF90,EDF60,EFDE tan 602 .3315.如图,四边形ABCD中,ABCADC90,BD平分ABC,DCB60,ABBC8,则AC的长是_ _.86316.如图,ABC 是边长为4 cm的等边三角形,边AB在射线OM上,且OA6 cm.点D从O点出发,沿OM的方向以1 cm/s的速度运动.当D不与点A重合时,将ACD绕点C逆时针方向旋转60得到BCE,连结DE.(1)求证:CDE是等边三角形;(2)如图,当6t10时,BDE的周长是否存在最 小值?若存在,求出BDE的最小周长;若不存在,请 说明理由;(3)如图,当点D在射线OM上运动时,是否存在

6、以D、E、B为顶点的三角形是直 角三角形?若存在,求出此时t的值;若不存在,请 说明理由.(1)证明:将ACD绕点C逆时针方向旋转60得到BCE,DCE60,DCEC,CDE是等边三角形;(2)存在.当6t10时,由旋转得BEAD,C BDE BEDBDEABDE4DE.由(1)知,CDE是等边三角形,DECD,C BDE CD4.由垂线段最短可知,当CDAB时,BDE的周长最小,此时C D2 cm,3BDE的最小周长为(2 4) cm;3(3)存在.当点D与点B重合时,D、 B、E不能构成三角形,当点D与点B重合时,不符合题意;当0t6时,由旋转可知,ABE60,BDE60,BED90.由(

7、1)可知,CDE是等边三角形,DEC60,CEB30.CEBCDA,CDA30.CAB60,ACDADC30,DACA4 cm,ODOADA642 cm,t212;当6t10时,DBE12090,此时不存在;当t10时 ,由旋转可得DBE60.又由(1)知CDE60, BDECDEBDC60BDC,而BDC0,BDE60,只能BDE90.4从而BDCBCD30, BDBC4 cm,OD14 cm,t14114.综上所述,当t2或14时,以D、E、B为顶点的三角形是直角三角形.17.如图,已知ABA 1B,A1B1A 1A2,A2B2A 2A3,A3B3A 3A4,若A70,则A n1 AnBn1 的度数为( C )A. B. C. D.702n 702n 1 702n 1 702n 2

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