苏科版数学八年级上册期中复习试卷09（含答案）

这是一份苏科版数学八年级上册期中复习试卷09（含答案），共8页。试卷主要包含了选择题世纪教育网，填空题，解答题等内容，欢迎下载使用。
苏科版数学八年级上册期中复习试卷一、选择题世纪教育网版权所有1． 下列手机软件图标中，是轴对称图形的是······························【 ▲ 】   2． 如图，∠BAD＝∠BCD＝90°，AB＝CB，据此可以证明△BAD≌△BCD，证明的依据是【 ▲ 】A．AAS B．ASA C．SAS D．HL      3． 下列线段长中，能构成直角三角形的一组是····························【 ▲ 】A．1，2，3 B．2，3，4 C．3，4，5 D．4，5，64． 在实数0，，，，1. 010 010 001中，无理数有·························【 ▲ 】A．1个 B．2个 C．3个 D．4个5． 如图，在数轴上表示实数的点可能是·····························【 ▲ 】A．点P B．点Q C．点M D．点N6． 如图是5×5的正方形网格，以点D、E为两个顶点作位置不同的格点三角形，使所作的格点三角形与△ABC全等，这样的格点三角形最多可以画出              【 ▲ 】A．2个 B．4个 C．6个 D．8个二、填空题（本大题共10小题，每小题2分，共20分．不需写出解答过程，请将答案直接写在答题卡相应位置上）7． 4的平方根是   ▲   ．8． 在△ABC中，AB＝AC，∠A＝100°，则∠B＝   ▲   °．9． 如图，△ABC与△A′B′C′关于直线对称，则∠B的度数为   ▲   °．      10．等腰三角形的两边长分别为3和6，则这个等腰三角形的周长为   ▲   ．11．如图，△ABC中，∠BAC的平分线交BC于点D，过点D作DE⊥AC于E，DE＝5，则点D到AB的距离是   ▲   ．12．我国“辽宁号”航空母舰的满载排水量为67 500吨，将数据67 500精确到千位的近似值为   ▲   ．（结果用科学计数法表示）13．如图是一株美丽的“勾股树”，其中所有的四边形都是正方形，所有的三角形都是直角三角形，若正方形A、B、C、D的面积分别为9、4、4、1，则最大的正方形E的面积是   ▲   ．14．如图，在△ABC中，AB、AC的垂直平分线分别交BC于E、F两点，若∠BAC＝120°，则∠EAF的度数为   ▲   °．15．若正数的两个平方根分别为和，则正数＝   ▲   ．16．如图，在△ABC中，ABAC＝10，BC＝12，AD是角平分线，P、Q分别是AD、AB边上的动点，则BP＋PQ的最小值为   ▲   ．三、解答题（本大题共10小题，共82分．请在答题卡指定区域内作答，解答时应写出文字说明，推理过程或演算步骤）17．（本题满分5分）如图，点B、F、C、E在同一条直线上，且FB＝CE，AC＝DF，∠ACB＝∠DFE．求证：△ACB≌△DFE．   18．（本题满分5分）尺规作图．如图，已知∠AOB与点M、N．求作：点P，使点P到OA、OB的距离相等，且到点M与点N的距离也相等．（不写作法与证明，保留作图痕迹）   19．（本题满分8分，每小题4分）（1）计算：；（2）求x的值：＝9．    20．（本题满分6分）方格纸中每个小方格都是边长为1的正方形，我们把以格点连线为边的多边形称为“格点多边形”．如图①，△ABC是格点三角形．（1）试在图②中确定格点D，画一个以A、B、C、D为顶点的四边形，使其为轴对称图形；（画出一个即可）（2）试在图③中画一个“格点正方形”，使其面积等于10．          21．（本题满分6分）如图，在四边形ABCD中，AB＝AD，∠ABC＝∠ADC．求证：BC＝DC．    22．（本题满分8分）如图，在△ABC中，AC＝21，BC＝13，D是AC边上一点，BD＝12，AD＝16，E是边AB的中点，求线段DE的长．     23．（本题满分10分）如图，在等边△ABC中，E，F分别在边AC、BC上，满足AE＝CF，连接BE，AF交于点P．（1）求证：△ABE≌△CAF；（2）求∠APB的度数．     24．（本题满分10分）一架长2.5米的梯子AB如图所示斜靠在一面墙上，这时梯足B离墙底C（∠C＝90°）的距离BC为0.7米．（1）求此时梯顶A距地面的高度AC；（2）如果梯顶A下滑0．9米，那么梯足B在水平方向，向右滑动了多少米？  25．（本题满分10分）如图，△ABC中，∠C＝90°，AB＝10 cm，BC＝6 cm，动点P从点C出发，以每秒2 cm的速度按C→A的路径运动，设运动时间为t秒．（1）出发2秒时，△ABP的面积为   ▲   cm2；（2）当t为何值时，BP恰好平分∠ABC？  26．（本题满分14分）【问题情境】课外兴趣小组活动时，老师提出了如下问题：如图①，△ABC中，若AB＝12，AC＝8，求BC边上的中线AD的取值范围．小明在组内经过合作交流，得到了如下的解决方法：延长AD至点E，使DE＝AD，连接BE．请根据小明的方法思考：       （1）由已知和作图能得到△ADC≌△EDB，依据是   ▲   ．A．SSS B．SAS C．AAS D．HL（2）由“三角形的三边关系”可求得AD的取值范围是   ▲   ．解后反思：题目中出现“中点”、“中线”等条件，可考虑延长中线构造全等三角形，把分散的已知条件和所求证的结论集中到同一个三角形之中．【初步运用】如图②，AD是△ABC的中线，BE交AC于E，交AD于F，且AE＝EF．若EF＝3，EC＝2，求线段BF的长．【灵活运用】如图③，在△ABC中， ∠A＝90°，D为BC中点， DE⊥DF，DE交AB于点E，DF交AC于点F，连接EF．试猜想线段BE、CF、EF三者之间的等量关系，并证明你的结论．
参考答案一、选择题（每小题3分，共18分）题号123456答案ADCBCB二、填空题（每小题2分，共20分）7． ±2． 8． 40． 9． 100． 10．15．11．5． 12．6.8×104． 13．18． 14．60．15．25． 16．9.6．三、解答题17．（本题满分5分）证明：∵FB＝CE，∴BC＝EF．······················································1分在△ACB和△DFE中，······························································4分∴△ACB≌△DFE．················································5分（说明：全等条件中每写对一个给1分）18．（1）作∠AOB的平分线；···········································2分（2）连接MN，作线段MN的垂直平分线，标出∠AOB的平分线与线段MN的垂直平分线的交点P．              4分如图，点P就是所要求作的点．·······································5分（说明：不交待“点P就是所要求作的点”不扣分）19．（本题满分8分，每小题4分）解：（1）原式＝·················································3分＝．···························································4分（说明：每化简正确一个给1分）解：（2）＝3或＝－3．············································2分∴＝1或＝－5．··················································4分（写出一个一元一次方程给1分，一个正确答案给1分）20．（本题满分6分）（1）如图①所示．················································ 3分（2）如图②所示．················································6分         21．（本题满分6分）证明：连接BD．∵AB＝AD，∴∠ABD＝∠ADB．················································2分∵∠ABC＝∠ADC，∴∠CBD＝∠CDB．················································4分∴BC＝DC． ………………………………………………………………………………6分22．（本题满分8分）解：CD＝21－16＝5．∵DC2＋BD2＝52＋122＝169，BC＝132＝169，∴DC2＋BD2＝BC2．················································2分∴△BCD是直角三角形，且∠BDC＝90°．·······························3分∴∠ADB＝90°···················································4分在Rt△ADB中，由勾股定理，得AB＝＝20································6分∵∠ADB＝90°，E为斜边AB的中点， ∴DE＝AB＝×20＝10．·············································8分23．（本题满分10分）（1）证明：∵△ABC是等边三角形，∴AB＝AC，∠BAE＝∠C＝60°，······································2分在△ABE和△CAF中，···························································5分∴△ABE≌△CAF（SAS）．··········································6分（2）∵△ABE≌△CAF，∴∠ABE＝∠CAF．················································7分在△ABP中，∵∠APB＋∠ABP＋∠BAP＝180°，·························8分∴∠APB＝180°－∠PBA－∠PAB＝180°－∠CAF－∠PAB＝180°－(∠CAF＋∠PAB)＝180°－∠BAC＝120°．              10分24．（本题满分10分）解：（1）在Rt△ABC中，由勾股定理，得AC2＋BC2＝AB2．··················2分∴AC2＋0.72＝2.52．解得AC＝2.4（m）．···············································3分答：此时梯顶距地面的高度AC为2.4m．·································4分（2）由题意得：A′C＝2.4－0.9＝1.5，A′B′＝2.5．·····················5分在Rt△A′B′C中，由勾股定理得A′C2＋B′C2＝A′B′2．·················7分∴1.52＋B′C2＝2.52．解得B′C ＝2（m）·················································8分∴BB′＝ B′C－BC＝2－0.7＝1.3（m）．······························9分答：梯足在水平方向向右滑动了1.3m．··································10分25．（本题满分10分）（1）12．·························································3分（2）解：过点P作PG⊥AB于G，则∠BGP＝90°．∵∠C＝90°，∴∠BGP＝∠C．··················································4分∵BP平分∠ABC，∴∠CBP＝∠ABP．················································5分又∵BP＝BP，∴△BCP≌△BGP．················································6分∴BG＝BC＝6，PG＝PC＝2t．∴PA＝8－2t，AG＝10－6＝4．·······································8分在Rt△APG中， AG2＋PG2＝AP2．∴42＋(2t)2＝(8－2t)2 ···········································9分解得t＝．·······················································10分（说明：用面积法求解类似给分）26．（本题满分14分）解：【问题提出】（1）B．··························································3分（2）2＜AD＜10．···················································6分【初步运用】如图①，延长AD到M，使DM＝AD，连接BM．∵AD是△ABC中线，∴BD＝DC．又∵∠ADC＝∠MDB，∴△ADC≌△MDB．∴BM＝AC，∠CAD＝∠M．·············································8分∵AE＝EF，∴∠CAD＝∠AFE．∵∠AFE＝∠BFD，∴∠BFD＝∠CAD＝∠M．∴BF＝BM＝AC＝3＋2＝5．···········································10分         【灵活运用】猜想：BE2＋CF2＝EF2···············································11分理由：如图②，延长FD至G，使得DG＝DF，连接BG、EG，则△FDC≌△GDB．∴CF＝BG，∠FCD＝∠GBD．∵DF＝DG，DE⊥DF，∴EF＝EG．·······················································12分在△ABC中，∵∠A＝90°，∴∠EBC＋∠FCB＝90°．∴∠EBC＋∠GBD＝90°，即∠EBG＝90°．································13分∴在Rt△EBG中，BE2＋BG2＝EG2．∴BE2＋CF2＝EF2．·················································14分