七升八暑假预习自学辅助作业(范围:14.2 三角形全等的判定03)2026-2027学年人教版八年级数学上册

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2026年秋七升八暑假预习自学辅助作业

(14.2 三角形全等的判定03)

 

知识点1 倍长中线模型

1.如图,eqIdb9f79d7939c88e9702962e5917cad290,eqId8c4eccda4edcee29a5f15609e85106df,eqIde1ffb98f1e3c1317c0db403d3af04bdc分别是eqId15c0dbe3c080c4c4636c64803e5c1f76的边eqId764509115979e9958101808383672ec0,eqIde1a9c6a736e6eac98a676fa3232db5a5,eqIdb79dd200766db27fb90d6bd1992cf658上的中线,三条中线相交于点eqId1dde8112e8eb968fd042418dd632759e,下列说法错误的是(   )

@@@9cdf98e9-7560-4aa6-a6f8-02a1fc6d65ba

A.点eqId1dde8112e8eb968fd042418dd632759e是eqId15c0dbe3c080c4c4636c64803e5c1f76的重心 B.eqId372629a8666de1e9bac3e7daadcac7b6和eqIdd01b9de38c02e28c475bf639727b59e6的面积相等

C.eqIdc1d91a5ee0e67b1719db31fe001b9a5a D.eqIdd2163ddb29f0b2f9a3f6171376982ceb

2.如图,在eqId15c0dbe3c080c4c4636c64803e5c1f76中,eqIddb0d67544e300cae6311f1401f5d0b29,eqId03902478df1a55bc99703210bccab910是eqId0dc5c9827dfd0be5a9c85962d6ccbfb1边上的中线,则eqId03902478df1a55bc99703210bccab910的取值范围是(   )

@@@c479a142-aefd-4c12-9897-a5b6548e7579

A.eqIda3aba7a32868ec1c3b504f00a660d935 B.eqId721d975f416f854c1bafbb7f9ae31f60 C.eqId2da5d969021ae69303cd01c91065eb3f D.无法确定

3.如图,eqId03902478df1a55bc99703210bccab910是eqId15c0dbe3c080c4c4636c64803e5c1f76中eqId0dc5c9827dfd0be5a9c85962d6ccbfb1边上的中线,若eqId592d7fe6e1e7e8f053654a0d8c1b5155,则eqId03902478df1a55bc99703210bccab910的取值范围为_____.

@@@b6bd5a69-df80-468b-bff9-242bf374c347

4.课外兴趣小组活动时,老师提出了如下问题:如图,在eqId15c0dbe3c080c4c4636c64803e5c1f76中,eqId54275b7e571660d0a9e0370fbfe5050b,eqIdf1682d306c38087d9e6f7efb9cec596a,求eqId0dc5c9827dfd0be5a9c85962d6ccbfb1边上的中线eqId03902478df1a55bc99703210bccab910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eqId03902478df1a55bc99703210bccab910至点eqId2a30f3a8b673cc28bd90c50cf1a35281,使eqId0ad98ad714864041a632ca949308e417,连接eqId4eedae8d316c76e3d0b451256de03fb9.依据小明的方法,可求得eqId03902478df1a55bc99703210bccab910的取值范围是____________.page number 0

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@@@dfea7d4b-e706-450d-8f96-ba4b0d4a8b5f

5.如图,在eqId15c0dbe3c080c4c4636c64803e5c1f76中,eqId03902478df1a55bc99703210bccab910是eqId0dc5c9827dfd0be5a9c85962d6ccbfb1边上的中线,eqIda8a7b5adfcac0f46a4cd19da4ebb4a2b,eqIdecf7679c8b4b1e442ce4286d4b0e9c32,eqIdf52a58fbaf4fea03567e88a9f0f6e37e的长度为偶数,则eqIdf52a58fbaf4fea03567e88a9f0f6e37e的所有可能值为___________.

@@@ba8cb13b-4277-4f58-be45-40496fadf018

6.如图已知eqId15c0dbe3c080c4c4636c64803e5c1f76中,eqId03902478df1a55bc99703210bccab910为eqId0dc5c9827dfd0be5a9c85962d6ccbfb1边上的中线,eqId68a83fdd2ba72a2dba0b6b10bb3e06b9平分eqId5157b42da58d55daad27d98b2fec15ff交eqId0dc5c9827dfd0be5a9c85962d6ccbfb1边于点eqId2a30f3a8b673cc28bd90c50cf1a35281,eqId222cd46b96ef8ccf47c397abffdda18f,eqIdecf7679c8b4b1e442ce4286d4b0e9c32,则eqIdae248960d8c1677cf948f8251275e863_____.

@@@91ad6299-acce-488e-a30c-cdc46d4cf59a

7.已知eqId15c0dbe3c080c4c4636c64803e5c1f76的eqIdb79dd200766db27fb90d6bd1992cf658边长为4,eqIde1a9c6a736e6eac98a676fa3232db5a5边长为8,则eqId764509115979e9958101808383672ec0边上的中线eqId88929f4ba0851730d5f941d426b87548的长度的取值范围(   )

A.eqId721d975f416f854c1bafbb7f9ae31f60 B.eqId1436741b99b9e88229fee53557d639fd C.eqId05d6f650ef60dcab4b27597e2963a4fa D.eqId87f80ab9dec197557826666d93ce56c4

8.如图,已知eqIdb9f79d7939c88e9702962e5917cad290为eqIde428e7a09732be85c1224e9c8f6a71c5的中线,eqId6ea6fde14ffe93582739d371bcfaf955,eqId6a8b0d7e169313c2a2d5c278cf968b9a的周长为eqIdfbe9f7abf7bcf4e1aa2579cd191d7761,则

@@@ae0ea397-1f6e-4be5-afca-b40b70c3d675

(1)eqId5fa485cf3776f36aaf4abaadaf30fb85的周长为多少?

(2)eqIdb9f79d7939c88e9702962e5917cad290的取值范围是多少?(直接写出答案)

9.【学习问题】课外兴趣小组活动时,老师提出了如下问题:

如图eqIdbdaa19de263700a15fcf213d64a8cd57,eqIde428e7a09732be85c1224e9c8f6a71c5中,若eqIde62676bcf5f9783c0457fce09fe561a8,eqId6ddeb9efb488b31ae67efd29efd56b3f,求eqId764509115979e9958101808383672ec0边上的中线eqId88929f4ba0851730d5f941d426b87548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:如图eqId61128ab996360a038e6e64d82fcba004,延长eqId88929f4ba0851730d5f941d426b87548至点eqId2a30f3a8b673cc28bd90c50cf1a35281,使eqIde491964b08c1ab19a71bc1b04dfe2b3e,连接eqIdd0213c5787a5a6b38d11bceca5567f67.请根据小明的方法思考并解答:page number 1

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@@@d7b9d639-e696-4f72-a8ec-c07d2b434aa0

(eqIdbdaa19de263700a15fcf213d64a8cd57)①由已知和作图能得到eqId9a67eb0a2dc3beba12211fdc6a16ff39,依据是 ;

eqIde4760d3a8ea9a5e8ebec21e7befaecd8  eqId44c69fa1b3e368014ebeec70e5e44c29    eqId207d5f241f897e34cd5fa18aab9ee284    eqIde17b028e454a732ac6e7acdf0a4f8f88

②由“三角形的三边关系”可求得eqId88929f4ba0851730d5f941d426b87548的取值范围是 ;

【学习反思】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、平行线、平移线段,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.

【类比运用】

(eqId61128ab996360a038e6e64d82fcba004)如图eqId5ca7d1107389675d32b56ec097464c14,eqId2a30f3a8b673cc28bd90c50cf1a35281是eqId764509115979e9958101808383672ec0中点,点eqId5963abe8f421bd99a2aaa94831a951e9在eqId1a6c6e7c025362c46a64a8956761f08e上,且eqId428332abfb633be91245061f94619b3e,求证:eqId7159d113d1301b49171e736384f00131;

【拓展运用】

(eqId5ca7d1107389675d32b56ec097464c14)如图eqIdb8860d9787671b53b1ab68b3d526f5ca,已知直线eqId14202336f061d96de67ddda55479ec49,点eqId5963abe8f421bd99a2aaa94831a951e9、eqId8455657dde27aabe6adb7b188e031c11是直线eqId294f5ba74cdf695fc9a8a8e52f421328上两点,点eqId7f9e8449aad35c5d840a3395ea86df6d、eqIdc5db41a1f31d6baee7c69990811edb9f是直线eqIdb6a24198bd04c29321ae5dc5a28fe421上两点,点eqIddad2a36927223bd70f426ba06aea4b45是线段eqIde1ffb98f1e3c1317c0db403d3af04bdc中点,且eqId4b68a2a20111bce06eebdb6a6fbb4b6a,两平行线eqId294f5ba74cdf695fc9a8a8e52f421328、eqIdb6a24198bd04c29321ae5dc5a28fe421间的距离为eqIdb8860d9787671b53b1ab68b3d526f5ca.求证:eqIdaddd924eb8d4a7eec717012aed9454d8.

 

知识点2 旋转模型

10.如图,在△ABC中,∠CAB=62°,将△ABC绕点A旋转到△AB'C'的位置,使得CC'∥AB,则∠BAB'的大小为(  )

@@@c21af965-5066-4165-90d8-39b471913d66

A.64° B.52° C.62° D.56°

11.如图,将eqIdcfa8cee7d2463f6f7d352e8b65f47cf3的斜边eqId0dc5c9827dfd0be5a9c85962d6ccbfb1绕点eqId7f9e8449aad35c5d840a3395ea86df6d顺时针旋转eqIdc02b54dc6b3e1bb6544f47d4c8743fcf得到线段eqIdd40b319212a7e7528b053e1c7097e966,过点eqId8455657dde27aabe6adb7b188e031c11作eqIdf52a58fbaf4fea03567e88a9f0f6e37e的垂线,交eqIdf52a58fbaf4fea03567e88a9f0f6e37e延长线于点eqId2a30f3a8b673cc28bd90c50cf1a35281.求证:eqId0a510a3eb906b3ef0760b0d1723d11ae.

@@@c32b8616-354c-4606-995b-b0862ee792e5

12.综合与实践

数学模型可以用来解决一类问题,是数学应用的基本途径.通过探究图形的变化规律,再结page number 2

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