勾股定理典型例题【含答案】免费.pdf

relationship,established equivalent relationship 14,and subject:application problem(4)-scores and percentage application problem review content overview answers scores,and percentage application problem of key is:according to meaning,(1)determine standard volume(units 1)(2)find associate volume rate corresponds to relationship,Then in-line solution.Category fraction multiplication word problem score Division applications engineering problem problem XV,a subject:review of the measurement of the amount of capacity,measurement and units of measurement of common units of measurement and their significance in rate 1,currency,length,area,volume,unit size,volume,weight and rate.(Omitted)2,commonly used time units and their relationships.(Slightly)with a measurement units Zhijian of of poly 1,and of method 2,and poly method 3,and of method and poly method of relationship measurement distance of method 1,and tool measurement 2,and estimates 16,and subject:geometry preliminary knowledge(1)-line and angle review content line,and segment,and Ray,and vertical,and parallel,and angle angle of classification(slightly)17,and subject:geometry preliminary knowledge(2)-plane graphics review content triangle,and edges shaped,and round,and fan axisymmetric graphics perimeter and area combination graphics of area subject:Preliminary knowledge(3)-review of solid content category 1-d shapes are divided into:cylinder and cone 2,column is divided into:cuboid,square 3,cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly solid surface area and volume 1,size 2,table.和和勾股定理复习一、知识要点：一、知识要点：1、勾股定理、勾股定理勾股定理：直角三角形两直角边的平方和等于斜边的平方。也就是说：如果直角三角形的两直角边为 a、b，斜边为 c，那么 a2+b2=c2。公式的变形：a2=c2-b2，b2=c2-a2。勾股定理在西方叫毕达哥拉斯定理，也叫百牛定理。它是直角三角形的一条重要性质，揭示的是三边之间的数量关系。它的主要作用是已知直角三角形的两边求第三边。勾股定理是一个基本的几何定理，它是用代数思想解决几何问题的最重要的工具之一，是数形结合的纽带之一。2、勾股定理的逆定理、勾股定理的逆定理如果三角形 ABC 的三边长分别是 a，b，c，且满足 a2+b2=c2，那么三角形 ABC 是直角三角形。这个定理叫做勾股定理的逆定理.该定理在应用时，同学们要注意处理好如下几个要点：已知的条件：某三角形的三条边的长度.满足的条件：最大边的平方=最小边的平方+中间边的平方.得到的结论：这个三角形是直角三角形，并且最大边的对角是直角.如果不满足条件，就说明这个三角形不是直角三角形。3、勾股数、勾股数满足 a2+b2=c2的三个正整数，称为勾股数。注意：勾股数必须是正整数，不能是分数或小数。一组勾股数扩大相同的正整数倍后，仍是勾股数。4、最短距离最短距离问题问题：主要运用的依据是两点之两点之间线间线段最短。段最短。二、二、知识结构：知识结构：relationship,established equivalent relationship 14,and subject:application problem(4)-scores and percentage application problem review content overview answers scores,and percentage application problem of key is:according to meaning,(1)determine standard volume(units 1)(2)find associate volume rate corresponds to relationship,Then in-line solution.Category fraction multiplication word problem score Division applications engineering problem problem XV,a subject:review of the measurement of the amount of capacity,measurement and units of measurement of common units of measurement and their significance in rate 1,currency,length,area,volume,unit size,volume,weight and rate.(Omitted)2,commonly used time units and their relationships.(Slightly)with a measurement units Zhijian of of poly 1,and of method 2,and poly method 3,and of method and poly method of relationship measurement distance of method 1,and tool measurement 2,and estimates 16,and subject:geometry preliminary knowledge(1)-line and angle review content line,and segment,and Ray,and vertical,and parallel,and angle angle of classification(slightly)17,and subject:geometry preliminary knowledge(2)-plane graphics review content triangle,and edges shaped,and round,and fan axisymmetric graphics perimeter and area combination graphics of area subject:Preliminary knowledge(3)-review of solid content category 1-d shapes are divided into:cylinder and cone 2,column is divided into:cuboid,square 3,cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly solid surface area and volume 1,size 2,table.和和2 直角三角形勾股定理应用判定直角三角形的一种方法三、考点剖析三、考点剖析考点一：利用勾股定理求面考点一：利用勾股定理求面积积求：（1）阴影部分是正方形；（2）阴影部分是长方形；（3）阴影部分是半圆2.如图，以 RtABC 的三边为直径分别向外作三个半圆，试探索三个半圆的面积之间的关系考点二：在直角三角形中，已知两考点二：在直角三角形中，已知两边边求第三求第三边边例如图 2，已知ABC 中，AB17，AC10，BC 边上的高，AD8，则边 BC 的长为（）A21 B15 C6 D以上答案都不对【强化训练】：1在直角三角形中,若两直角边的长分别为 1cm，2cm，则斜边长为 2（易错题、注意分类的思想）已知直角三角形的两边长为 3、2，则另一条边长的平方是 relationship,established equivalent relationship 14,and subject:application problem(4)-scores and percentage application problem review content overview answers scores,and percentage application problem of key is:according to meaning,(1)determine standard volume(units 1)(2)find associate volume rate corresponds to relationship,Then in-line solution.Category fraction multiplication word problem score Division applications engineering problem problem XV,a subject:review of the measurement of the amount of capacity,measurement and units of measurement of common units of measurement and their significance in rate 1,currency,length,area,volume,unit size,volume,weight and rate.(Omitted)2,commonly used time units and their relationships.(Slightly)with a measurement units Zhijian of of poly 1,and of method 2,and poly method 3,and of method and poly method of relationship measurement distance of method 1,and tool measurement 2,and estimates 16,and subject:geometry preliminary knowledge(1)-line and angle review content line,and segment,and Ray,and vertical,and parallel,and angle angle of classification(slightly)17,and subject:geometry preliminary knowledge(2)-plane graphics review content triangle,and edges shaped,and round,and fan axisymmetric graphics perimeter and area combination graphics of area subject:Preliminary knowledge(3)-review of solid content category 1-d shapes are divided into:cylinder and cone 2,column is divided into:cuboid,square 3,cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly solid surface area and volume 1,size 2,table.和和33、已知直角三角形两直角边长分别为5和12，求斜边上的高（结论：直角三角形的两条直角边的积等于斜边与其高的积，ab=ch）考点三：考点三：应应用勾股定理在等腰三角形中求底用勾股定理在等腰三角形中求底边边上的高上的高例、如图 1 所示，等腰中，是底边上的高，若，求 AD 的长；ABC 的面积 考点四考点四:应应用勾股定理解决楼梯上用勾股定理解决楼梯上铺铺地毯地毯问题问题例、某楼梯的侧面视图如图 3 所示，其中米，因某种活动要求铺设红色地毯，则在 AB 段楼梯所铺地毯的长度应为 分析：如何利用所学知识，把折线问题转化成直线问题，是问题解决的关键。仔细观察图形，不难发现，所有台阶的高度之和恰好是直角三角形 ABC 的直角边 BC 的长度，所有台阶的宽度之和恰好是直角三角形 ABC 的直角边 AC 的长度，只需利用勾股定理，求得这两条线段的长即可。考点五、利用列方程求考点五、利用列方程求线线段的段的长长（方程思想）（方程思想）、小强想知道学校旗杆的高，他发现旗杆顶端的绳子垂到地面还多 1 米，当他把绳子的下端拉开 5 米后，发现下端刚好接触地面，你能帮他算出来吗？relationship,established equivalent relationship 14,and subject:application problem(4)-scores and percentage application problem review content overview answers scores,and percentage application problem of key is:according to meaning,(1)determine standard volume(units 1)(2)find associate volume rate corresponds to relationship,Then in-line solution.Category fraction multiplication word problem score Division applications engineering problem problem XV,a subject:review of the measurement of the amount of capacity,measurement and units of measurement of common units of measurement and their significance in rate 1,currency,length,area,volume,unit size,volume,weight and rate.(Omitted)2,commonly used time units and their relationships.(Slightly)with a measurement units Zhijian of of poly 1,and of method 2,and poly method 3,and of method and poly method of relationship measurement distance of method 1,and tool measurement 2,and estimates 16,and subject:geometry preliminary knowledge(1)-line and angle review content line,and segment,and Ray,and vertical,and parallel,and angle angle of classification(slightly)17,and subject:geometry preliminary knowledge(2)-plane graphics review content triangle,and edges shaped,and round,and fan axisymmetric graphics perimeter and area combination graphics of area subject:Preliminary knowledge(3)-review of solid content category 1-d shapes are divided into:cylinder and cone 2,column is divided into:cuboid,square 3,cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly solid surface area and volume 1,size 2,table.和和4ABC【强化训练】：折叠矩形 ABCD 的一边 AD,点 D 落在 BC 边上的点 F 处,已知AB=8CM,BC=10CM,求 CF 和 EC。.ABCEFD考点六：考点六：应应用勾股定理解决勾股用勾股定理解决勾股树问题树问题例、如右图所示的图形中，所有的四边形都是正方形，所有的三角形都是直角三角形，其中最大的正方形的边长为 5，则正方形 A，B，C，D 的面积的和为 分析：勾股树问题中，处理好两个方面的问题，一个是正方形的边长与面积的关系，另一个是正方形的面积与直角三角形直角边与斜边的关系。点评：请同学们自己把其内在的一般变化规律总结一下。考点七：考点七：应应用勾股定理解决数学用勾股定理解决数学风车问题风车问题relationship,established equivalent relationship 14,and subject:application problem(4)-scores and percentage application problem review content overview answers scores,and percentage application problem of key is:according to meaning,(1)determine standard volume(units 1)(2)find associate volume rate corresponds to relationship,Then in-line solution.Category fraction multiplication word problem score Division applications engineering problem problem XV,a subject:review of the measurement of the amount of capacity,measurement and units of measurement of common units of measurement and their significance in rate 1,currency,length,area,volume,unit size,volume,weight and rate.(Omitted)2,commonly used time units and their relationships.(Slightly)with a measurement units Zhijian of of poly 1,and of method 2,and poly method 3,and of method and poly method of relationship measurement distance of method 1,and tool measurement 2,and estimates 16,and subject:geometry preliminary knowledge(1)-line and angle review content line,and segment,and Ray,and vertical,and parallel,and angle angle of classification(slightly)17,and subject:geometry preliminary knowledge(2)-plane graphics review content triangle,and edges shaped,and round,and fan axisymmetric graphics perimeter and area combination graphics of area subject:Preliminary knowledge(3)-review of solid content category 1-d shapes are divided into:cylinder and cone 2,column is divided into:cuboid,square 3,cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly solid surface area and volume 1,size 2,table.和和5例 7、(09 年安顺)图甲是我国古代著名的“赵爽弦图”的示意图，它是由四个全等的直角三角形围成的。在 RtABC 中，若直角边 AC6，BC5，将四个直角三角形中边长为 6 的直角边分别向外延长一倍，得到图乙所示的“数学风车”，则这个风车的外围周长（图乙中的实线）是_。分析：因为，直角边 AC6，BC5，当将四个直角三角形中边长为 6 的直角边分别向外延长一倍后，得到四个直角边分别是 12 和 5 的直角三角形，所求的最长实边恰好是这些直角三角形的斜边长，因此，斜边长为：=13，较短的实边长是 6，所以，这个风车的外围周长为：413+46=76。解：这个风车的外围周长为 76。考点八：判考点八：判别别一个三角形是否是直角三角形一个三角形是否是直角三角形例 1：分别以下列四组数为一个三角形的边长：（1）3、4、5（2）5、12、13（3）8、15、17（4）4、5、6，其中能够成直角三角形的有 【强化训练】：已知ABC 中，三条边长分别为 an，b2n，cn（n1）试判断22该三角形是否是直角三角形，若是，请指出哪一条边所对的角是直角考点九：其他考点九：其他图图形与直角三角形形与直角三角形例：如图是一块地，已知 AD=8m，CD=6m，D=90，AB=26m，BC=24m，求这块地的面积。relationship,established equivalent relationship 14,and subject:application problem(4)-scores and percentage application problem review content overview answers scores,and percentage application problem of key is:according to meaning,(1)determine standard volume(units 1)(2)find associate volume rate corresponds to relationship,Then in-line solution.Category fraction multiplication word problem score Division applications engineering problem problem XV,a subject:review of the measurement of the amount of capacity,measurement and units of measurement of common units of measurement and their significance in rate 1,currency,length,area,volume,unit size,volume,weight and rate.(Omitted)2,commonly used time units and their relationships.(Slightly)with a measurement units Zhijian of of poly 1,and of method 2,and poly method 3,and of method and poly method of relationship measurement distance of method 1,and tool measurement 2,and estimates 16,and subject:geometry preliminary knowledge(1)-line and angle review content line,and segment,and Ray,and vertical,and parallel,and angle angle of classification(slightly)17,and subject:geometry preliminary knowledge(2)-plane graphics review content triangle,and edges shaped,and round,and fan axisymmetric graphics perimeter and area combination graphics of area subject:Preliminary knowledge(3)-review of solid content category 1-d shapes are divided into:cylinder and cone 2,column is divided into:cuboid,square 3,cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly solid surface area and volume 1,size 2,table.和和6考点十：构造直角三角形解决考点十：构造直角三角形解决实际问题实际问题在某一平地上，有一棵树高 8 米的大树，一棵树高 2 米的小树，两树之间相距 8 米。今一只小鸟在其中一棵树的树梢上，要飞到另一棵树的树梢上，问它飞行的最短距离是多少？（画出草图然后解答）考点十一：与展开考点十一：与展开图图有关的有关的计计算算例、如图，在棱长为 1 的正方体 ABCDABCD的表面上，求从顶点 A 到顶点 C的最短距离【强化训练】：如图一个圆柱，底圆周长 6cm，高 4cm，一只蚂蚁沿外壁爬行，要从 A 点爬到 B点，则最少要爬行 cm四、课时作业优化设计四、课时作业优化设计 1设直角三角形的三条边长为连续自然数，则这个直角三角形的面积是_2直角三角形的两直角边分别为 5cm，12cm，其中斜边上的高为（）A6cm B8.5cm Ccm Dcm30136013 【提升提升“学力学力”】ABrelationship,established equivalent relationship 14,and subject:application problem(4)-scores and percentage application problem review content overview answers scores,and percentage application problem of key is:according to meaning,(1)determine standard volume(units 1)(2)find associate volume rate corresponds to relationship,Then in-line solution.Category fraction multiplication word problem score Division applications engineering problem problem XV,a subject:review of the measurement of the amount of capacity,measurement and units of measurement of common units of measurement and their significance in rate 1,currency,length,area,volume,unit size,volume,weight and rate.(Omitted)2,commonly used time units and their relationships.(Slightly)with a measurement units Zhijian of of poly 1,and of method 2,and poly method 3,and of method and poly method of relationship measurement distance of method 1,and tool measurement 2,and estimates 16,and subject:geometry preliminary knowledge(1)-line and angle review content line,and segment,and Ray,and vertical,and parallel,and angle angle of classification(slightly)17,and subject:geometry preliminary knowledge(2)-plane graphics review content triangle,and edges shaped,and round,and fan axisymmetric graphics perimeter and area combination graphics of area subject:Preliminary knowledge(3)-review of solid content category 1-d shapes are divided into:cylinder and cone 2,column is divided into:cuboid,square 3,cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly solid surface area and volume 1,size 2,table.和和73如图，ABC 的三边分别为 AC=5，BC=12，AB=13，将ABC 沿 AD 折叠，使 AC落在 AB 上，求 DC 的长4如图，一只鸭子要从边长分别为 16m 和 6m 的长方形水池一角 M游到水池另一边中点 N，那么这只鸭子游的最短路程应为多少米？【聚焦聚焦“中考中考”】5如图，铁路上 A、B 两点相距 25km，C、D 为两村庄，DA垂直 AB 于 A，CB 垂直 AB于 B，已知 AD=15km，BC=10km，现在要在铁路 AB 上建一个土特产品收购站 E，使得 C、D两村到 E 站的距离相等，则 E 站建在距 A 站多少千米处？