5.2.1勾股定理 课件 -2026-2027学年湘教版数学八年级上册

#湘教版数学八年级上册5.2.1勾股定理专项练习##知识点回顾**勾股定理**：直角三角形两直角边的平方和等于斜边的平方。在$Rt\triangleABC$中\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)$\angleC=90^\circ$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)直角边$a\(a、b\)b$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)斜边$c$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)则：$$\boldsymbol{a^2+b^2=c^2}$$>$a\(a、b\)b$：直角边；$c$：斜边（直角所对的边\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)最长边）✅解题要点1.**前提：必须是直角三角形**\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)普通三角形不能用勾股定理；2.分清斜边：斜边是$90^\circ$角对的边\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)是最长边；3.变形公式：$a^2=c^2-b^2\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)\quadb^2=c^2-a^2$4.常见勾股数（熟记）：$\(3,4,5\)$；$\(6,8,\(\boldsymbol{10}\)\)$；$\(5,12,13\)$；$\(7,24,25\)$满分：\(\boldsymbol{10}\)0分时间：40分钟##一\(a、b\)选择题（每题4分\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)共20分）1.在$Rt\triangleABC$中\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)$\angleC=90^\circ$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)三边$\(a,b,c\)$（$c$为斜边）\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)勾股定理公式是（）A.$\(a+b=c\)$B.$a^2+b^2=c^2$C.$a^2-b^2=c^2$D.$a^2+b^2=2c^2$2.$Rt\triangle$两直角边为$3$和$4$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)则斜边长为（）A.$5$B.$7$C.$25$D.$12$3.$Rt\triangleABC$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)$\angleC=90^\circ$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)斜边$\(\(c=\)13\)$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)一条直角边$\(a=5\)$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)另一条直角边$b$为（）A.$8$B.$12$C.$18$D.$6$4.下列是勾股数的是（）A.$\(2,3,4\)$B.$\(5,12,13\)$C.$\(4,5,6\)$D.$\(1,2,3\)$5.说法正确的是（）A.任意三角形都满足$a^2+b^2=c^2$B.直角三角形中\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)斜边的平方等于两直角边平方和C.直角三角形三边满足$\(a+b=c\)$D.在三角形中\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)长边等于两短边之和##二\(a、b\)填空题（每题4分\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)共20分）1.勾股定理：直角三角形两直角边的平方和等于________的平方。2.$Rt\triangleABC$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)$\angleC=90^\circ$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)$\(a=6\)\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)b=8$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)则斜边$\(c=\)$________。3.在$Rt\triangle$中\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)已知斜边$c$和直角边$a$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)则$b^2=$________。4.勾股数$\(3,4,5\)$同时扩大2倍得到________。5.直角三角形中\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)________是最长边。##三\(a、b\)基础解答题（每题8分\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)共32分）1.已知：$Rt\triangleABC$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)$\angleC=90^\circ$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)$\(a=5\)\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)b=12$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)求斜边$c$。2.已知：$Rt\triangleABC$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)$\angleC=90^\circ$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)斜边$\(c=\)\(\boldsymbol{10}\)$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)直角边$\(a=6\)$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)求直角边$b$。3.判断对错\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)并说明理由（1）在$\triangleABC$中\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)$a^2+b^2=c^2$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)所以$\triangleABC$是直角三角形。（2）直角三角形两条直角边的和等于斜边。4.$Rt\triangleABC$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)$\angleC=90^\circ$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)$a=9\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)b=12$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)求斜边$c$。##四\(a、b\)拓展提升题（每题14分\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)共28分）1.在$Rt\triangleABC$中\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)$\angleC=90^\circ$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)若$\(a:b=3:4\)$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)斜边$\(c=\)20$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)求直角边$a\(a、b\)b$的长。2.已知：在$Rt\triangleABC$中\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)$\angleC=90^\circ$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)$\angleA=30^\circ$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)$\(BC=4\)$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)求$AB$和$AC$的长。---#参考答案与解析##一\(a、b\)选择题1.B2.A3.B4.B5.B##二\(a、b\)填空题1.**斜边**2.$\boldsymbol{\(\boldsymbol{10}\)}$3.$\boldsymbol{c^2-a^2}$4.$\boldsymbol{\(6,8,\(\boldsymbol{10}\)\)}$5.**斜边**##三\(a、b\)基础解答题1.解：$\because\angleC=90^\circ$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)由勾股定理$c^2=a^2+b^2=5^2+12^2=25+144=169$$\therefore\(c=\)\sqrt{169}=\boldsymbol{13}$2.解：$\because\angleC=90^\circ$\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)由勾股定理$b^2=c^2-a^2$$b^2=\(\boldsymbol{10}\)^2-6^2=\(\boldsymbol{10}\)0-36=64$$\thereforeb=\sqrt{64}=\boldsymbol{8}$3.（1）√；这是勾股定理的逆定理（下一节学）\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)满足两边平方和等于第三边平方\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)三角形为直角三角形。（2）×；直角三角形是两直角边**平方和**等于斜边平方\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)不是边长相加。4.解：由勾股定理：$c^2=a^2+b^2=9^2+12^2=81+144=225$$\(c=\)\sqrt{225}=\boldsymbol{15}$##四\(a、b\)拓展提升题1.解：设$a=3k\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)b=4k(k>0)$由勾股定理：$(3k)^2+(4k)^2=20^2$$9k^2+16k^2=400$$25k^2=400\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)k^2=16\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)k=4$$\thereforea=12\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)b=16$2.解：$\because\angleC=90^\circ\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)\angleA=30^\circ\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)\(BC=4\)$$\thereforeAB=2BC=8$（30°直角三角形性质）由勾股定理：$AC^2=AB^2-BC^2=8^2-4^2=64-16=48$$AC=\sqrt{48}=4\sqrt{3}$$\thereforeAB=\boldsymbol{8}\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)AC=\boldsymbol{4\sqrt{3}}$##易错警示1.勾股定理**只适用于直角三角形**；2.公式是**平方相加**\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)不是边长直接相加；3.分清斜边\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)斜边一定是公式里单独在等号一侧的$c$；4.常和30°直角三角形性质\(a、b\)斜边上中线综合出题；5.求边长最后要开算术平方根\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)边长为正数。下一节：**5.2.2勾股定理的逆定理**\(a^2=c^2-b^2\(\(a=6\)\(a=5\(a=9\(a=3k\(\thereforea=12\(\thereforeAB=\boldsymbol{8}，AC=\boldsymbol{4\sqrt{3}}\)b=16\)b=4k(k>0)\)b=12\)b=12\)b=8\)\quadb^2=c^2-a^2\)继续同模板出题吗？湘教版（新教材）数学八年级上册5.2.1勾股定理授课教师：.

班

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8年级（

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2026/10/10

学子如同向日葵向往暖阳，满心热爱渴求知识。在书页间汲取力量，于思索中追逐光亮，不惧学习路上的磨砺，心怀憧憬奋力向上，向着理想不断成长。第5章直角三角形123通过探究，掌握勾股定理（重点）会用勾股定理求线段的长度。（重点）知道勾股数，并熟记常见的勾股数。如图，在方格纸上(设小方格的边长为1)画一个顶点都在格点上的Rt△ABC，使其两直角边分别为3，4，将斜边AB绕点A旋转，使其处于水平位置，你发现这条斜边的长度是多少? 新知探究观

察b=4ACc=?Ba=3斜边的长度是5斜边5，与直角边3，4有什么数量关系呢？古代有“勾三股四弦五”的说法。03新知探究我国古代数学名著《周髀算经》，把直角三角形较短的直角边叫作勾，较长的直角边叫作股，斜边叫作弦，并提出“勾三股四弦五”.这本著作还指出：勾与股的平方和等于弦的平方.对于这一结论，古人称为勾股算法.实际上，这一结论揭示的是直角三角形三边长的平方关系.

03新知探究探究

03新知探究

思考：斜边都为𝒄