22.3.5相似三角形的应用1 课件 -2026-2027学年华东师大版数学九年级上册

#华东师大版九年级上册22.3.5相似三角形的应用1（测高\(B、C\)测距）##一\(B、C\)知识点梳理核心思路：**利用平行光线/影子\(B、C\)视线构造相似三角形\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)列比例式求解**两大基础模型：###模型1：阳光下的影子（平行光）同一时刻\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)太阳光线平行。物体高度之比=影子长度之比$$\frac{h_1}{h_2}=\frac{l_1}{l_2}$$原理：垂直地面→两个直角\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)太阳光线平行得到一组同位角相等\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)**AA相似**。>注意：必须**同一时刻**！不同时刻太阳角度不同\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)不能直接比。###模型2：平面镜反射（光的反射）入射角=反射角\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)得到两组角相等\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)AA证相似。$$\frac{\text{物高}}{\text{人高}}=\frac{\text{物体到镜子距离}}{\text{人到镜子距离}}$$###解题步骤（固定模板\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)考试直接套用）1.找两个直角三角形；2.证明相似（AA）；3.写出对应边比例式；4.代入数值解方程；5.写答。⚠易错点1.对应边不要写反；2.单位统一（米\(B、C\)厘米要换成一样）；3.影子题\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)看清是**全部影子**还是**影子一部分落在墙上**（难题会考）。##二\(B、C\)选择题1.同一时刻\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)高2m竹竿影长1.5m\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)一棵树影长\(\boldsymbol{6}\)m\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)则树高（）A.\(\boldsymbol{6}\)mB.8mC.9mD.10m2.利用镜子反射测树高\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)人眼高1.\(\boldsymbol{6}\)m\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)人离镜子2m\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)树离镜子8m\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)则树高（）A.\(\boldsymbol{6}\).4mB.8mC.9.\(\boldsymbol{6}\)mD.12.8m3.同一时刻\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)甲\(B、C\)乙两物体高度比2:3\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)则它们影子长之比为（）A.2:3B.4:9C.3:2D.9:44.小明身高1.7米\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)影子长2米\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)同一时刻大楼影子长20米\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)则大楼高（）A.17mB.20mC.34mD.40m5.下面不能用相似三角形测量的是（）A.旗杆高度B.河宽C.地图距离D.无法直接到达的大树高度##三\(B、C\)填空题1.同一时刻物体高度与影长成______比例。2.镜子反射测高\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)依据______相等\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)证明三角形相似。3.竹竿高1.8m\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)影长0.9m；同一时刻楼房影长12m\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)楼房高____m。4.树高$h$\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)树影长9m；1.5m标杆影长3m\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)则树高=____m。5.平面镜模型：人高1.5m\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)人到镜3m\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)树到镜12m\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)树高____m。##四\(B、C\)解答题1.阳光下\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)一根长3m的竹竿\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)影子长2m。同一时刻\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)测得旗杆影子长10m\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)求旗杆高度。2.用平面镜测大树高度：镜子放在地面\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)人站在离镜子3m处\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)眼睛离地面1.\(\boldsymbol{6}\)m；大树距离镜子9m\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)刚好在镜子中看到树顶\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)求大树高度。3.小明身高1.\(\boldsymbol{6}\)m\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)站立时影长2.4m；同一时刻\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)一座古塔影长3\(\boldsymbol{6}\)m\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)求古塔高度。4.拓展：为测河宽$AB$\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)在对岸选点$A$\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)这边取$B\(B、C\)C$\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)$BC\perpAB$；取点$D$\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)$CD\perpBC$\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)$AD$交$BC$于$E$。已知$BE=12m\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)EC=\(\boldsymbol{6}\)m\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)CD=8m$\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)求河宽$AB$。##五\(B、C\)参考答案与解析###选择题1.B解析：$\dfrac{2}{1.5}=\dfrac{h}{\(\boldsymbol{6}\)}\Rightarrowh=8$2.A解析：$\dfrac{h}{1.\(\boldsymbol{6}\)}=\dfrac{8}{2}\Rightarrowh=\(\boldsymbol{6}\).4$3.A解析：同一时刻\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)高之比=影长之比4.A解析：$\dfrac{1.7}{2}=\dfrac{H}{20}\RightarrowH=17$5.C###填空题1.正2.入射角与反射角3.$\boldsymbol{\(\boldsymbol{24}\)}$解析：$\dfrac{1.8}{0.9}=2$\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)$12\times2=\(\boldsymbol{24}\)$4.$\boldsymbol{\(\boldsymbol{4.5}\)}$解析：$\dfrac{h}{9}=\dfrac{1.5}{3}\Rightarrowh=\(\boldsymbol{4.5}\)$5.$\boldsymbol{\(\boldsymbol{6}\)}$解析：$\dfrac{h}{1.5}=\dfrac{12}{3}\Rightarrowh=\(\boldsymbol{6}\)$###解答题1.解：设旗杆高$x$m同一时刻\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)$\dfrac{\text{竹竿高}}{\text{竹竿影长}}=\dfrac{\text{旗杆高}}{\text{旗杆影长}}$$\dfrac{3}{2}=\dfrac{x}{10}$$\(2x=30\)$\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)$\(x=15\)$答：旗杆高15米。2.解：由反射原理\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)$\angleAEB=\angleCED$\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)$\angleB=\angleD=90^\circ$$\therefore\triangleABE\backsim\triangleCDE$（AA）$\dfrac{AB}{CD}=\dfrac{BE}{DE}$$\dfrac{AB}{1.\(\boldsymbol{6}\)}=\dfrac{9}{3}$$\(AB=4.8\)$答：大树高4.8米。3.解：设古塔高$x$m$\dfrac{1.\(\boldsymbol{6}\)}{2.4}=\dfrac{x}{3\(\boldsymbol{6}\)}$$2.4x=1.\(\boldsymbol{6}\)\times3\(\boldsymbol{6}\)$$x=\(\boldsymbol{24}\)$答：古塔高\(\boldsymbol{24}\)m。4.解：$\becauseAB\perpBC\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)CD\perpBC$$\thereforeAB\parallelCD\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)\angleB=\angleC=90^\circ$$\angleAEB=\angleDEC$（对顶角）$\therefore\triangleABE\backsim\triangleDCE$（AA）$\dfrac{AB}{CD}=\dfrac{BE}{EC}$$\dfrac{AB}{8}=\dfrac{12}{\(\boldsymbol{6}\)}$$\dfrac{AB}{8}=2$\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)$AB=1\(\boldsymbol{6}\)$答：河宽$AB=1\(\boldsymbol{6}\)\mathrm{m}$。---下一节：**相似三角形的应用2（盲区\(B、C\)遮挡类）**\(\becauseAB\perpBC\(\thereforeAB\parallelCD，\angleB=\angleC=90^\circ\)CD\perpBC\)继续吗？华东师大版（新教材）数学九年级上册22.3.5相似三角形的应用1授课教师：.

班

级：

9年级（

）班.

时

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2026/10/11

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22章图形的相似新知探究

利用相似三角形测量高度

新知探究

利用相似三角形测量高度

你还有其他方法吗？新知探究利用相似三角形测量高度AFEBO┐┐利用物理中平面镜反射的原理：OBEF=OAAF

OB=OA·EFAF平面镜利用相似三角形测量高度的一般步骤：1.找垂直：明确物体垂直地面，得到两组直角；太阳光线平行，得到一组相等锐角。2.证相似：依据两角分别相等，证明两个直角三角形相似。3.列比例：根据相似三角形对应边成比例列出等式。4.代值计算：代入已知长度，求解未知高度，写答。归纳总结归纳总结利用相似三角形测量高度