21.3.1几何问题 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数学《一元二次方程的应用——几何问题》专项练习（全文1444字）##一、知识点梳理一元二次方程在几何中的应用\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)核心思路：**利用几何图形的面积、周长、勾股定理等性质列出等量关系式\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)建立一元二次方程求解\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)最后检验解是否符合几何实际意义**。常见题型分类：1.**面积问题**：矩形、正方形\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)道路、边框、裁剪拼接。常设宽度为未知数\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)用“总面积减去空白面积=阴影面积”或者“平移法”合并道路\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)简化列式。2.**勾股定理问题**：直角三角形\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)两直角边的平方和等于斜边平方\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)建立方程。3.**动点几何**：点在线段上移动\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)用含$t$的代数式表示线段长度\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)再结合面积或勾股定理列方程。解题通用步骤：①审题\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)画出示意图；②设未知数；③用未知数表示相关线段长度；④根据面积公式或勾股定理列一元二次方程；##五、参考答案与解析###选择题1.B。解析：左右、上下都加宽\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)长和宽各加$2x$。2.A。解析：设两边$\(a,b\)$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)$a+b=7\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)\dfrac12ab=6$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)解得两边3、4\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)斜边5。3.A。解析：平移道路\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)草坪拼成新矩形\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)长$30-x$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)宽$20-x$。4.B。解析：长+宽=12\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)设长$x$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)$x(12-x)=32$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)$x_1=8\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)x_2=4$。5.A。解析：底边一半4\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)勾股定理$4^2+(x-2)^2=x^2$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)解得$\(x=5\)$。###填空题1.$6$；解析：$\(x(x+2)=48\)$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)$x^2+2x-48=0$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)$\(x=6\)$（舍去负根）。2.$\(3,4,5\)$3.$5$；解析：$(x+3)^2-x^2=39$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)$6x+9=39\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)\(x=5\)$。4.$2$；解析：$(40-2x)(26-x)=864$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)解得$\(x=2\)$。5.$8$；解析：$\dfrac12\times6\times\(x=2\)4$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)$\(x=8\)$。###解答题1.解：折盒后底面长$(100-2x)$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)宽$(50-2x)$$(100-2x)(50-2x)=3600$整理：$x^2-75x+350=0$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)$(x-5)(x-70)=0$$x_1=5\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)x_2=70$。$\(x=70\)$时\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)$50-2x<0$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)舍去。$\therefore\(x=5\)\\text{cm}$。2.解：设较短直角边$x\\text{cm}$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)另一直角边$(x+3)\\text{cm}$$\dfrac12x(x+3)=18$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)$x^2+3x-36=0$解得$x_1=\dfrac{-3+3\sqrt{17}}{2}$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)$x_2$负数舍去。两直角边：$\dfrac{-3+3\sqrt{17}}{2}\\text{cm}$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)$\dfrac{3+3\sqrt{17}}{2}\\text{cm}$。3.解：设运动时间$t$秒\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)$PB=12-2t$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)$\(BQ=t\)$$S_{\trianglePBQ}=\dfrac12(12-2t)\cdott=9$整理：$t^2-6t+9=0$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)$(t-3)^2=0$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)$t_1=t_2=3$。检验：$\(t=3\)$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)$\(PB=6>0\)$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)符合条件。$\therefore3$秒。4.解：设垂直墙的边长$x\\text{m}$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)平行墙边长$(40-2x)\\text{m}$$x(40-2x)=150$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)$x^2-20x+75=0$解得$x_1=5\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)x_2=15$。墙长15m\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)$40-2x\le15$。当$\(x=5\)$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)平行墙长$\(30>15\)$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)超出墙长\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)舍去；当$\(x=15\)$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)平行墙长$10\le15$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)符合。$\therefore$垂直墙边长15m。##六、易错总结几何应用题**检验是必不可少的一步**。求出方程的根后\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)一定要看线段长度大于0\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)不能超过原图边长；靠墙问题还要注意平行墙一边不能超过墙的最大长度\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)不符合的根直接舍去。道路面积题优先平移法\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)把分散草坪拼成完整矩形\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)减少计算错误\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)注意是减去$2x$还是$x$。动点问题\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)用时间$t$表示线段长度\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)时刻注意取值范围\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)线段不能为负数。勾股定理类\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)分清斜边\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)不要把直角边、斜边弄混。面积公式不要漏乘$\dfrac12$\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)三角形面积是高频丢分点。画图非常重要\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)根据题意画出简易图\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)能快速理清长、宽关系。几何应用题的方程一般有两个解\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)一个符合实际\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)一个不符合\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)千万不要直接两个都写。这类大题在中考常考\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)和一元二次方程的解法结合\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)既要会列方程\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)也要会解方程\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)最后落实实际意义检验。（全文1444字）下一节：**一元二次方程应用——增长率问题**\(8+x\(8+2x\(8-x\(8-2x\(a+b=7\(x_1=8\(6x+9=39\(x_1=5\(x_1=5，x_2=15\)x_2=70\)x=5\)x_2=4\)\dfrac12ab=6\)5-2x\)5-x\)5+2x\)5+x\)继续同格式1444字专项练习吗？华东师大版（新教材）数学九年级上册21.3.1几何问题授课教师：.

班

级：

9年级（

）班.

时

间：.

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21章一元二次方程新知探究一元二次方程解几何问题

学校生物小组有一块长32m、宽20m的矩形试验田，为了管理方便，准备沿平行于两边的方向纵、横各开辟一条等宽的小道.要使种植面积为540m2，小道的宽应是多少？分析：问题中没有明确小道在试验田中的位置，我们可先设计画出草图，数形结合。问题1和同桌交换看一看，是否设计得一样？新知探究一元二次方程解几何问题

学校生物小组有一块长32m、宽20m的矩形试验田，为了管理方便，准备沿平行于两边的方向纵、横各开辟一条等宽的小道.要使种植面积为540m2，小道的宽应是多少？问题12032你发现了什么？小道的占地面积与位置无关.你是怎样计算小道的宽度的？新知探究一元二次方程解几何问题

学校生物小组有一块长32m、宽20m的矩形试验田，为了管理方便，准备沿平行于两边的方向纵、横各开辟一条等宽的小道.要使种植面积为540m2，小道的宽应是多少？问题12032xx解：设小道宽为xm，

根据题意，得

新知探究

如果设想把小道平移到两边，如图所示，那么小道所占面积是否保持不变？一元二次方程解几何问题试一试32m20mxmxm解：设道路的宽为x

m.

则

当

x=50

时，32−x=−18，不合题意，舍去.∴

取

x=2.答：道路的宽为

2m.利用“图形经过平移，它的面积大小不会改变”的道理，纵、横两条路移动一下，使列方程容易些（目的是求出路面的宽，至于实施施工，仍可按原图的位置修路）在应用一元二次方程解决实际问题时，要注意：1.分析题意，抓住等量关系；2.列出方程，把实际问题转化为数学问题来解决；3.求得方程的根之后，要注意检验是否符合题意，最后得到实际问题的解答.归纳总结归纳总结一元二次方程解几何问题典例分析例1如图，要利用一面墙(墙足够长)建羊圈，用58m的围栏围成面积为200m2的矩形羊圈，则羊圈的边

AB和

BC的长各是多少米？

DCBA一元二次方程解几何问题应用1

边框问题

返回2.

软笔书法承载着中华五千年的灿烂文化，如图①，它是李叔叔的软笔作品，

返回新知探究一元二