名校
解题方法
1 . 材料1.类比是获取数学知识的重要思想之一,很多优美的数学结论就是利用类比思想获得的.例如:若
,
,则
,当且仅当
时,取等号,我们称为二元均值不等式.类比二元均值不等式得到三元均值不等式:
,
,
,则
,当且仅当
时,取等号.我们经常用它们求相关代数式或几何问题的最值,某同学做下面几何问题就是用三元均值不等式圆满完成解答的.
题:将边长为
的正方形硬纸片(如图1)的四个角裁去四个相同的小正方形后,折成如图2的无盖长方体小纸盒,求纸盒容积的最大值.
,则纸盒容积
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/71c081a56a12c5d11c9b4f31008a65ec.png)
当且仅当
,即
时取等号.所以纸金的容积取得最大值
.在求
的最大值中,用均值不等式求最值时,遵循“一正二定三相等”的规则.你也可以将
变形为
求解.
你还可以设纸盒的底面边长为
,高为
,则
,则纸盒容积
.
当且仅当
,即
,
时取等号,所以纸盒的容积取得最大值
.
材料2.《数学必修二》第八章8.3节习题8.3设置了如下第4题:
如图1,圆锥的底面直径和高均为
,过
的中点
作平行于底面的截面,以该截面为底的面挖去一个圆柱,求剩下几何体的表面积和体积.我们称圆柱为圆锥的内接圆柱.
根据材料1与材料2完成下列问题.
如图2,底面直径和高均为
的圆锥有一个底面半径为
,高为
的内接圆柱.
与
的关系式;
(2)求圆柱侧面积的最大值;
(3)求圆柱体积的最大值.
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/94440d3e4c073f94f2b266ff99d50e74.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/67ca5fd57c2c2fcc3c7a574fdd1467d9.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/689f982af451283289255c87593ec338.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/1f22fec5a381ae8aca93d876e54c79de.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/94440d3e4c073f94f2b266ff99d50e74.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/67ca5fd57c2c2fcc3c7a574fdd1467d9.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/8cec12441802f71e803efaf2c62ee588.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/8d936ea1443a8c881633d5e04fdd3434.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/44acc0ee22dc4b7750e8be825e7c1355.png)
题:将边长为
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/689ff84e2d7f52c7446ef789a54557da.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e4e3c92be4b3f494e7d03c67819632c0.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/71c081a56a12c5d11c9b4f31008a65ec.png)
当且仅当
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/efaf86a31a17f80098a020b74d5282bc.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/707ea658f3a9359f5740d5aab48f7948.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b50995580ef9cbc240041c2f8d00d79d.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/be54e84508decfcce6d2fcbe6c8c1a92.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/dbb2757026c0f75d4f1ea56349b177b7.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/ab79a858ff360048fb4f1f7784cbfe8d.png)
你还可以设纸盒的底面边长为
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0a6936d370d6a238a608ca56f87198de.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2c94bb12cee76221e13f9ef955b0aab1.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/493dbbbcf8aecaf1b586774ad7846f64.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/db442d96d27b4c73a3dc684756b7a0b2.png)
当且仅当
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/3527a89afa5fbd67781a204d3954a02e.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/36e15cbd7c42d7b15d7ba8d2b28ab8df.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/03837b3769eda7f0d3804cc5ad4a6d60.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/b50995580ef9cbc240041c2f8d00d79d.png)
材料2.《数学必修二》第八章8.3节习题8.3设置了如下第4题:
如图1,圆锥的底面直径和高均为
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0a6936d370d6a238a608ca56f87198de.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/ef49a3ca580a144cc65a609c167facc1.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/23f919bd3dde10dbbc076f7ec5149699.png)
根据材料1与材料2完成下列问题.
如图2,底面直径和高均为
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/4dd6f4250ca6b1b9bce234a01f00d44d.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/4aa0df7f1e45f9de29e802c7f19a4f64.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/73465a1f9aa03481295bf6bd3c6903ac.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/4aa0df7f1e45f9de29e802c7f19a4f64.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/73465a1f9aa03481295bf6bd3c6903ac.png)
(2)求圆柱侧面积的最大值;
(3)求圆柱体积的最大值.
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2 . 为了求一个棱长为
的正四面体的体积,某同学设计如下解法.
解:构造一个棱长为1的正方体,如图1:则四面体
为棱长是
的正四面体,且有
.
![](https://img.xkw.com/dksih/QBM/editorImg/2022/6/16/6db5d8bf-a942-4eb1-b74e-0d41be5b6734.png?resizew=583)
(1)类似此解法,如图2,一个相对棱长都相等的四面体,其三组棱长分别为
,
,
,求此四面体的体积;
(2)对棱分别相等的四面体
中,
,
,
.求证:这个四面体的四个面都是锐角三角形;
(3)有4条长为2的线段和2条长为
的线段,用这6条线段作为棱且长度为
的线段不相邻,构成一个三棱锥,问
为何值时,构成三棱锥体积最大,最大值为多少?
[参考公式:三元均值不等式
及变形
,当且仅当
时取得等号]
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/cf298f00799cbf34b4db26f5f63af92f.png)
解:构造一个棱长为1的正方体,如图1:则四面体
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/68ac02c2f91cadb1e328bc6ab9b9c491.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/cf298f00799cbf34b4db26f5f63af92f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/f6f878ffcff2ca25a434cbeea7d5c841.png)
![](https://img.xkw.com/dksih/QBM/editorImg/2022/6/16/6db5d8bf-a942-4eb1-b74e-0d41be5b6734.png?resizew=583)
(1)类似此解法,如图2,一个相对棱长都相等的四面体,其三组棱长分别为
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2967337e3fcb228dded64ab0c41a17e0.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/50690dab38f4512eb72e18b7f86cf6f6.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/d4056761b8f826eeb6ad8c9a151d3c9c.png)
(2)对棱分别相等的四面体
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/411b38a18046fea8e9fab1f9f9b80a5f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c220eadc312101e2fb89dfe920f7b30d.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/7de966c316db1013defc56372fcf814e.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e8d2530e7023b2345c651e8f53629ff1.png)
(3)有4条长为2的线段和2条长为
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/294f5ba74cdf695fc9a8a8e52f421328.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/294f5ba74cdf695fc9a8a8e52f421328.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/294f5ba74cdf695fc9a8a8e52f421328.png)
[参考公式:三元均值不等式
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/1ffb6b373d2e672bb2afc8de547861a2.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/4849ff71159df2bb9099b26065d81e1f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/44acc0ee22dc4b7750e8be825e7c1355.png)
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2021-07-15更新
|
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2卷引用:重庆市西南大学附属中学2020-2021学年高一下学期期末数学试题
3 . 《九章算术商功》:“斜解立方,得两堑堵.斜解堑堵,其一为阳马,一为鳖臑.”其中,阳马是底面为矩形,且有一条侧棱与底面垂直的四棱锥.如图,在阳马
中,侧棱PA垂直于底面ABCD,且
,则该阳马的外接球的表面积等于______ .
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0585b6c0f156eecf9662b9846d4eb693.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/be5d1bfdfd201164a16ee8cc4644a984.png)
![](https://img.xkw.com/dksih/QBM/editorImg/2022/12/15/80b9dad1-df02-4f7a-81bc-3df40603215e.png?resizew=121)
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4 . 《九章算术·商功》:“斜解立方,得两堑(qiàn)堵(dǔ).斜解堑堵,其一为阳马,一为鳖(biē)臑(nào).阳马居二,鳖臑居一,不易之率也.合两鳖臑三而一,验之以棊,其形露矣.”刘徽注:“此术臑者,背节也,或曰半阳马,其形有似鳖肘,故以名云·中破阳马,得两鳖臑,鳖臑之起数,数同而实据半,故云六而一即得.”阳马和鳖臑是我国古代对一些特殊锥体的称谓,取一长方体,按下图斜割一分为二,得两个一模一样的三棱柱,称为堑堵,再沿堑堵的一顶点与相对的棱剖开,得四棱锥和三棱锥各一个,以矩形为底,另有一棱与底面垂直的四棱锥,称为阳马,余下的三棱锥是由四个直角三角形组成的四面体,称为鳖臑.
:
①在右图中,求三棱锥
的高.
②求三棱锥
外接球的表面积.
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/0c8cf5d11389d101e9ebf87764d0f8dd.png)
①在右图中,求三棱锥
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/ac7a3e0ef4980cc0ca102f733d357263.png)
②求三棱锥
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/ac7a3e0ef4980cc0ca102f733d357263.png)
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5 . (1)求一个棱长为
的正四面体的体积,有如下未完成的解法,请你将它补充完成.解:构造一个棱长为1的正方体—我们称之为该四面体的“生成正方体”,如左下图:则四面体
为棱长是___________的正四面体,且有
___________.
![](https://img.xkw.com/dksih/QBM/2021/5/3/2712963070132224/2799670116392960/STEM/b3a43d08-3809-4282-8b2f-46b52950fd10.png?resizew=380)
(2)模仿(1),对一个已知四面体,构造它的“生成平行六面体”,记两者的体积依次为
和
,试给出这两个体积之间的一个关系式,不必证明;
(3)如1图,一个相对棱长都相等的四面体(通常称之为等腰四面体),其三组棱长分别为
,
,
,类比(1)(2)中的方法或结论,求此四面体的体积.
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/cf298f00799cbf34b4db26f5f63af92f.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/68ac02c2f91cadb1e328bc6ab9b9c491.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/dafefd79de97043ba8a070428467e285.png)
![](https://img.xkw.com/dksih/QBM/2021/5/3/2712963070132224/2799670116392960/STEM/b3a43d08-3809-4282-8b2f-46b52950fd10.png?resizew=380)
(2)模仿(1),对一个已知四面体,构造它的“生成平行六面体”,记两者的体积依次为
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e6cefccb97d4ec7d785b9db04ea196a4.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/9f37c08f1fd1b7f1d7a1052b9fd8c60e.png)
(3)如1图,一个相对棱长都相等的四面体(通常称之为等腰四面体),其三组棱长分别为
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2967337e3fcb228dded64ab0c41a17e0.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/50690dab38f4512eb72e18b7f86cf6f6.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/d4056761b8f826eeb6ad8c9a151d3c9c.png)
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6卷引用:8.3简单几何体的表面积与体积C卷
(已下线)8.3简单几何体的表面积与体积C卷上海市复旦大学附属中学2020-2021学年高二下学期期中数学试题沪教版(2020) 必修第三册 同步跟踪练习 第11章 11.3.1 多面体(已下线)第02讲 简单几何体(核心考点讲与练)(2)(已下线)11.3 多面体与旋转体(作业)(夯实基础+能力提升)-【教材配套课件+作业】2022-2023学年高二数学精品教学课件(沪教版2020必修第三册)(已下线)专题08多面体与旋转体(2个知识点3种题型1种高考考法)-【倍速学习法】2023-2024学年高二数学核心知识点与常见题型通关讲解练(沪教版2020必修第三册)
2021·上海浦东新·三模
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6 . 某工厂承接制作各种弯管的业务,其中一类弯管由两节圆管组成,且两节圆管是形状、大小均相同的斜截圆柱,其尺寸如图1所示(单位:
),其中斜截面与底面所成的角为
,将其中一个斜截圆柱的侧面沿
剪开并摊平,可以证明由截口展开而成的曲线
是函数
的图像,其中
,
,如图2所示.
![](https://img.xkw.com/dksih/QBM/2021/5/20/2725319832076288/2730995936878592/STEM/306e80d3-1869-484b-b0c3-2bec5b669829.png?resizew=228)
![](https://img.xkw.com/dksih/QBM/2021/5/20/2725319832076288/2730995936878592/STEM/6a824180-fff5-4dc0-822f-7fb64b201efd.png?resizew=240)
(1)若
,求
的解析式;
(2)已知函数
的图像与x轴围成区域的面积可由公式
计算,若制作该种该类弯管的一截圆管所用材料面积(即斜截圆柱的侧面积)等于与之底面相同且高为
的圆柱的面积,求
的值(结果精确到
).
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/9efa9fbcfb9595e2f031aa691db4564b.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e170f206fdbbd834aad7580c727e2cc6.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2777840758e70e7dbbc18cef8f3d6d2b.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/7e58750ec6571eaa9f2ac3ca6f0a6ce5.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e4a63412a8e10dca8d002978e17c45a4.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/2480f87a11c4cd450bc9454ea7276722.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/a4456675a5dbe545462a22cef9aca8fe.png)
![](https://img.xkw.com/dksih/QBM/2021/5/20/2725319832076288/2730995936878592/STEM/306e80d3-1869-484b-b0c3-2bec5b669829.png?resizew=228)
![](https://img.xkw.com/dksih/QBM/2021/5/20/2725319832076288/2730995936878592/STEM/6a824180-fff5-4dc0-822f-7fb64b201efd.png?resizew=240)
(1)若
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/c6e2e859e649b43a21b623f63472122a.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/51c530f4b7491b95acb8ce3eef9aa09d.png)
(2)已知函数
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/51c530f4b7491b95acb8ce3eef9aa09d.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e58a22ac4aca667f4363d3526feb8f25.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/18703d241f6fe0a0dedcc815603322fc.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/e170f206fdbbd834aad7580c727e2cc6.png)
![](https://staticzujuan.xkw.com/quesimg/Upload/formula/1008a3fc217ce647e16fa09e42ceadb1.png)
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