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【324209】2024八年级数学下册 专题5.5 四边形中的最值问题专项训练(含解析)(新版)浙教版

时间:2025-01-15 21:41:51 作者: 字数:27418字
简介:


 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 专题5.5四边形中的最值问题专项训练(30道)

考卷信息:

本套训练卷共30题,选择15题,填空15题,题型针对性较高,覆盖面广,选题有深度,可强化学生对四边形中最值问题模型的记忆与理解!

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

1.(德阳期末)如图,将矩形ABCD放置在平面直角坐标系的第一象限内,使顶点AB分别在x轴、y轴上滑动,矩形的形状保持不变,若AB2BC1,则顶点C到坐标原点O的最大距离为(  )

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

A1 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> B1 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> C3 D <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解题思路】取AD的中点E,连接OECEOC,求得CE <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> OE1,再根据OCCE+OE1 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> ,即可得到点C到原点O距离的最大值是1 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解答过程】解:如图,取AB的中点E,连接OECEOC

∵∠AOB90°

Rt△AOB中,OE <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> AB1

又∵∠ABC90°AEBECB1

Rt△CBE中,CE <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

又∵OCCE+OE1 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

OC的最大值为1 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

即点C到原点O距离的最大值是1 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

故选:A

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

2.(西岗区期末)如图,在Rt△ABC中,∠BAC90°AB3AC4P为边BC上一动点,PEABEPFACFMEF的中点,则AM的最小值是(  )

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

A2.4 B2 C1.5 D1.2

【解题思路】AM <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> EF <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> AP,所以当AP最小时,AM最小,根据垂线段最短解答.

【解答过程】解:由题意知,四边形AFPE是矩形,

M是矩形对角线EF的中点,则延长AM应过点P

AP为直角三角形ABC的斜边上的高时,即APBC时,AM有最小值,

此时AM <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> AP,由勾股定理知BC <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 5

SABC <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> ABAC <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> BCAP

AP <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

AM <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> AP <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 1.2

故选:D

3.(龙口市期末)如图,在边长为6的正方形ABCD中,点P为对角线AC上一动点,PEABEPFBCF,则EF的最小值为(  )

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

A <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> B <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> C4 D3

【解题思路】连接BP,根据PEABPFBC得到四边形PEBF为矩形,得EFBPBP最短时即BPAC,即可求解.

【解答过程】解:连接BP,如图,

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

四边形ABCD是正方形,

∴∠ABC90°ABBC6

PEABPFBC

四边形PEBF为矩形,

EFBP

BPACBP最短,

Rt△BPC中,BPPCBC6

根据勾股定理可解得BP3 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

EF得最小值为3 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

故选:B

4.(重庆期末)如图,以边长为4的正方形ABCD的中心O为端点,引两条互相垂直的射线,分别与正方形的边交于EF两点,则线段EF的最小值是(  )

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

A <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> B2 C <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> D4

【解题思路】根据正方形的性质得到∠EAO=∠FDO45°AODO,证得△AOE≌△DOF,根据全等三角形的性质得到OEOF,求出OE的范围,借助勾股定理即可解决问题.

【解答过程】解:∵四边形ABCD为正方形,

∴∠EAO=∠FDO45°AODO

∵∠EOF90°,∠AOD90°

∴∠AOE=∠DOF

在△AOE与△DOF中,

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

∴△AOE≌△DOFASA),

OEOF(设为λ);

∴△EOF是等腰直角三角形,

由勾股定理得:

EF2OE2+OF22

EF <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> OE <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> λ

正方形ABCD的边长是4

OA2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> OAB的距离等于2OAB的垂线段的长度),

由题意可得:2≤λ≤2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> EF≤4

所以线段EF的最小值为2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

故选:C

5.(马鞍山期末)如图,在菱形ABCD中,∠B45° <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> EF分别是边CDBC上的动点,连接AEEFGH分别为AEEF的中点,连接GH,则GH的最小值为(  )

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

A <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> B <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> C <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> D1

【解题思路】连接AF,利用三角形中位线定理,可知GH <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> AF,求出AF的最小值即可解决问题.

【解答过程】解:连接AF,如图所示:

四边形ABCD是菱形,

ABBC2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

GH分别为AEEF的中点,

GH是△AEF的中位线,

GH <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> AF

AFBC时,AF最小,GH得到最小值,

则∠AFB90°

∵∠B45°

∴△ABF是等腰直角三角形,

AF <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> AB <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

GH <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

GH的最小值为 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

故选:B

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

6.(潜山市期末)如图,点E是边长为8的正方形ABCD的对角线BD上的动点,以AE为边向左侧作正方形AEFG,点PAD的中点,连接PG,在点E运动过程中,线段PG的最小值是(  )

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

A2 B <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> C2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> D4 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解题思路】连接DG,可证△AGD≌△AEB,得到G点轨迹,利用点到直线的最短距离进行求解.

【解答过程】解:连接DG,如图,

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

四边形ABCD、四边形AEFG均为正方形,

∴∠DAB=∠GAE90°ABADAGAE

∵∠GAD+∠DAE=∠DAE+∠AE

∴∠GAD=∠BAE

ABADAGAE

∴△AEB≌△AGDSAS),

∴∠PDG=∠ABE45°

G点轨迹为线段DH

PGDH时,PG最短,

Rt△PDG中,∠PDG45°PAD中点,DP4

PGx,则DGx,由勾股定理得,

x2+x242

解得x <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

故选:C

7.(蚌埠期末)如图,矩形ABCD中,ABAD21,点EAB的中点,点FEC上一个动点,点PDF的中点,连接PB.若PB的最小值为5 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> ,则AD的值为(  )

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

A5 B6 C7 D8

【解题思路】F点在运动时,P点轨迹为平行EC的线段,BP最短为点到直线的最短距离.

【解答过程】解:当F运动时,P点轨迹为GH,如图,

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

ABAD21

ADAEEBBC

∴∠ADE=∠DEA=∠CEB=∠ECB45°

∴∠DEC90°

BP的最距离为BPGH时,此时P点与H点重合,F点与C点重合.

HCD中点,

CHCB,∠GHB90°

Rt△HCB中,BH5 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

CHCB5

故选:A

8.(南安市期末)如图,在矩形ABCD中,AB4AD6,点PAD上,点QBC上,且APCQ,连接CPQD,则PC+QD的最小值为(  )

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

A8 B10 C12 D20

【解题思路】连接BP,则PC+QD的最小值转化为PC+PB的最小值,在BA的延长线上截取AEAB4,连接PECE,则PC+QDPC+PBPC+PECE,再根据勾股定理求解即可.

【解答过程】解:如图,连接BP

在矩形ABCD中,ADBCADBC6

APCQ

ADAPBCCQ

DPQBDPBQ

四边形DPBQ是平行四边形,

PBDQPBDQ

PC+QDPC+PB,则PC+QD的最小值转化为PC+PB的最小值,

BA的延长线上截取AEAB4,连接PE

BE2AB8

PABE

PABE的垂直平分线,

PBPE

PC+PBPC+PE

连接CE,则PC+QDPC+PBPC+PECE

CE <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 10

PC+PB的最小值为10

PC+QD的最小值为10

故选:B

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

9.(连云港期末)如图,线段AB的长为8,点DAB上,△ACD是边长为3的等边三角形,过点D作与CD垂直的射线DP,过DP上一动点G(不与D重合)作矩形CDGH,记矩形CDGH的对角线交点为O,连接OB,则线段BO的最小值为(  )

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

A5 B4 C <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> D <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解题思路】连接AO,根据矩形对角线相等且互相平分得:OCOD,再证明△ACO≌△ADO,则∠OAB30°;点O一定在∠CAB的平分线上运动,根据垂线段最短得:当OBAO时,OB的长最小,根据直角三角形30度角所对的直角边是斜边的一半得出结论.

【解答过程】解:连接AO

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

四边形CDGH是矩形,

CGDHOC <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> CGOD <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> DH

OCOD

∵△ACD是等边三角形,

ACAD,∠CAD60°

在△ACO和△ADO中,

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

∴△ACO≌△ADOSSS),

∴∠OAB=∠CAO30°

O一定在∠CAB的平分线上运动,

OBAO时,OB的长度最小,

∵∠OAB30°,∠AOB90°

OB <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> AB <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 84

OB的最小值为4

故选:B

10.(惠山区期中)如图,平面内三点ABCAB5AC4,以BC为对角线作正方形BDCE,连接AD,则AD的最大值是(  )

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

A5 B9 C9 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> D <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解题思路】如图将△BDA绕点D顺时针旋转90°得到△CDM.由旋转不变性可知:ABCM5DADM.∠ADM90°,得出△ADM是等腰直角三角形,推出AD <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> AM,当AM的值最大时,AD的值最大,根据三角形的三边关系求出AM的最大值即可解决问题.

【解答过程】解:如图,

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

将△BDA绕点D顺时针旋转90°得到△CDM

由旋转不变性可知:ABCM5DADM,∠ADM90°

∴△ADM是等腰直角三角形,

AD <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> AM

AM的值最大时,AD的值最大,

AMAC+CM

AM≤9

AM的最大值为9

AD的最大值为 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

故选:D

11.(邗江区期末)如图,以边长为4的正方形ABCD的中心O为端点,引两条相互垂直的射线,分别与正方形的边交于EF两点,则线段EF的最小值为(  )

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

A2 B4 C <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> D2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解题思路】如图,作辅助线;证明△AOE≌△DOF,进而得到OEOF,此为解决该题的关键性结论;求出OE的范围,借助勾股定理即可解决问题.

【解答过程】解:如图,连接EF

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

四边形ABCD为正方形,

∴∠EAO=∠FDO45°AODO

∵∠EOF90°,∠AOD90°

∴∠AOE=∠DOF

在△AOE与△DOF中,

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

∴△AOE≌△DOFASA),

OEOF(设为λ);

∴△EOF是等腰直角三角形,

由勾股定理得:

EF2OE2+OF22

EF <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> OE <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> λ

正方形ABCD的边长是4

OA2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> OAB的距离等于2OAB的垂线段的长度),

由题意可得:2≤λ≤2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> EF≤4

所以线段EF的最小值为2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

故选:D

12.(宁蒗县模拟)如图,菱形ABCD的的边长为6,∠ABC60°,对角线BD上有两个动点EF(点E在点F的左侧),若EF2,则AE+CF的最小值为(  )

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

A2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> B4 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> C6 D8

【解题思路】作AMAC,连接CMBDF,根据菱形的性质和等边三角形的判定和性质以及勾股定理解答即可.

【解答过程】解:如图,连接AC,作AMAC,使得AMEF2,连接CMBDF

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

ACBD是菱形ABCD的对角线,

BDAC

AMAC

AMBD

AMEF

AMEFAMEF

四边形AEFM是平行四边形,

AEFM

AE+CFFM+FCCM

根据两点之间线段最短可知,此时AE+FC最短,

四边形ABCD是菱形,AB6,∠ABC60°

BCAB

∴△ABC是等边三角形,

ACAB6

Rt△CAM中,CM <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

AE+CF的最小值为2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

故选:A

13.(宜兴市期中)如图,在边长为4的正方形ABCD中,点EF分别是边BCCD上的动点,且BECF,连接BFDE,则BF+DE的最小值为(  )

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

A <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> B <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> C <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> D <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解题思路】连接AE,利用△ABE≌△BCF转化线段BF得到BF+DEAE+DE,则通过作A点关于BC对称点H,连接DHBCE点,利用勾股定理求出DH长即可.

【解答过程】解:连接AE,如图1

四边形ABCD是正方形,

ABBC,∠ABE=∠BCF90°

BECF

∴△ABE≌△BCFSAS).

AEBF

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

所以BF+DE最小值等于AE+DE最小值.

作点A关于BC的对称点H点,如图2

连接BH,则ABH三点共线,

连接DHDHBC的交点即为所求的E点.

根据对称性可知AEHE

所以AE+DEDH

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

Rt△ADH中,DH <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

BF+DE最小值为 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

故选:D

14.(重庆期末)如图,矩形ABCD中,AB2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> BC6P为矩形内一点,连接PAPBPC,则PA+PB+PC的最小值是(  )

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

A4 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 3 B2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> C2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 6 D4 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解题思路】将△BPC绕点C逆时针旋转60°,得到△EFC,连接PFAEAC,则AE的长即为所求.

【解答过程】解:将△BPC绕点C逆时针旋转60°,得到△EFC,连接PFAEAC,则AE的长即为所求.

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

由旋转的性质可知:△PFC是等边三角形,

PCPF

PBEF

PA+PB+PCPA+PF+EF

APFE共线时,PA+PB+PC的值最小,

四边形ABCD是矩形,

∴∠ABC90°

tan∠ACB <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

∴∠ACB30°AC2AB4 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

∵∠BCE60°

∴∠ACE90°

AE <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

故选:B

15.(江阴市模拟)如图,在边长为6的正方形ABCD中,点EFG分别在边ABADCD上,EGBF交于点IAE2BFEGDGAE,则DI的最小值等于(  )

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

A <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 3 B2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 2 C2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> D2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 3

【解题思路】过点EEMCD于点M,取BE的中点O,连接OIOD,根据HL证明Rt△BAF≌Rt△EMG,可得∠ABF=∠MEG,所以再证明∠EPF90°,由直角三角形斜边上的中线等于斜边的一半可得OI <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> BE,由ODOIDI,当ODI共线时,DI有最小值,即可求DI的最小值.

【解答过程】解:如图,过点EEMCD于点M,取BE的中点O,连接OIOD

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

四边形ABCD是正方形,

ABAD,∠A=∠D=∠DME90°ABCD

四边形ADME是矩形,

EMADAB

BFEG

Rt△BAF≌Rt△EMGHL),

∴∠ABF=∠MEG,∠AFB=∠EGM

ABCD

∴∠MGE=∠BEG=∠AFB

∵∠ABF+∠AFB90°

∴∠ABF+∠BEG90°

∴∠EIF90°

BFEG

∵△EIB是直角三角形,

OI <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> BE

AB6AE2

BE6﹣24OBOE2

ODOIDI

ODI共线时,DI有最小值,

IO <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> BE2

OD <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

ID2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 2,即DI的最小值为2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 2

故选:B

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

16.如图,菱形ABCD的两条对角线长分别为AC6BD8,点PBC边上的一动点,则AP的最小值为 4.8 

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解题思路】由垂线段最短,可得APBC时,AP有最小值,由菱形的性质和勾股定理可求BC的长,由菱形的面积公式可求解.

【解答过程】解:设ACBD的交点为O

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

PBC边上的一动点,

APBC时,AP有最小值,

四边形ABCD是菱形,

ACBDAOCO <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> AC3BODO <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> BD4

BC <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 5

S菱形ABCD <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> AC×BDBC×AP

AP <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 4.8

故答案为:4.8

17.(椒江区期末)如图,矩形ABCD中,AB8AD6,连接BDEBD上一动点,PCE中点,连接PA,则PA的最小值是 2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>  

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解题思路】P点运动轨迹为△CDB的中位线,即求A点到这条中位线的最短距离.

【解答过程】解:当点E运动时,P点轨迹为△CBD中位线GH,如图,

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

A到直线GH的最短距离为AF,但是E点在运动中,P点轨迹为GH

A到线段GH的最短距离为AG

GCD中点,

DG4

Rt△ADG中,AD6DG4

AG <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

故答案为2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

18.(宁德期末)如图,在矩形ABCD中,AB4AD3,点ECD上一个动点,点FG分别是ABAE的中点,则线段FG的最小值是  <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>  

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解题思路】连接BE,可得FG是△ABE的中位线,要使线段FG最小,需BE最小,当点E与点C重合时,BE最小为3,进而可得线段FG的最小值.

【解答过程】解:如图,连接BE

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

FG分别是ABAE的中点,

FG是△ABE的中位线,

FG <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> BE

要使线段FG最小,

BE最小,

当点E与点C重合时,BE最小为3

则线段FG的最小值是 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

故答案为: <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

19.(东海县期末)如图,在菱形ABCD中,AC24BD10,对角线交于点O,点EAD上,且DE <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> AD,点FOB的中点,点G为对角线AC上的一动点,则GEGF的最大值为  <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>  

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解题思路】由菱形的性质可得AOCO12BODO5ACBD,在Rt△AOD中,由勾股定理可求AD的长,作点F关于AC的对称点F',连接GF',取AD中点H,连接OH,可得GFGF'OFOF',则GEGFGEGF'≤EF',即当点GEF'的延长线时,GEGF有最大值为EF'的长,由直角三角形的性质和三角形中位线定理可求解.

【解答过程】解:∵四边形ABCD是菱形,

AOCO12BODO5ACBD

AD <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 13

如图,作点F关于AC的对称点F',连接GF',取AD中点H,连接OH

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

ACBD,点HAD中点,

OHHD <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> AD <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

F与点F'关于AC对称,

GFGF'OFOF'

GEGFGEGF'≤EF'

当点GEF'的延长线时,GEGF有最大值为EF'的长,

DE <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> ADHD <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> AD

DEEH

FOB的中点,

OF <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> OBOF' <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> DO

EF' <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> OH <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

故答案为: <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

20.(淄博)两张宽为3cm的纸条交叉重叠成四边形ABCD,如图所示.若∠α30°,则对角线BD上的动点PABC三点距离之和的最小值是 6 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> cm 

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解题思路】作DEBCE,解直角三角形求得ABBC6cm,把△ABP绕点B逆时针旋转60°得到△A'BP,由旋转的性质,ABAB6cmBPBPA'PAP,∠PBP60°A'BA60°,所以△PBP是等边三角形,根据两点间线段距离最短,可知当PA+PB+PCA'C时最短,连接A'C,利用勾股定理求出A'C的长度,即求得点PABC三点距离之和的最小值.

【解答过程】解:如图,作DEBCE,把△ABP绕点B逆时针旋转60°得到△A'BP

∵∠α30°DE3cm

CD2DE6cm

同理:BCAD6cm

由旋转的性质,ABABCD6mBPBPA'PAP,∠PBP60°,∠A'BA60°

∴△PBP是等边三角形,

BPPP'

PA+PB+PCA'P′+PP'+PC

根据两点间线段距离最短,可知当PA+PB+PCA'C时最短,连接A'C,与BD的交点即为P点,即点PABC三点距离之和的最小值是AC

∵∠ABC=∠DCE=∠α30°,∠ABA60°

∴∠ABC90°

AC <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 6 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> cm),

因此点PABC三点距离之和的最小值是6 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> cm

故答案为6 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> cm

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

21.(龙岩期末)如图,正三角形ABC与正方形CDEF的顶点BCD三点共线,动点P沿着CACA运动.连接EP,若AC10CF8.则EP的最小值是 4 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 4 

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解题思路】过点EEPAC,交FC于点G,当EPAC时,EP取得最小值,然后根据含30度角的直角三角形列式计算即可求出EP的最小值.

【解答过程】解:如图,过点EEPAC,交FC于点G

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

EPAC时,EP取得最小值,

正三角形ABC与正方形CDEF的顶点BCD三点共线,

∴∠ACB60°,∠FCD90°

∴∠ACF30°

∴∠CGP=∠EGF60°

∵∠F90°

∴∠FEG30°

PGx,则CG2x

FGCFCG8﹣2x

EG2FG28﹣2x),

FG <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> EF

8﹣2x8 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

x4 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

EPEG+PG28﹣2x+x16﹣3x4 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 4

故答案为:4 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 4

22.(茅箭区校级期末)如图,已知线段AB12,点C在线段AB上,且△ACD是边长为4的等边三角形,以CD为边在CD的右侧作矩形CDEF,连接DF,点MDF的中点,连接MB,则线段MB的最小值为 6 

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解题思路】连接AMCMEM,根据四边形CDEF是矩形,和△ACD是等边三角形,证明△ADM≌△ACM,从而求出∠CAM30°,当BMAM时,MB有最小值,然后用含有30°角的直角三角形的性质求出MB

【解答过程】解:连接AMCMEM,如图:

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

矩形CDEFMDF的中点,

CME共线,

DM <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> DF <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> CECM

∵△ACD是等边三角形,

∴∠DAC60°ADAC

在△ADM和△ACM中,

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

∴△ADM≌△ACMSSS),

∴∠DAM=∠CAM

∵∠DAC60°

∴∠CAM30°

BMAM时,MB有最小值,

此时,BM <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> AB <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 126

故答案为:6

23.(北仑区二模)如图,△ABC的边AB3AB边上的中线CM1,分别以ACBC为边向外作正方形ACGH与正方形BCDE,连接GD,取GD中点N.则点N到线段AB的距离最大值为  <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>  

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解题思路】当GDAB时,N点到AB的距离最大,则ACBC,∴NCM三点共线且MNAB,通过证明△AMC≌△GOC,可以求出AM,然后再证明出OCNG是矩形,从而求出MN

【解答过程】解:∵点NAB的距离介于GDAB的距离之间,

GDAB时,N点到AB的距离最大,

ACBC

NCM三点共线且MNAB

过点CCPAB,作GOCPO为垂足,

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

PCAB

∴∠PCA=∠CAM,∠PCA+∠OCG90°,∠OGC+∠OCG90°

∴∠OGC=∠PCA=∠CAM

在△AMC和△GOC中,

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

∴△AMC≌△GOCAAS),

GOAM <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> AB <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

GOPCMNABPCAB

PCMNMNGD

四边形GDCN是矩形,

GONC

MNCM+CN

CM1GONC <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

MN1 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

故答案为: <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

24.(眉山)如图,在菱形ABCD中,ABAC10,对角线ACBD相交于点O,点M在线段AC上,且AM3,点P为线段BD上的一个动点,则MP <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> PB的最小值是  <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>  

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解题思路】过点PPEBCE,由菱形的性质可得ABBCAC10,∠ABD=∠CBD,可证△ABC是等边三角形,可求∠CBD30°,由直角三角形的性质可得PE <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> PB,则MP <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> PBPM+PE,即当点M,点P,点E共线且MEBC时,PM+PE有最小值为ME,由锐角三角函数可求解.

【解答过程】解:如图,过点PPEBCE

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

四边形ABCD是菱形,ABAC10

ABBCAC10,∠ABD=∠CBD

∴△ABC是等边三角形,

∴∠ABC=∠ACB60°

∴∠CBD30°

PEBC

PE <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> PB

MP <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> PBPM+PE

当点M,点P,点E共线且MEBC时,PM+PE有最小值为ME

AM3

MC7

sin∠ACB <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

ME <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

MP <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> PB的最小值为 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

故答案为 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

25.(海安市二模)如图,矩形ABCD中,AB2BC4E在边BC上运动,MN在对角线BD上运动,且MN <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> ,连接CMEN,则CM+EN的最小值为  <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>  

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解题思路】先作C点关于BD的对称点F,然后再把F左移2个单位,下移1个单位,得到Q,再过QQEBCE,交BDN,连接BF,过FFPBCP,以B为原点建立平面直角坐标系,求出F的坐标,再求出Q的坐标,即可得出答案.

【解答过程】解:先作C点关于BD的对称点F,然后再把F左移2个单位,下移1个单位,得到Q,再过QQEBCE,交BDN,连接BF,过FFPBCP,以B为原点建立平面直角坐标系,如图所示,

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>


AB2CDBC4

C40),BFBC4

由勾股定理得:BD <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

由三角形面积公式得: <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> CR×BD <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> BC×CD

CR <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

CF2CR <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

由勾股定理得:BF2BP2CF2CP2

42BP2=( <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 24﹣BP2

解得:BP <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

FP <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

F的坐标是( <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>  <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> ),

Q的坐标是( <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>  <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> ),

CM+EN的最小值为 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

故答案为: <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

26.(浙江自主招生)如图,正方形ABCD的边长为1,点P为边BC上任意一点(可与B点或C点重合),分别过BCD作射线AP的垂线,垂足分别是BCD,则BB′+CC′+DD的最大值为 2 ,最小值为  <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>  

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解题思路】连接ACDP,根据三角形的面积公式得出SDPCSAPC <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> AP×CC,根据S正方形ABCDSABP+SADP+SDPC,推出BB′+DD′+CC <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> ,根据已知得出1≤AP <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

代入求出即可.

【解答过程】解: <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

连接ACDP

S正方形ABCD1×11

由勾股定理得:AC <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

AB1

1≤AP <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

∵△DPC和△APC的边CP上的高DCAB

SDPCSAPC <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> AP×CC

1S正方形ABCDSABP+SADP+SDPC <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> APBB′+DD′+CC),

BB′+DD′+CC <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

1≤AP <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> BB′+CC′+DD′≤2

故答案为:2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

27.(乾县一模)如图,在菱形ABCD中,∠BAD120°,点E为边AB的中点,点P在对角线BD上且PE+PA6,则AB长的最大值为  <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>  

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解题思路】连接PCCEAC;由已知条件可以得出PE+PCPE+PA6≥CE(当PAEDB的交点时取等号),再利用等边三角形的性质得出CE <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> AB,进而求出AB长的最大值.

【解答过程】解:连接PCCEAC,如图所示:

四边形ABCD是菱形,

ABBCAPPC

PE+PCPE+PA6≥CE

∵∠DAB120°

∴∠ABC60°

∴△ABC是等边三角形,

E为线段AB的中点,

AEBE

∴∠AEC90°,∠BCE30°

CE <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> BC <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> AB≤6

所以AB <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

AB长的最大值是 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

故答案为: <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

28.(寿光市二模)如图所示,四边形ABCD中,ACBD于点OAOCO4BODO3,点P为线段AC上的一个动点.过点P分别作PMAD于点M,作PNDC于点N.连接PB,在点P运动过程中,PM+PN+PB的最小值等于 7.8 

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解题思路】证四边形ABCD是菱形,得CDAD5,连接PD,由三角形面积关系求出PM+PN4.8,得当PB最短时,PM+PN+PB有最小值,则当BPAC时,PB最短,即可得出答案.

【解答过程】解:∵AOCO4BODO3

AC8,四边形ABCD是平行四边形,

ACBD于点O

平行四边形ABCD是菱形,AD <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 5

CDAD5

连接PD,如图所示:

SADP+SCDPSADC

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> ADPM <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> DCPN <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> ACOD

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> PM <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> PN <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 8×3

PM+PN)=8×3

PM+PN4.8

PB最短时,PM+PN+PB有最小值,

由垂线段最短可知:当BPAC时,PB最短,

当点P与点O重合时,PM+PN+PB有最小值,最小值=4.8+37.8

故答案为:7.8

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

29.(河西区二模)已知正方形ABCD的边长为2EF分别是边BCCD上的两个动点,且满足BECF,连接AEAF,则AE+AF的最小值为 2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>  

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解题思路】连接DE,作点A关于BC的对称点A,连接BAEA,易得AE+AFAE+DEA'E+DE,当DEA在同一直线时,AE+AF最小,利用勾股定理求解即可.

【解答过程】解:连接DE,作点A关于BC的对称点A,连接BAEA

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

四边形ABCD为正方形,

ADCDBC,∠ADC=∠BCD90°

BECF

DFCE

在△DCE与△ADF中,

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

∴△DCE≌△ADFSAS),

DEAF

AE+AFAE+DE

作点A关于BC的对称点A,连接BAEA

AEAE

AE+AFAE+DEA'E+DE

DEA在同一直线时,AE+AF最小,

AA2AB4

此时,在Rt△ADA中,DA <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

AE+AF的最小值为2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

故答案为:2 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

30.(鹿城区校级期中)学习新知:如图1、图2P是矩形ABCD所在平面内任意一点,则有以下重要结论:AP2+CP2BP2+DP2.该结论的证明不难,同学们通过勾股定理即可证明.

应用新知:如图3,在△ABC中,CA4CB6D是△ABC内一点,且CD2,∠ADB90°,则AB的最小值为 4 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 2 

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

【解题思路】以ADBD为边作矩形ADBE,连接CEDE,由矩形的性质得出ABDE,由题意得CD2+CE2CA2+CB2,求出CE4 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> ,当CDE三点共线时,DE最小,得出AB的最小值=DE的最小值=CECD4 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 2

【解答过程】解:以ADBD为边作矩形ADBE,连接CEDE,如图所示:

ABDE

由题意得:CD2+CE2CA2+CB2

22+CE242+62

解得:CE4 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>

CDE三点共线时,DE最小,

AB的最小值=DE的最小值=CECD4 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 2

故答案为:4 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a> 2

 <a href="/tags/55/" title="数学" class="c1" target="_blank">数学</a> <a href="/tags/129/" title="四边形" class="c1" target="_blank">四边形</a> <a href="/tags/245/" title="训练" class="c1" target="_blank">训练</a> <a href="/tags/282/" title="专题" class="c1" target="_blank">专题</a>


11