1．在实验室中，通常将金属钠保存在

A．水中　　　 B. 煤油中　　　　　 C. 四氯化碳中　　 D.汽油中

102.(08广东佛山25题)25．我们所学的几何知识可以理解为对“构图”的研究：根据给定的(或构造的)几何图形提出相关的概念和问题(或者根据问题构造图形)，并加以研究.

例如：在平面上根据两条直线的各种构图，可以提出“两条直线平行”、“两条直线相交”的概念；若增加第三条直线，则可以提出并研究“两条直线平行的判定和性质”等问题(包括研究的思想和方法).

请你用上面的思想和方法对下面关于圆的问题进行研究：

(1) 如图1，在圆O所在平面上，放置一条直线(和圆O分别交于点A、B)，根据这个图形可以提出的概念或问题有哪些(直接写出两个即可)？

(2) 如图2，在圆O所在平面上，请你放置与圆O都相交且不同时经过圆心的两条直线和(与圆O分别交于点A、B，与圆O分别交于点C、D).

请你根据所构造的图形提出一个结论，并证明之.

(3) 如图3，其中AB是圆O的直径，AC是弦，D是的中点，弦DE⊥AB于点F. 请找出点C和点E重合的条件，并说明理由.

(08广东佛山25题解答)解：(1) 弦(图中线段AB)、弧(图中的ACB弧)、弓形、求弓形的面积(因为是封闭图形)等.　 (写对一个给1分，写对两个给2分)

(2) 情形1　 如图21，AB为弦，CD为垂直于弦AB的直径.　 …………………………3分

结论：(垂径定理的结论之一).　 …………………………………………………………4分

证明：略(对照课本的证明过程给分).　 …………………………………………………7分

情形2　 如图22，AB为弦，CD为弦，且AB与CD在圆内相交于点P.

结论：.

证明：略.

情形3 (图略)AB为弦，CD为弦，且与在圆外相交于点P.

结论：.

证明：略.

情形4　 如图23，AB为弦，CD为弦，且AB∥CD.

结论：　 =　　 .

证明：略.

(上面四种情形中做一个即可，图1分，结论1分，证明3分；

其它正确的情形参照给分；若提出的是错误的结论，则需证明结论是错误的)

(3) 若点C和点E重合，

则由圆的对称性，知点C和点D关于直径AB对称. …………………………………8分

设，则，.………………………………9分

又D是　　 的中点，所以，

即.………………………………………………………10分

解得.……………………………………………………………11分

(若求得或等也可，评分可参照上面的标准；也可以先直觉猜测点B、C是圆的十二等分点，然后说明)

101.(08山东聊城25题)25．(本题满分12分)如图，把一张长10cm，宽8cm的矩形硬纸板的四周各剪去一个同样大小的正方形，再折合成一个无盖的长方体盒子(纸板的厚度忽略不计)．

(1)要使长方体盒子的底面积为48cm²，那么剪去的正方形的边长为多少？

(2)你感到折合而成的长方体盒子的侧面积会不会有更大的情况？如果有，请你求出最大值和此时剪去的正方形的边长；如果没有，请你说明理由；

(3)如果把矩形硬纸板的四周分别剪去2个同样大小的正方形和2个同样形状、同样大小的矩形，然后折合成一个有盖的长方体盒子，是否有侧面积最大的情况；如果有，请你求出最大值和此时剪去的正方形的边长；如果没有，请你说明理由．

(08山东聊城25题解答)(本题满分12分)

解：(1)设正方形的边长为cm，则

．························································································ 1分

即．

解得(不合题意，舍去)，．

剪去的正方形的边长为1cm．·············································································· 3分

(注：通过观察、验证直接写出正确结果给3分)

(2)有侧面积最大的情况．

设正方形的边长为cm，盒子的侧面积为cm²，

则与的函数关系式为：

．

即．······························································································· 5分

改写为．

当时，．

即当剪去的正方形的边长为2.25cm时，长方体盒子的侧面积最大为40.5cm²．········ 7分

(3)有侧面积最大的情况．

设正方形的边长为cm，盒子的侧面积为cm²．

若按图1所示的方法剪折，则与的函数关系式为：

．

即．

当时，．····························· 9分

若按图2所示的方法剪折，则与的函数关系式为：

．

即．

当时，．················································································· 11分

比较以上两种剪折方法可以看出，按图2所示的方法剪折得到的盒子侧面积最大，即当剪去的正方形的边长为cm时，折成的有盖长方体盒子的侧面积最大，最大面积为cm²．

说明：解答题各小题只给了一种解答及评分说明，其他解法只要步骤合理，解答正确，均应给出相应分数．

100.(08广东梅州23题)23．本题满分11分．

如图11所示，在梯形ABCD中，已知AB∥CD， AD⊥DB，AD=DC=CB，AB=4．以AB所在直线为轴，过D且垂直于AB的直线为轴建立平面直角坐标系．

(1)求∠DAB的度数及A、D、C三点的坐标；

(2)求过A、D、C三点的抛物线的解析式及其对称轴L．

(3)若P是抛物线的对称轴L上的点，那么使PDB为等腰三角形的点P有几个?(不必求点P的坐标，只需说明理由)

(08广东梅州23题解答)解： (1) DC∥AB，AD=DC=CB，

　∠CDB=∠CBD=∠DBA，··············································································· 0.5分

　　 ∠DAB=∠CBA， ∠DAB=2∠DBA， ············ 1分

∠DAB+∠DBA=90， ∠DAB=60， ·········· 1.5分

　 ∠DBA=30，AB=4， DC=AD=2，　 ········· 2分

RtAOD，OA=1，OD=，··························· 2.5分

A(-1，0)，D(0， )，C(2， )．　 · 4分

(2)根据抛物线和等腰梯形的对称性知，满足条件的抛物线必过点A(－1，0)，B(3，0)，

故可设所求为　 = (+1)( -3) ······························································ 6分

将点D(0， )的坐标代入上式得， =．

所求抛物线的解析式为　 =　　 ·········································· 7分

其对称轴L为直线=1．······················································································ 8分

(3) PDB为等腰三角形，有以下三种情况：

①因直线L与DB不平行，DB的垂直平分线与L仅有一个交点P₁，P₁D=P₁B，

P₁DB为等腰三角形；　 ················································································· 9分

②因为以D为圆心，DB为半径的圆与直线L有两个交点P₂、P₃，DB=DP₂，DB=DP₃， P₂DB， P₃DB为等腰三角形；

③与②同理，L上也有两个点P₄、P₅，使得 BD=BP₄，BD=BP₅．　 ···················· 10分

由于以上各点互不重合，所以在直线L上，使PDB为等腰三角形的点P有5个．

99.(08福建南平26题)26．(14分)

(1)如图1，图2，图3，在中，分别以为边，向外作正三角形，正四边形，正五边形，相交于点．

①如图1，求证：；

②探究：如图1，　　　　 ；

如图2，　　　　 ；

如图3，　　　　 ．

(2)如图4，已知：是以为边向外所作正边形的一组邻边；是以为边向外所作正边形的一组邻边．的延长相交于点．

①猜想：如图4，　　　　 (用含的式子表示)；

②根据图4证明你的猜想．

(08福建南平26题解答)(1)①证法一：与均为等边三角形，

，························································································ 2分

且··············································· 3分

，

即························································ 4分

．··················································· 5分

证法二：与均为等边三角形，

，························································································ 2分

且························································································ 3分

可由绕着点按顺时针方向旋转得到··································· 4分

．··························································································· 5分

②，，．········································································ 8分(每空1分)

(2)①········································································································ 10分

②证法一：依题意，知和都是正边形的内角，，，

，即．····························· 11分

．·························································································· 12分

，，······ 13分

，

········································ 14分

证法二：同上可证　 ．··························································· 12分

，如图，延长交于，

，

································ 13分

················· 14分

证法三：同上可证　 ．··························································· 12分

．

，

························································ 13分

即········································································ 14分

证法四：同上可证　 ．··························································· 12分

．如图，连接，

．···································· 13分

即······························· 14分

注意：此题还有其它证法，可相应评分．

98.(08四川资阳24题)24．(本小题满分12分)

如图10，已知点A的坐标是(－1，0)，点B的坐标是(9，0)，以AB为直径作⊙O′，交y轴的负半轴于点C，连接AC、BC，过A、B、C三点作抛物线．

(1)求抛物线的解析式；

(2)点E是AC延长线上一点，∠BCE的平分线CD交⊙O′于点D，连结BD，求直线BD的解析式；

(3)在(2)的条件下，抛物线上是否存在点P，使得∠PDB＝∠CBD?如果存在，请求出点P的坐标；如果不存在，请说明理由．

(08四川资阳24题解答)(1) ∵以AB为直径作⊙O′，交y轴的负半轴于点C，

∴∠OCA+∠OCB=90°，

又∵∠OCB+∠OBC=90°，

∴∠OCA=∠OBC，

又∵∠AOC= ∠COB=90°，

∴ΔAOC∽ ΔCOB，························································································ 1分

∴．

又∵A(–1，0)，B(9，0)，

∴，解得OC=3(负值舍去)．

∴C(0，–3)，

······················································································································ 3分

设抛物线解析式为y=a(x+1)(x–9)，

∴–3=a(0+1)(0–9)，解得a=，

∴二次函数的解析式为y=(x+1)(x–9)，即y=x²–x–3．···························· 4分

(2) ∵AB为O′的直径，且A(–1，0)，B(9，0)，

∴OO′=4，O′(4，0)，····················································································· 5分

∵点E是AC延长线上一点，∠BCE的平分线CD交⊙O′于点D，

∴∠BCD=∠BCE=×90°=45°，

连结O′D交BC于点M，则∠BO′D=2∠BCD=2×45°=90°，OO′=4，O′D=AB=5．

∴D(4，–5)．································································································· 6分

∴设直线BD的解析式为y=kx+b(k≠0)

∴··························································· 7分

解得

∴直线BD的解析式为y=x–9.····································· 8分

(3) 假设在抛物线上存在点P，使得∠PDB=∠CBD，

解法一：设射线DP交⊙O′于点Q，则．

分两种情况(如答案图1所示)：

①∵O′(4，0)，D(4，–5)，B(9，0)，C(0，–3)．

∴把点C、D绕点O′逆时针旋转90°，使点D与点B重合，则点C与点Q₁重合，

因此，点Q₁(7，–4)符合，

∵D(4，–5)，Q₁(7，–4)，

∴用待定系数法可求出直线DQ₁解析式为y=x–．··································· 9分

解方程组得

∴点P₁坐标为(，)，[坐标为(，)不符合题意，舍去]．

······················································································································ 10分

②∵Q₁(7，–4)，

∴点Q₁关于x轴对称的点的坐标为Q₂(7，4)也符合．

∵D(4，–5)，Q₂(7，4)．

∴用待定系数法可求出直线DQ₂解析式为y=3x–17．······································ 11分

解方程组得

∴点P₂坐标为(14，25)，[坐标为(3，–8)不符合题意，舍去]．

······················································································································ 12分

∴符合条件的点P有两个：P₁(，)，P₂(14，25)．

解法二：分两种情况(如答案图2所示)：

①当DP₁∥CB时，能使∠PDB=∠CBD．

∵B(9，0)，C(0，–3)．

∴用待定系数法可求出直线BC解析式为y=x–3．

又∵DP₁∥CB，∴设直线DP₁的解析式为y=x+n．

把D(4，–5)代入可求n= –，

∴直线DP₁解析式为y=x–．························· 9分

解方程组得

∴点P₁坐标为(，)，[坐标为(，)不符合题意，舍去]．

······················································································································ 10分

②在线段O′B上取一点N，使BN=DM时，得ΔNBD≌ΔMDB(SAS)，∴∠NDB=∠CBD．

由①知，直线BC解析式为y=x–3．

取x=4，得y= –，∴M(4，–)，∴O′N=O′M=，∴N(，0)，

又∵D(4，–5)，

∴直线DN解析式为y=3x–17．······································································ 11分

解方程组得

∴点P₂坐标为(14，25)，[坐标为(3，–8)不符合题意，舍去]．

······················································································································ 12分

∴符合条件的点P有两个：P₁(，)，P₂(14，25)．

解法三：分两种情况(如答案图3所示)：

①求点P₁坐标同解法二．··············································································· 10分

②过C点作BD的平行线,交圆O′于G,

此时，∠GDB=∠GCB=∠CBD．

由(2)题知直线BD的解析式为y=x–9,

又∵ C(0，–3)

∴可求得CG的解析式为y=x–3,

设G(m,m–3)，作GH⊥x轴交与x轴与H，

连结O′G,在Rt△O′GH中，利用勾股定理可得，m=7，

由D(4，–5)与G(7，4)可得，

DG的解析式为，··········································································· 11分

解方程组得

∴点P₂坐标为(14，25)，[坐标为(3，–8)不符合题意，舍去]．························ 12分

∴符合条件的点P有两个：P₁(，)，P₂(14，25)．

说明：本题解法较多，如有不同的正确解法，请按此步骤给分.

97.(08新疆自治区24题)(10分)某工厂要赶制一批抗震救灾用的大型活动板房．如图，板房一面的形状是由矩形和抛物线的一部分组成，矩形长为12m，抛物线拱高为5.6m．

(1)在如图所示的平面直角坐标系中，求抛物线的表达式．

(2)现需在抛物线AOB的区域内安装几扇窗户，窗户的底边在AB上，每扇窗户宽1.5m，高1.6m，相邻窗户之间的间距均为0.8m，左右两边窗户的窗角所在的点到抛物线的水平距离至少为0.8m．请计算最多可安装几扇这样的窗户？

(08新疆自治区24题解析)24．(10分)解：(1)设抛物线的表达式为　 1分

点在抛物线的图象上．

∴

······························································ 3分

∴抛物线的表达式为············································································· 4分

(2)设窗户上边所在直线交抛物线于C、D两点，D点坐标为(k，t)

已知窗户高1.6m，∴··························································· 5分

(舍去)············································································ 6分

∴(m)·············································································· 7分

又设最多可安装n扇窗户

∴····················································································· 9分

．

答：最多可安装4扇窗户．···················································································· 10分

(本题不要求学生画出4个表示窗户的小矩形

96.(08四川自贡26题)抛物线的顶点为M，与轴的交点为A、B(点B在点A的右侧)，△ABM的三个内角∠M、∠A、∠B所对的边分别为m、a、b。若关于的一元二次方程有两个相等的实数根。

(1)判断△ABM的形状，并说明理由。

(2)当顶点M的坐标为(－2，－1)时，求抛物线的解析式，并画出该抛物线的大致图形。

(3)若平行于轴的直线与抛物线交于C、D两点，以CD为直径的圆恰好与轴相切，求该圆的圆心坐标。

(08四川自贡26题解析)解：(1)令

　　　　　　 得

　　　　 由勾股定理的逆定理和抛物线的对称性知

△ABM是一个以、为直角边的等腰直角三角形

　　　　 (2)设

∵△ABM是等腰直角三角形

∴斜边上的中线等于斜边的一半

又顶点M(－2，－1)

∴，即AB＝2

∴A(－3，0)，B(－1，0)

将B(－1，0) 代入中得

∴抛物线的解析式为，即

图略

(3)设平行于轴的直线为

解方程组错误！不能通过编辑域代码创建对象。

得，　 (

∴线段CD的长为

∵以CD为直径的圆与轴相切

据题意得

∴

解得

∴圆心坐标为和

95.(08四川巴中30题)(12分)30．已知：如图14，抛物线与轴交于点，点，与直线相交于点，点，直线与轴交于点．

(1)写出直线的解析式．

(2)求的面积．

(3)若点在线段上以每秒1个单位长度的速度从向运动(不与重合)，同时，点在射线上以每秒2个单位长度的速度从向运动．设运动时间为秒，请写出的面积与的函数关系式，并求出点运动多少时间时，的面积最大，最大面积是多少？

(08四川巴中30题解析)解：(1)在中，令

，

，··············································· 1分

又点在上

的解析式为·············································································· 2分

(2)由，得　 ···················································· 4分

，

，······························································································· 5分

························································································· 6分

(3)过点作于点

······························································································· 7分

·········································································································· 8分

由直线可得：

在中，，，则

，······················································································· 9分

···················································································· 10分

····························································································· 11分

此抛物线开口向下，当时，

当点运动2秒时，的面积达到最大，最大为．···························· 12分

94.(08山东济宁26题)(12分)

中，，，cm．长为1cm的线段在的边上沿方向以1cm/s的速度向点运动(运动前点与点重合)．过分别作的垂线交直角边于两点，线段运动的时间为s．

(1)若的面积为，写出与的函数关系式(写出自变量的取值范围)；

(2)线段运动过程中，四边形有可能成为矩形吗？若有可能，求出此时的值；若不可能，说明理由；

(3)为何值时，以为顶点的三角形与相似？

(08山东济宁26题解析)解：(1)当点在上时，，．

．········································································ 2分

当点在上时，．

．·················································· 4分

(2)，．．

．········································································ 6分

由条件知，若四边形为矩形，需，即，

．

当s时，四边形为矩形．································································· 8分

(3)由(2)知，当s时，四边形为矩形，此时，

．··························································································· 9分

除此之外，当时，，此时．

，．．····························· 10分

，．

又，．········································ 11分

，．

当s或s时，以为顶点的三角形与相似．··················· 12分