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2008年全国中考数学压轴题精选(十)

91.（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？

2（08新疆自治区24题解析）24．（10分）解：（1）设抛物线的表达式为y?ax 1分 点B(6，?5.6)在抛物线的图象上． ∴?5.6?36a

7 ····················································· 3分 4572∴抛物线的表达式为y??································································· 4分 x ·

45a??（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，∴t??5.6?(?1.6)??4 ·················································· 5分

?4??72k 45k1≈5.07，k2≈?5.07（舍去） ································································· 6分

∴CD?5.07?2≈10.14（m） ··································································· 7分 又设最多可安装n扇窗户

∴1.5n?0.8(n?1)≤10.14 ········································································· 9分

n≤4.06．

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）

92.（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．

图10

（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 OAOC∴． ?OCOB又∵A(–1，0)，B(9，0)，

1OC∴，解得OC=3(负值舍去)． ?OC9∴C(0，–3)， ····································································································· 3分 设抛物线解析式为y=a(x+1)(x–9)，

1∴–3=a(0+1)(0–9)，解得a=，

3118∴二次函数的解析式为y=(x+1)(x–9)，即y=x2–x–3． ························ 4分

333(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，

11∴∠BCD=∠BCE=×90°=45°，

221连结O′D交BC于点M，则∠BO′D=2∠BCD=2×45°=90°，OO′=4，O′D=AB=5．

2∴D(4，–5)． ··················································································· 6分 ∴设直线BD的解析式为y=kx+b（k≠0） ?9k?b?0,∴? ·················································· 7分

4k?b??5.??k?1,解得?

b??9.?∴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BQ?CD． 分两种情况(如答案图1所示)：

图10答案图1 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BQ?CD， ∵D(4，–5)，Q1(7，–4)，

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

33??1199?419?41?y?x?，x?，x?，?1?2????3322解方程组?得? ?18?29?41??29?41?y?x2?x?3.?y?；y?.12???33?66??9?41?29?419?41?29?41，)，[坐标为(，)不符合题意，舍去]． 2626 ····································································································· 10分 ②∵Q1(7，–4)，

∴点P1坐标为(∴点Q1关于x轴对称的点的坐标为Q2(7，4)也符合BQ?CD． ∵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

?y?3x?17，?x?3，?x2?14，?解方程组?得?1 ?128y??8；y?25.y?x?x?3.?1?2?33?∴点P2坐标为(14，25)，[坐标为(3，–8)不符合题意，舍去]． ····································································································· 12分

9?41?29?41∴符合条件的点P有两个：P1(，)，P2(14，25)．

26

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)．

1∴用待定系数法可求出直线BC解析式为y=x–3．

31又∵DP1∥CB，∴设直线DP1的解析式为y=x+n．

319把D(4，–5)代入可求n= –，

3119∴直线DP1解析式为y=x–． ····················· 9分 图10答案图2 33??1199?419?41?y?x?，x?，x?，??12????3322解方程组?得? ?18?29?41??29?41?y?x2?x?3.?y?；y?.12???33?66??9?41?29?419?41?29?41，)，[坐标为(，)不符合题意，舍去]． 2626 ····································································································· 10分

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

1由①知，直线BC解析式为y=x–3．

355517取x=4，得y= –，∴M(4，–)，∴O′N=O′M=，∴N(，0)，

3333又∵D(4，–5)，

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

?y?3x?17，?x1?3，?x2?14，?解方程组?得 ??128y??8；y?25.y?x?x?3.?1?2?33?∴点P2坐标为(14，25)，[坐标为(3，–8)不符合题意，舍去]．

∴点P1坐标为( ····································································································· 12分

9?41?29?41∴符合条件的点P有两个：P1(，)，P2(14，25)．

26

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y?3x?17， ································································ 11分 图10答案?y?3x?17，?x1?3，?x2?14，?解方程组?得 ??128y??8；y?25.y?x?x?3.?1?2?33?∴点P2坐标为(14，25)，[坐标为(3，–8)不符合题意，舍去]． ···················· 12分

9?41?29?41∴符合条件的点P有两个：P1(，)，P2(14，25)．

26说明：本题解法较多，如有不同的正确解法，请按此步骤给分.

93.（08福建南平26题）26．（14分）

（1）如图1，图2，图3，在△ABC中，分别以AB，AC为边，向△ABC外作正三角形，正四边形，正五边形，BE，CD相交于点O．

①如图1，求证：△ABE≌△ADC； ②探究：如图1，?BOC? ； 如图2，?BOC? ； 如图3，?BOC? ．

（2）如图4，已知：AB，AD是以AB为边向△ABC外所作正n边形的一组邻边；AC，AE是以AC为边向△ABC外所作正n边形的一组邻边．BE，CD的延长相交于点O．

①猜想：如图4，?BOC? （用含n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：△ABD与△ACE均为等边三角形，

·········································································· 2分 ?AD?AB，AC?AE ·

且?BAD??CAE?60 ········································ 3分

??BAD??BAC??CAE??BAC，

即?DAC??BAE ················································ 4分

·········································· 5分 ?△ABE≌△ADC． ·

证法二：△ABD与△ACE均为等边三角形，

·········································································· 2分 ?AD?AB，AC?AE ·

且?BAD??CAE?60 ··········································································· 3分 ····························· 4分 ?△ADC可由△ABE绕着点A按顺时针方向旋转60得到 ·

············································································· 5分 ?△ABE≌△ADC． ·

②120，90，72． ······························································ 8分（每空1分）

（2）①

360 ·························································································· 10分 n②证法一：依题意，知?BAD和?CAE都是正n边形的内角，AB?AD，AE?AC，

??BAD??CAE?(n?2)180

n························ 11分 ??BAD??DAE??CAE??DAE，即?BAE??DAC． ·

············································································ 12分 ?△ABE≌△ADC． ·

······ 13分 ??ABE??ADC，?ADC??ODA?180，??ABO??ODA?180

?ABO??ODA??DAB??BOC?360，??BOC??DAB?180 ??BOC?180??DAB?180?(n?2)180360? ··································· 14分

nn证法二：同上可证 △ABE··················································· 12分 ≌△AD．C ·

??ABE??ADC，如图，延长BA交CO于F，

?AFD??ABE??BOC?180，

?AFD??ADC??DAF?180 ···························· 13分 ??BOC??DAF?180??BAD?360 ·············· 14分 n证法三：同上可证 △ABE··················································· 12分 ≌△AD．C ·

??ABE??ADC．?BOC?180?(?ABE??ABC??ACB??ACD)

??BOC?180?(?ADC??ABC??ACB??ACD)