中考数学压轴题详解—圆
更新时间:2023-05-26 02:01:01 阅读量: 实用文档 文档下载
复习圆
京翰提示:圆作为初中数学中重要的知识点,在历年高考题中都出现在重要的得分点高的部分,尤其是压轴题中,有些同学往往认为压轴题一定是很难很难得到分数的部分,其实在题目中往往前一到两个小题都是考察大家的基础知识,只要正确列出公式就能得到相应的分数。要学好圆的部分,不仅要靠平时的练习,最重要的还是回归课本,把基础知识参透,只有基础牢固了,才能进一步对圆的认识进行延伸和扩展。
1 如图,将△AOB置于平面直角坐标系中,其中点O为坐标原点,点A的坐标为(3,0),∠ABO=60°. (1)若△AOB的外接圆与y轴交于点D,求D点坐标.
(2)若点C的坐标为(-1,0),试猜想过D、C的直线与△AOB的外接圆的位置关系,并加以说明. (3)二次函数的图象经过点O和A且顶点在圆上,求此函数的解析式.
O为圆心、OA1为半径作扇形OAC2 如图(4),正方形OA与OB1相交于点B2,设正AC1B1C1的边长为1,以11,11
O为圆心,、方形OA1B1C1与扇形OAC11之间的阴影部分的面积为S1;然后以OB2为对角线作正方形OA2B2C2,又以
OA2为半径作扇形OA2C2, A2C2与OB1相交于点B3,设正方形OA2B2C2与扇形OA2C2之间的阴影部分面积为S2;
按此规律继续作下去,设正方形OAnBnCn与扇形OAnCn之间的阴影部分面积为Sn.
1 C(1)求S1,S2,S3; (2)写出S2008; C(3)试猜想Sn(用含n的代数式表示,n为正整数). C
O
A3 A2 图4
A1
3 (10分)如图,点I是△ABC的内心,线段AI的延长线交△ ABC的外接圆于点D,交BC边于点E.
(1)求证:ID=BD;
(2)设△ABC的外接圆的半径为5,ID=6,AD x,DE y,当点A在优弧
并指出自变量x的取值范围.
上运动时,求y与x的函数关系式,
的中点,AC交BD于点E, AE=2, EC=1.
4 如图,点A,B,C,D是直径为AB的⊙O上四个点,C是劣弧BD
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/
复习圆
(1)求证:△DEC∽△ADC; (3分) (2)试探究四边形ABCD是否是梯形?若是,请你给予 证明并求出它的面积;若不是,请说明理由. (4分) (3)延长AB到H,使BH =OB. 求证:CH是⊙O的切线. (3分)
(第4题图)
上的一动点. 5 如图10,半圆O为△ABC的外接半圆,AC为直径,D为BC
(1)问添加一个什么条件后,能使得
BDBE
?请说明理由; BCBD
(2)若AB∥OD,点D所在的位置应满足什么条件?请说明理由;
(3)如图11,在 (1)和(2)的条件下,四边形AODB是什么特殊的四边形?证明你的结论.
6 如图1M是P不与M,D重合),以AB为直径作⊙O,过点P作⊙O的切线交BC于点F,切点为E.
(1)除正方形ABCD的四边和⊙O中的半径外,图中还有哪些相等的线段(不能添加字母和辅助线)? (2)求四边形CDPF的周长;
(3)延长CD,FP相交于点G,如图2所示. 是否存在点P,使BF*FG=CF*OF?如果存在,试求此时AP的长;
G 如果不存在,请说明理由.
A D M D
O O
C C O F B F
图2
7 如图,在平面直角坐标系xoy中,M是x轴正半轴上一点, M与x轴的正半轴交于A,B两点,A在B的左侧,且OA,OB的长是方程x 12x 27 0的两根,ON是 M的切线,N为切点,N在第四象限. (1)求 M的直径.
(2)求直线ON的解析式.
(3)在x轴上是否存在一点T,使△OTN是等腰三角形,若存在请在图2中标出T点所在位置,并画出△OTN(要求尺规作图,保留作图痕迹,不写作法,不证明,不求T的坐标)若不存在,请说明理由. 2
B 图1
复习圆
1 解:(1)连结AD. ∵∠ABO=60°,
∴∠ADO=60°…..1分
由点A的坐标为(3,0)得OA=3. ∵在Rt△ADO中有 cot∠ADO=
OD
,…………….2分 OA
∴OD=OA·cot∠ADO=3·cot60°=3
N F
∴点D的坐标为(0
3分
(2)DC与△AOB的外接圆相切于点D,理由如下: 由(1)得
∴AD
又∵C点坐标是(-1,0), ∴OC=1.
∴CD 2………………4分 ∵AC=OA+OC=3+1=4,
∴CD+AD=2=4=AC…………………5分 ∴∠ADC=90°,即AD⊥DC.
由∠AOD=90°得AD为圆的直径.
∴DC与△AOB的外接圆相切于点D……………6分
(说明:也可用解直角三角形或相似三角形等知识求解.) (3)由二次函数图象过点O(0,0)和A(3,0), 可设它的解析式为 y=ax(x-3)(a≠0).
如图,作线段OA的中垂线交△AOB的外接圆于E、F两点,交AD于M点,交OA于N点.
2
2
2
2
2
2
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复习圆
由抛物线的对称性及它的顶点在圆上可知,抛物线的顶点就是点E或F. ∵EF垂直平分OA, ∴EF是圆的直径. 又∵AD是圆的直径,
∴EF与AD的交点M是圆的圆心………….7分 由(1)、(2)得OA=3,
∴AN=
12OA=32,AM=FM=EM=1
2
∴MN ∴
2
=2
2
=2. ∴点E的坐标是(32
, 2,点F的坐标是(32
, -2
)……..8分
当点E为抛物线顶点时,
有
32(32
a=
∴y=
即y=
2
…………………………9分
当点F为抛物线顶点时,
有
3
32(2
-3)a=-2,
a=
9.
∴
x(x-3).
即
y=2
9x 3
x.
故二次函数的解析式为y=
3x2
或
y=29x 3
x ….10分
2 (1)S2
12
1 1 4
π
1 1 π
4
;
·····················S 2
1 2
1π2 2 4 π 2 2 8;
···················· 2
2
S 1 1π
3 22 4 π ················ 22
4 16;
(2)S1π
2008 22007 2
2009; ························京翰教育1对1家教 /
2分
4分
6分8分
复习圆
(3)Sn
1π (n为正整数). ··················· 10分 2n 12n 1
3 (1) 证明: 如图,
∵ 点I是△ABC的内心, ∴ ∠BAD=∠CAD,∠ABI=∠CBI. ………………2分 ∵ ∠CBD=∠CAD, ∴ ∠BAD=∠CBD. ……………………………3分 ∴ ∠BID=∠ABI+∠BAD =∠CBI+∠CBD=∠IBD. ∴ ID=BD. ………………………5分 (2)解:如图, ∵∠BAD=∠CBD=∠EBD, ∠D=∠D, ∴ △ABD∽△BED. …………………………7分
BDAD∴ . ∴ AD DE BD2 ID2. …………………8分
DEBD∵ ID=6,AD=x,DE=y,∴ xy=36. ………………9分 又∵ x=AD>ID=6, AD不大于圆的直径10, ∴ 6<x≤10.
36
∴ y与x的函数关系式是y .(6 x≤10) …………………………10分
x
说明:只要求对xy=36与6<x≤10,不写最后一步,不扣分.
的中点, 4 (1)证明:∵C是劣弧BD
∴ DAC CDB. ·············································· 1分 而 ACD公共,
∴△DEC∽△ADC. ··········································· 3分
DCEC
, ACDC
∵CE 1.AC AE EC 2 1 3,
2
∴DC AC EC 3 1 3 .
∴DC . ··················································································································· 4分
由已知BC DC AB是⊙O的直径, ∴ ACB 90 ,
(2)证明:连结OD,由⑴得∴AB AC CB 3
2
2
2
2
2
12.
∴AB
∴OD OB BC DC , ∴四边形OBCD是菱形. ∴DC∥AB,DC AB, ∴四边形ABCD是梯形. ···················································· 5分 法一:
过C作CF垂直AB于F,连结OC
,则OB BC OC ∴ OBC 60 . ··············································································································· 6分
CF3
,CF BC sin60 , BC2113∴S梯形ABCD=CF
AB+DC = ·········································· 7分
222∴sin60
法二:(接上证得四边形ABCD是梯形)
又DC∥AB ∴AD BC,连结OC,则△AOD,△DOC和△
OBC 6分
∴△AOD≌△DOC≌△OBC,
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复习圆
(3)证明:连结OC交BD于G由(2)得四边形OBCD是菱形, ∴OC BD且OG GC. ···························································································· 8分 又已知OB=BH , ∴BG∥CH. ········································································ 9分 ∴ OCH OGB 90 , ∴CH是⊙O的切线. ················································· 10分
5 解: (1)添加 AB=BD ····································································································· 2分
∴∠BDE =∠BCD ·∵AB=BD ∴ ············································································ 3分 AB=BD又∵∠DBE =∠DBC ∴△BDE∽△BCD
BDBE
································································································································ 4分 BCBD
的中点 ·(2)若AB∥DO,点D所在的位置是BC···························································· 5分
∵AB∥DO ∴∠ADO =∠BAD ·················································································· 6分
=DC ·∵∠ADO =∠OAD ∴∠OAD =∠BAD ∴DB······················································· 7分
(3)在(1)和(2)的条件下,.
=DC ∴∠∵ BDA =∠DAC ∴ BD∥OA AB=BD
又∵AB∥DO ∴四边形AODB是平行四边形 ················································· 9分 ∵OA=OD ∴平行四边形AODB是菱形 ······················································ 10分
6 解:(1)FB=FE ,PE=PA ···················································································· 2分
(2)四边形CDPF的周长为
FC+CD+DP+PE+EF=FC+CD+DP+PA+BF ·········································· 3分
=BF+FC+CD+DP+PA ········································ 4分 =BC+CD+DA ··············································· 5
分 =
3= ················································· 6分
(3)存在.
7分
OF,则 若BF FG CF
BFCF
OFFG
BFCF
∵ cos∠OFB= ,cos∠GFC=
OFFG
∴ ∠OFB=∠GFC
又 ∵ ∠OFB=∠OFE
∴ ∠OFE=∠OFB=∠GFC=60 ······································································· 8分 ∴ 在Rt△OFB中 FE=FB=∴ 在Rt△GFC中
CG
=CF
tan GFC CF tan60 1tan60 6∴ DG CG CD
6 ∴ DP DG ··········································· 9
分 tan PGD DG tan30 3 ∴
AP AD DP 3 3 ··············································· 10分
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OB
=1
tan60
复习圆
7 解:(1)解方程x2
12x 27 0,得x1 9,x2 3
A在B的左侧
OA 3,OB 9 AB OB OA 6 OM的直径为6 ················································································································· 1分 (2)过N作NC⊥OM,垂足为C, 连结MN,则MN⊥ON
sin∠MON MNOM 31
6 2
∠MON 30
又cos∠
MON
ON
OM
ON OM cos30 在Rt△OCN中
OC ON
cos30 9
2
CN ON sin30 12
2
N的坐标为 9 2, 2 ··································································································· 3分
(用其它方法求N的坐标,只要方法合理,结论正确,均可给分.) 设直线ON的解析式为y
kx
9
2x k 直线ON
的解析式为y x ······················································································ 4分 (3)如图2,T1,T2,T3,T4为所求作的点,△OT1N,△OT2N,△OT3N,△OT4N为所求等腰三角形.(每作出一种图形给一分) ·········································································································· 8分
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