上海市各区2018届中考二模数学分类汇编:压轴题专题(含答案) 下载本文

'.

青浦区

25.(本题满分14分,第(1)小题4分,第(2)小题6分,第(3)小题4分)

如图9-1,已知扇形MON的半径为2,∠MON=90,点B在弧MN上移动,联结BM,作OD?BM,垂足为点D,C为线段OD上一点,且OC=BM,联结BC并延长交半径OM于点A,设OA= x,∠COM的正切值为y.

(1)如图9-2,当AB?OM时,求证:AM =AC; (2)求y关于x的函数关系式,并写出定义域; (3)当△OAC为等腰三角形时,求x的值.

25.解:(1)∵OD⊥BM,AB⊥OM,∴∠ODM =∠BAM =90°. ······································· (1分)

∵∠ABM +∠M =∠DOM +∠M,∴∠ABM =∠DOM. ································· (1分) ∵∠OAC=∠BAM,OC =BM,

∴△OAC≌△ABM, ························································································· (1分) ∴AC =AM. ······································································································· (1分) (2)过点D作DE//AB,交OM于点E. ······························································· (1分)

∵OB=OM,OD⊥BM,∴BD=DM. ···························································· (1分) ∵DE//AB, ∴

OCADMOCDMOMNBNBN

A图9-1 图9-2

备用图

MDME,∴AE=EM, ?DMAE1∵OM=2,∴AE=······························································· (1分) 2?x. ·

2??∵DE//AB, ∴

OAOC2DM, ················································································ (1分) ??OEODOD;.

'.

DMOA, ?OD2OEx.(0?x?2) ······································································ (2分)

x?2∴y?(3)(i) 当OA=OC时, ∵DM?111BM?OC?x, 2222?在Rt△ODM中,OD?OM2?DM2?DM12x.∵y?, 4OD1x14?2?14?2x2∴.解得x?,或x?(舍).(2分) ?221x?22?x24(ii)当AO=AC时,则∠AOC =∠ACO,

∵∠ACO >∠COB,∠COB =∠AOC,∴∠ACO >∠AOC,

∴此种情况不存在. ························································································ (1分) (ⅲ)当CO=CA时,

则∠COA =∠CAO=?,

∵∠CAO >∠M,∠M=90???,∴?>90???,∴?>45?,

∴?BOA?2??90?,∵?BOA?90?,∴此种情况不存在. ··········· (1分)

松江区

25.(本题满分14分,第(1)小题4分,第(2)小题每个小题各5分)

如图,已知Rt△ABC 中,∠ACB=90°,BC=2,AC=3,以点C为圆心、CB为半径的圆交AB于点D,过点A作AE∥CD,交BC延长线于点E. (1)求CE的长;

(2)P是 CE延长线上一点,直线AP、CD交于点Q.

① 如果△ACQ ∽△CPQ,求CP的长;

② 如果以点A为圆心,AQ为半径的圆与⊙C相切,求CP的长.

D A

D A

;.

B

C

E B

C

E '.

25.(本题满分14分,第(1)小题4分,第(2)小题每个小题各5分) 解:(1)∵AE∥CD

∴BCDCBE?AE…………………………………1分 ∵BC=DC

∴BE=AE …………………………………1分 设CE=x 则AE=BE=x+2 ∵ ∠ACB=90°, ∴AC2?CE2?AE2

即9?x2?(x?2)2………………………1分

∴x?54 即CE?54…………………………………1分

(2)①

∵△ACQ ∽△CPQ,∠QAC>∠P ∴∠ACQ=∠P…………………………………1分 又∵AE∥CD ∴∠ACQ=∠CAE

∴∠CAE=∠P………………………………1分 ∴△ACE ∽△PCA,…………………………1分 ∴AC2?CE?CP…………………………1分

即32?54?CP ∴CP?365 ……………………………1分

;.

A D B

C

E (第25题图)

Q A D B

C E P

'.

②设CP=t,则PE?t?∵∠ACB=90°, ∴AP?9?t2 ∵AE∥CD

5 4AQEC……………………………1分 ?APEP5AQ5即 ?4?254t?5t?9t?4∴

5t2?9∴AQ?……………………………1分

4t?55t2?9?1 若两圆外切,那么AQ?4t?5此时方程无实数解……………………………1分

5t2?9?5 若两圆内切切,那么AQ?4t?5∴15t?40t?16?0 解之得t?又∵t?220?410………………………1分

155 4∴t?20?410………………………1分

15徐汇区

25. 已知四边形ABCD是边长为10的菱形,对角线AC、BD相交于点E,过点C作CF∥

DB交AB延长线于点F,联结EF交BC于点H.

(1)如图1,当EF?BC时,求AE的长;

(2)如图2,以EF为直径作⊙O,⊙O经过点C交边CD于点G(点C、G不重合),设AE的长为x,EH的长为y;

① 求y关于x的函数关系式,并写出定义域;

;.