发布时间 : 星期四 文章上海市普陀区2019年初三二模数学试卷(含答案)更新完毕开始阅读060684dfba68a98271fe910ef12d2af90342a881
21.解:
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2)∵
S?DE?DE1S△ADE1?S△ADE?3S△ABC?27AAHBCH?DEBCADEABC△ADE???S△ABC?BC?BC3S△ABC91?BC?AH?27BC?9AH?6AHBC?AHB??AHC?90?ABH22BH2AH6?BH?4CH?5ACH?AHC?90?tanC??DE?AHB?90?cotB?3AH3HC5S△ABC?2766BC?AED??Ctan?AED??AEDyxy?kx?b(k?0)55?6?20k?b,??5.6?28k?b.1??k??,20???b?7.111x?7y?4.84.8??x?7x?44x(?x?7)?200x1?40x2?100202020EDEC??ACE??BCD?DCE??BCAEC2?ED?EA?EEDCECAECEAyxy???DCE??CAE?BCA??CAEADBCAD?BCABCDABCDEDCECAECCD?EAACECAB?EAACAB?DCABCD?B??DCBADBC?EDC??DCBEDDCAB2?ED?BCCCHOBH?ABBC?EDC??B?ECD??ACBEDCABC2y??x?4mxyABA?6m,0?B?0,4m?(2分)
3∴OA?6m,OB?4m. ∵CH⊥OB,∴CHOACHBHBC??AB?2BCCH?3mBH?2m∴点C的坐OAOBAB标是??3m,6m?. ···················································································································· (1分) (2) ∵抛物线y??x2?bx?10经过点A、点C,
13?12??(6m)?6m?b?10?0,??3可得 ?1··································································· (2分)
???(?3m)2?3m?b?10?6m.??3?m?1,?∵m?0,解得 ?························································································ (1分) 1. ·
b??3?∴抛物线的表达式是y??x2?x?10.····························································· (1分)
1313(3)过点P分别作PQ⊥OA、垂足为点Q.
设点P的坐标为(n,?n2?n?10).可得OQ?n,PQ?n2?n?10. ∵S△PAB?2S△OBC,AB?2BC.
∴△PAB与△OBC等高,∴OPAB?BAO??POQtan?BAO?tan?POQ.
13131313121n?n?10233?. ·∴································································································· (1分)
n3解得 n1?3?1293?129,n2?(舍去). ····················································· (1分) 22?3?1293?129?∴点P的坐标是?.·································································· (1分) ,????23??25.解:
(1)在Rt△ABC中,?ACB?90?,cos?BAC?4AC4,∴?. 5AB5∵AB?5,∴AC?4. ··························································································· (1分) 由勾股定理得 BC?3. ···························································································· (1分) ∵OB?OA?x,∴CO?4?x. 在Rt△BCO中,?C?90?,
由勾股定理得 32?(4?x)2?x2. ············································································ (1分) 解得x?25. ············································································································· (1分) 8(2)过点O、C分别作OH⊥AB、CG⊥AB,垂足为点H、G.
∵OH⊥AB,∴AH?DH. ·················································································· (1分) 同理 DG?EG. ∵cos?BAC?∴AD?44,∴AH?x. 558·············································································································· (1分) x. ·5∵CG⊥AB,∴?AGC?90?. ∴?AGC??ACB?90?.
又∵?CAB是公共角,∴△AGC∽△ACB. ∴
AGAC16.∴AG?. ?ACAB5∵AE?y,∴GE?∴DG?16··················································································· (1分) ?y. ·
516?y. 516168∴?y??y?y?x. 55532825∴化简得 y?································································ (2分) ?x(2?x≤). ·
5587(3)0?x? ····················································································································· (2分)
8························································································································· (1分) x?2. ·
25················································································································ (2分) ?x?4. ·
8