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． ·
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