(word完整版)2019年广东省中考数学真题(Word版,含答案),推荐文档 下载本文

22.解:(1)AB?22?62?210,

AC?62?22?210,

BC?42?82?45.

(2)由(1)得AB2?AC2?BC2, ∴?BAC?90?.

连接AD,AD?22?42?25, ∴S阴=S?ABC?S扇形AEF

?11AB?AC???AD2 24?20?5?

.

五、解答题(三)

23.解:(1)x??1或0?x?4. (2)把A??1,4?代入y?∴y??k2,得k2??4. x4. x4上,∴n??1. x∵点B?4,n?在y??∴B?4,?1?.

把A??1,4?,B?4,?1?代入y?k1x?b1得

??k1?b?4,?k1??1, 解得??4k?b??1,b?3.??1∴y??x?3.

(3)设AB与y轴交于点C,

∵点C在直线y??x?3上,∴C?0,3?.

11S?AOB?OC??xA?xB???3??1?4??7.5,

22又S?AOD:S?BOP?1:2, ∴S?AOP?又S?AOC1?7.5?2.5,S?BOP?5. 31??3?1?1.5,∴点P在第一象限. 2∴S?COP?2.5?1.5?1.

12?3?xP?1,解得xP?. 2327把xP?代入y??x?3,得yP?.

33又OC?3,∴∴P??27?,?. 3?3?24.(1)证明:∵AB?AC,∴?ABC??ACB. 又∵?ACB??BCD,?ABC??ADC, ∴?BCD??ADC.∴ED?EC. (2)证明:连接OA, ∵AB?AC,∴?AB??AC. ∴OA?BC.

∵CA?CF,∴?CAF??CFA. ∴?ACD??CAF??CFA?2?CAF. ∵?ACB??BCD,∴?ACD?2?ACB. ∴?CAF??ACB.∴AF//BC. ∴OA?AF.∴AF为eO的切线.

(3)∵?ABE??CBA,?BAD??BCD??ACB, ∴?ABE:?CBA.∴∴AB2?BC?BE.

∵BC?BE?25,∴AB?5.

ABBE?. BCAB连接AG,∴?BAG??BAD??DAG,

?BGA??GAC??ACB.

∵点G为内心,∴?DAG??GAC. 又∵?BAD??BCD??ACB, ∴?BAD??DAG??GAC??ACB. ∴?BAG??BGA. ∴BG?AB?5.

25.(1)解:令323373x?x??0, 848解得x?1或?7. 故A?1,0?,B??7,0?.

配方得y?32?x?3??23,故D?3,?23. 8??(2)证明:∵CF?CA,OA?OF?1, 易证?DD1F:?COF.

D1DCO?. ∴

FD1OF∴OC?3.

∴CA?CF?FA?2,即?ACF为等边三角形. ∴?AFC??ECF?60?. ∴EC//BF.

又∵EC?DC?6,BF?6, ∴EC//BF.

∴四边形BFCE是平行四边形. (3)设点P的坐标为??x,??323373?x?x?, ??848?①当点P在B点左侧时,

DD1D1A?,∴x1?1(舍),x2??11. PMMADD1D1A372). ?,∴x1?1(舍),x2??PAAM3则1)

②当点P在A点右侧时, 因为?PAM与?DD1A相似,

PMDD1?则3),∴x1?1(舍),x2??3(舍). MAD1A4)

PMD1A5?,∴x1?1(舍),x2??(舍). MADD13③当点P在AB之间时, ∵?PAM与?DD1A相似, 则5)

PMDD1?,x1?1(舍),x2??3(舍). MAD1APMD1A5?6),x1?1(舍),x2??. MADD13综上所述,点P的横坐标为?,?11,?

5337,点共有3个. 3