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2019平谷一模数学试题及答案

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  • 2025/5/23 0:15:40

24.(1)证明:∵AC切⊙O于点A,

∴∠BAC=90°. ········································································· 1 连接AD.

∵点E是BD的中点,

CDE∴∠BAE=∠DAE. F∵AB是⊙O的直径, ∴∠ADB=90°. BAO∵∠CAD+∠DAB=∠DAB+∠B=90°, ∴∠CAD=∠B.

∵∠CAD+∠DAE =∠B+∠BAE, ∴∠CAF=∠CFA. ···································································· 2 ∴AC=CF. ············································································· 3

(2)解:∵AB=4,AC=3,

∴BC=5. ················································································ 4 ∵AC=CF=3, ∴BF=2.

∵cosB?∴BD=

BDAB4??, ABBC516. ············································································· 5 5126∴AD=,DF=.

551∴tan∠BAE= tan∠DAE = ························································· 6

225.(1)3.0; ····························································································· 1

(2)如图; ··························································································· 3

(3)1.2或1.6或3.0. ············································································· 6

26.(1)m; ······························································································· 1

(2)∵y?x2?2mx?m2?3??x?m??3,

∴抛物线顶点坐标为(m,-3). ························································ 2

∵抛物线经过点A,B时,且AB∥x轴, ∴抛物线对称轴为x=m=2. ······························································· 3

∴抛物线的表达式为y?x?4x?1; ················································· 4 (3)0?m?1. ···················································································· 6

2227.(1)∠BCD=120°-α. ············································································· 1 (2)解:

方法一:延长BA使AE=BC,连接DE. ······ 2

D 由(1)知△ADC是等边三角形,

∴AD=CD.

∵∠DAB+∠DCB=∠DAB+∠DAE=180°, ∴∠DAB=∠DAE.

C∴△ADE≌△CDB. ···················· 3

P∴BD=BE.

∴BD=AB+BC. ·························· 4

BEA方法二:延长AB使AF=BC,连接CF. ······ 2

∠BDC=∠ADE. ∵∠ABC=120°,

D∴∠CBF=60°.

∴△BCF是等边三角形. ∴BC=CF.

∵∠DCA=∠BCF=60°,

∴∠DCA+∠ACB=∠BCF+∠ACB. CP即∠DCB=∠ACF. ∵CA=CD,

BAF∴△ACF≌△DCB.····················· 3

∴BD=AF.

∴BD=AB+BC. ·························· 4 (3)AC,BD的数量关系是:AC?3BD; ········································· 5 2位置关系是:AC⊥BD于点P. ······································· 6

28.(1)22; ·························································································· 1

(2)22?r?4; ················································································ 3 (3)?25?2?t??5?2或6

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24.(1)证明:∵AC切⊙O于点A, ∴∠BAC=90°. ········································································· 1 连接AD. ∵点E是BD的中点, CDE∴∠BAE=∠DAE. F∵AB是⊙O的直径, ∴∠ADB=90°. BAO∵∠CAD+∠DAB=∠DAB+∠B=90°, ∴∠CAD=∠B. ∵∠CAD+∠DAE =∠B+∠BAE, ∴∠CAF=∠CFA. ···································································· 2 ∴AC=CF. ·························································

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