中考数学压轴题详解—圆1(5)

时间:2025-07-11

(3)Sn

12

n 1

π2

n 1

(n为正整数). ··················· 10分

3 (1) 证明: 如图,

∵ 点I是△ABC的内心, ∴ ∠BAD=∠CAD,∠ABI=∠CBI. ………………2分 ∵ ∠CBD=∠CAD, ∴ ∠BAD=∠CBD. ……………………………3分 ∴ ∠BID=∠ABI+∠BAD =∠CBI+∠CBD=∠IBD. ∴ ID=BD. ………………………5分 (2)解:如图, ∵∠BAD=∠CBD=∠EBD, ∠D=∠D, ∴ △ABD∽△BED. …………………………7分

BDDE

ADBD

. ∴

AD DE BD

2

ID

2

. …………………8分

∵ ID=6,AD=x,DE=y,∴ xy=36. ………………9分 又∵ x=AD>ID=6, AD不大于圆的直径10, ∴ 6<x≤10. ∴

的中点, 4 (1)证明:∵C是劣弧BD

∴ DAC CDB. ·········································· 1分 而 ACD公共,

∴△DEC∽△ADC. ······································· 3分

y

与x的函数关系式是y

36x

.(6

x≤10

) …………………………10分

说明:只要求对xy=36与6<x≤10,不写最后一步,不扣分.

(2)证明:连结OD,由⑴得

DC

ACDC∵CE 1.AC AE EC 2 1 3,

EC

∴DC2 AC EC 3 1 3 .

∴DC

.··········································································································· 4分

由已知BC DC AB是⊙O的直径,

∴ ACB 90 ,

∴AB AC CB 3

2

2

2

2

2

12.

∴AB

∴OD OB BC DC , ∴四边形OBCD是菱形. ∴DC∥AB,DC AB, ∴四边形ABCD是梯形. ················································ 5分 法一:

过C作CF垂直AB于F,连结OC

,则OB BC OC ∴ OBC 60 . ······································································································ 6分 ∴sin60

CFBC

,CF BC sin60

12 3

2

32

4

∴S梯形ABCD=

12

CF

AB+DC =

2

······································· 7分

法二:(接上证得四边形ABCD是梯形)

又DC∥AB ∴AD BC,连结OC,则△AOD,△DOC和△

OBC的等边三角形 6分 ∴△AOD≌△DOC≌△OBC,

京翰教育1对1家教 /

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