浙江省宁波市鄞州区2018学年九年级第一学期科学期末试卷(含答案)

19．解：3tan30cos6032sin245

233123232··································3分 2·32 ·

····················································6分 20．解：(1)摸到白球的概率是12 .······································· 3分 （2）图略.···················································· 6分

两次摸到红球的概率为

16. ································ 8分 21．解：(1) 过点B作BDAC交AC于点D.

BDACBDA90

tanBAC3BD3

4AD4AB3mBDABsinBAD3351.8m 答：点B离开地面的距离为1.8m.·································3分

（2）过E作EFAC交AC、AB于点F、G.

EFAEBG90,AGFEGB，GAFGEB，tanGEBGB3EB4.EB2,BGBEtanGEB2341.5.EG2.5.

EF3.1，FGEFEG0.6.AGFGsinBAC0.60.61，AB11.52.5m. 答：AB的长为2.5m. ·····································8分

22．解：（1）连结BC、OC.

AB为直径，∴∠ACB=90°. ∵AB=2BP, ∴AO=OB=BP.

∵AC=3BP=3OA,∴∠A=30°.······························2分

∴∠COB=2∠A=60°.∵OB=OC，∴△OCB为正三角形. ∴OB=OC=BC=BP，∴∠BCP=∠P=

1OBC=30°. 2∴∠OCP=∠OCB+∠PCB=90°,∴OCCP.·····················4分 ∵OC为半径，∴PC与⊙O相切.································5分 （2)∵SAOC19················7分 AOOCsin603， ·

24nr212032扇形OAC的面积为·················9分 3 ·

360360∴阴影部分弓形面积为39··························10分 3.· ·

423．解：（1）2，2 ···················································2分 （2）图略（扣分讨论） ······················ 6分

（3）B ···············································10分

24．解：（1）由题得：BC=x，AB=10-则sABBC1x. 212·······························2分 x10x·

2x的取值范围为0x4.····································3分 (2)s121·················4分 x10x(x10)250 ·

22 又 0x4

∴当0x4时，s随着x的增大而增大.

∴当x4时，s的值最大，且最大s=32.····················6分 答：当BC为4时，矩形花园ABCD的面积最大，最大值为32. (3) 由题得：BC=x，DE=x-4, AB=[20x(x4)]2112x. 2则sABBCx12x(4x12) ················8分 当x6时，s的值最大，且最大s=36. ···············10分 答：矩形花园ABCD的面积最大，面积为36.

25．解（1）画出点D的2个位置。

·················2分

（2）∵四边形ABCD为被BD分割的友谊四边形

∴△ABD与△DBC相似.

若△ABD∽△CBD则AB=BC=8. ···························4分 若△ABD∽△DBC则

ABBD∴AB=6 ·····················6分 BDBC综上所述：AB=6或8.

（3）①∵E是AC的中点，∴ABECBE∴CBFC150.

∵四边形ABCD内接于圆O，∴BADC180. ∵DAF30,BAFC150.∴BFCBAF. ∴△ABF∽△FBC.

∴四边形ABCF为友谊四边形. ···········9分 ②过点A作AG⊥BC交BC与G. ∵△ABF∽△FBC，∴

⌒1ABC30 2SABC∴

ABBF2∴FBABBC· BFBC11BCAGBCABsin6063 223ABBC63即ABBC24FB2··················11分 4∵FB0，∴FB26 ··································12分

26．（1)∵四边形ABCD是圆O的内接四边形，∴∠ABC=180°-∠ADC=∠CDE. ∵AB=AC， ∴∠ABC=∠ACB.

∴∠ADB=∠ACB=∠ABC=∠CDE. · ························4分 (2)①∵四边形ABCD内接于圆，∴∠BAD=180°-∠BCD=∠DCE. 又∠ADB＝∠CDE，∴△ADB∽△CDE. ∴

ADDB, ∴ADDE=BDCD=7×3=21. ·······················7分 CDDE②方法一：如图，过点A画AH┴CD于点H， ∵∠AFB=∠H=90°，∠ABD=∠ACH，AB=AC， ∴△ABF≌△ACH.∴BF=CH，AF=AH.

∵∠ADB=∠CDE=∠ADH，DF┴AC，DH┴AH， ∴DF=DH.（也可以根据全等证明得到或勾股定理计算）

∴DF=BD-BF=BD-CH=BD-(CD+DH),即2DF=BD-CD=7-3=4,DF=2,BF=5. ---9分 在Rt△CDF中，CF=CDDF325, ∴tan∠ACB=

22225······················11分 5. ·

5方法二：如图，连结AO并延长交圆于点G，连结BG,CG， ∵AG是直径，∴∠ACG=90°, 又BD┴AC，∴CG//BD，∴BGCD. ∵直径AG平分BAC,∴BGCG ∴CDBGCG,∴CG=BG=CD=3

在四边形BGCD中，CG//BD,BG=CG=CD=3,BD=7,∴DF=2. ······9分 同上可得tan∠ACB=5. ······11分 方法三：连结AO并延长交BD于点M，连结CM，

∵AM平分BAC,∴AM┴BC， ∴∠CAD=∠CBD=90°-∠ACB=∠MAF. ∴△MAF≌△DAF（ASA）.

∴MF=DF，即AC是线段MD的中垂线.

∴BM=CM=CD=3,∴MF=DF=2,同上可得同上可得tan∠ACB=5. ······11分 方法四：如图，在BD上截取BM＝CD，

∵∠ABM=∠ACD，AB=AC，BM=CD,∴△ABM≌△ACD（SAS） ∴AM=AD，又AC┴BD， ∴MF=DF=

A111DM=(BD-BM)=(BD-CD)=2， ······9分 D222O同上可得tan∠ACB=5. M ······11分

FBCE方法五：延长BD至点H，使得 DH=CD，则BH=10.

∵∠ADC=∠ADB+∠BDC=∠CDE+∠BDC=∠ADH，AD=AD,CD=HD, ∴△ADC≌△ADH(SAS)，∴AH=AC=AB， 又AC⊥BH，∴BF=

1BH=5，FD=2. A ······9分 2同理可得tan∠ACB＝5. ······11分

DHF21方法六：设AD＝m，由错误！未找到引用源。则DE＝, mB∴AB2＝ADAE＝m2＋21. ∴BF2-DF2＝AB2－AD2＝21.

设BF＝n，则DF＝7－n，即n2－（7－n）2＝21

OCE∵∠BAD=∠EAB，∠ADB=∠ACB=∠ABC，∴△ABD∽△AEB.

解得BF＝n＝5，所以DF＝2，所以CF＝5， ······9分 同理可得tan∠ACB＝5. ·····11分 (3)S102·····14分 x（下面提供详细解答过程，考试时不要求学生书写） ·

29

如图，过点A画AH┴CD于点H，AK┴BD于点K，

同上可证△ABK≌△ACH得BK=CH，AK=AH，再得DK=DH.故2DK=BD-CD.

SSABCSDBCSABFSDFCSABDSADC1111BDAKCDAHAK(BDCD)AK2DK2222102AKDKxsinADBxcosADBx29

（3）易证△ABD∽△AEB，∴AB2＝ADAE.

易证△ABD∽△DEC，∴BDCD＝ADDE.

11

S△ABC－ S△BCD＝ABACsin∠BAC－BDCDsin∠BDC 221

＝ sin∠BAC（ADAE－ ADDE） 21

＝x2sin∠BAC 2

5

又tan∠ABC＝tan∠CDE＝，

2

如图，设BM＝2a，则AM＝5a，AB＝29a，

2020

由面积法可得BN＝a,即sin∠BAC＝

2929

AODFCABE∴S△ABC－ S△BCD＝2x2×29＝29x2

12010

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