内容正文:
2018年中考模拟试卷(三)
数学参考答案及评分标准
一、选择题(每题3分,共30分)
1.C 2. B 3. C 4. D 5. B 6. A 7. C 8. A 9. B 10. C
二、填空题(每题3分,共15分)
11.
12.15
13.4
14.5
15.
或
.
三、解答题(本大题共8个小题,满分75分)
16.(8分)解:原式=
·····························2 分
=
······················································ 3分
=
·················································· 4 分
=
················································5 分
=
··························································· 6 分
∵x2+2x-15=0,
∴x2+2x=15,······················································· 7 分
∴原式=错误!未找到引用源。.··························································8 分
【或∵x2+2x-15=0,
∴x=3或x=-5,[来源:Z*xx*k.Com]
当 x =3 时,原式=
········································ 7 分
当 x =-5 时,原式=
】.································ 8 分
17.(9分)解:(1)b=50÷0.1=500,
a=500-(50+75+150)=225,
c=150÷500=0.3;
故答案为:225,500,0.3;………………………………3分
(2)m%=×100%=45%,
∴m=45,
“C”所对应的圆心角的度数是360°×0.3=108°,
故答案为:45,………………………………………………5分
108°(填108也正确);……………………………………………………………7分
(3)5000×0.45=2250,
答:估计成绩在95分及以上的学生大约有2250人.………………9分
18. 证明:(1)连接OC,如图①所示.
∵CD⊥AB,AE⊥CF,
∴∠AEC=∠ADC=90°,……………………………1分
∵CF是⊙O 的切线,OC是半径,
∴CO⊥CF,∴∠ECO=90°,
∴AE∥OC,……………………………2分
∴∠EAC=∠ACO,
∵OA=OC,
∴∠CAO=∠ACO,
∵∠EAC=∠CAO,………………………3分
在△CAE和△CAD中,
∴△CAE≌△CAD,……………………………4分
∴AE=AD;……………………………5分
(2) 连接CB,如图②所示.
在Rt△ACD中,
∵AD=AE=3,CD=4,
∴AC=5,……………………………6分
在Rt△AEC中,cos∠EAC=
=
,
∵AB是直径,∴∠ACB=90°,
∴cos∠CAB=
∵∠EAC=∠CAB,
∴
,……………………………8分
∴
……………………………9分
(2)解法二:设半径为r,
在Rt△CDA中,CD=4, OC=r,
∵AD=AE=3,
∴OD= r-3,
由勾股定理得:(r-3)2 +16=r2,∴r=
,
∴AB=2r=
19.(9分)解:如图,过点P作PE⊥AM于E,PF⊥AB于F………………`1分.
在Rt△PME中,∵∠PME=30°,PM=40,
∴PE=20.…………………………………………………2分
∵四边形AEPF是矩形,
∴FA=PE=20.……………………………………………3分
设BF=x米.∵∠FPB=45°,
∴FP=BF=x.………………………………………………4分
∵∠FPC=60°,
∴CF=PFtan60°=
x.……………………………………5分[来源:学#科#网Z#X#X#K]
∵CB=80,
∴80+x=
x.……………………………………………7分
解得x=40(
+1).…………………………………………………8分
∴AB=40(
+1)+20=60+40
≈129(米).
答:山高AB约为129米. ………………………………………………9分
20.(9分)(1)解:∵反比例函数(k>0)的图象经过点D(3,1),
∴k=3×1=3,[来源:学科网ZXXK]
∴反比例函数