湖北省宜昌市2018年中考数学试题及答案(Word版) 下载本文

?CDCB ?CECAx4 ??27?x?x2?7x?8?0

?x1?1或x2??8(舍去)

解法三:如图1,连接BD,

QAB为半径的直径,

??ADB?90o

可证?CDB∽?CEA

?CDCB ?CECAx4 ??27?x?x1?1或x2??8(舍去)

1?S半圆=???42=8?

2?BD?15,

?S菱形=815 22.解:(1)Q40n?12

?n?0.3

(2)Q40?40(1?m)?40(1?m)?190 解得:m1?217,m2??(舍去) 22∴第二年用乙方案治理的工厂数量为40(1?m)?40?(1?50%)?60(家) (3)设第一年用甲方案整理降低的Q值为x,

第二年Q值因乙方案治理降低了100n?100?0.3?30, 解法一:?30?a??2a?39.5

?a?9.5 ?x?20.5

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解法二:??x?a?30

?x?2a?39.5?x?20.5,a?9.5

23.(1)证明:在矩形ABCD中,?A??D?90,AB?DC, 如图1,又QAE?DE,

o图1

?ABE??DCE,

(2)如图2,

图2

①在矩形ABCD中,?ABC?90o,

Q?BPC沿PC折叠得到?GPC

??PGC??PBC?90o,?BPC??GPC

QBE?CG ?BE//PG, ??GPF??PFB

??BPF??BFP ?BP?BF

②当AD?25时,

Q?BEC?90o

??AEB??CED?90o,

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Q?AEB??ABE?90o,

??CED??ABE

又Q?A??D?90o,

??ABE∽?DEC

?ABDE ?AECD1225?x, ?x12∴设AE?x,则DE?25?x,

?解得x1?9,x2?16

QAE?DE ?AE?9,DE?16, ?CE?20,BE?15,

由折叠得BP?PG,

?BP?BF?PG, QBE//PG, ??ECF∽?GCP

?EFCE ?PGCG设BP?BF?PG?y,

?15?y20? y252525 则BP? 332510BC25310,cos?PCB? ??3PC2510103?y?在Rt?PBC中,PC?③若BP?9,

解法一:连接GF,(如图3)

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Q?GEF??BAE?90o, QBF//PG,BF?PG

∴四边形BPGF是平行四边形

QBP?BF,

?平行四边形BPGF是菱形

?BP//GF, ??GFE??ABE, ??GEF∽?EAB

?EFABGF?BE ?BEgEF?ABgGF?12?9?108解法二:如图2,

Q?FEC??PBC?90o,

?EFC??PFB??BPF, ??EFC∽?BPC

?EFCEBP?CB 又Q?BEC??A?90o, 由AD//BC得?AEB??EBC,

??AEB∽?EBC

?ABBE?CECB ?AEEFBE?BP ?BEgEF?AEgBP?12?9?108

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解法三:(如图4)过点F作FH?BC,垂足为HS?BPFS四边形PFEG?BFBF?

EF?PGBE图4

BFS?BFCEF?BCEF??? BES?BEC12?BC12?9EF ?BE12?BEgEF?12?9?108

24.解:(1)天空:OA?6,k??6,点E的坐标为??(2)①设直线MN,y?k1x?b1

?3?,4?; 2??3?12?t?5t??k1(t?1)?b1??22由题意得?

17??t2?3t??k(?t?3)?b11??22解得k1?1,b1??t2?4t?∴直线MN:y?x?∵抛物线y??121 2121t?4t? 2212x?bx?c过点M,N 231?122?t?5t???(t?1)?b(t?1)?c??22??2 ??1t2?3t?7??1(?t?3)2?b(?t?3)?c??222解得b??1,c?5t?2

12x?x?5t?2 23?顶点P(?1,5t?)

2∴抛物线y??15 / 17