江苏省宿迁市2019年中考数学试卷解析版

23523563232019年江苏宿迁市2352356323一选题本题8小题，小3分，分在小所出四选中有只一是合题要的请正选的母号涂答卡应置）2019A

BC

D2019Aa+Caa

BaDa437A3B3.5D7ABFBCBFCA105

B

C75

DA

B

C12

D9A1

B2

C3

D42A6

B6

2

C6+

D6πxOyABCDAOBx

222222BDMM

A

B

C2D二填题本题小题每题3分，分，不写解过，把案接写在题相应置）21120182750000000002750000000002.07S甲乙甲乙312x

+

160ABCBAM2BCA4FAB

00三解题本题10题，96分请答卡定域作，答应出必的字明证过或算骤

a+by1m12ABCD2EFCDDF121

m

1mn23eq\o\ac(△,Rt)ABCC1OABOABDBCACF22MAB相ABl32BCDBCEB15cm1E2ECD0.8

EEsin640.44tan64605021xy1yx2x3wABCABC30ACDα018010CEBEC2GAGC3Gy+bxcA10yC0312PPABP3QxDAQBQMDMDN

2352356323523563232326323年江省迁中数试参考答案与试题解析一选题本题8小题，小3分，分在小所出四选中有只一是合题要的请正选的母号涂答卡应置）2019A

BC

D2019BAa+Caa

BaDa+BCaaDaD437A3B3.5D7377

4ABFBCBFC

A105

B

C75

D30BBFCE30DEE45CFBBFC180BCF18045A30A

B

C12

D9

365BA1

B2

C3

D4xx3

2222x032222D2A6

B6

2

C6+

D6π636

AxOyABCDAOBxBDMM

A

B

C2DmM

ytODABtkMABDmtBDAMBM

m

m

22222222222222M

yk3mABCDm+32

mm

meq\o\ac(△,Rt)ABMtanMABA

ykk0xy二填题本题小题每题3分，分，不写解过，把案接写在题相应置）2442aaa2aa2a2

11n1111n22211n1111n2221120182750000000002750000000002.75101a10nn1012.7510a1anan2.07S甲乙甲乙S甲乙3

x+

2+4+2

1612c

a

13

2

cx

+

1a5a3a+2x2x505xa2aa53

11111122221211111122221260ABCBAM2BCABCCBCANBCAMCeq\o\ac(△,Rt)ABCAB60ABC30

2ACABeq\o\ac(△,Rt)ABCAB60ACAC2ABC

A4FAB

00FGGFGFG也EFBEEHGEBHHNHNCM+HE+EC1+三解题本题10题，96分请答卡定域作，答应出必的字明证过或算骤

2

1

2

a2

+by1m12B2OD1mn15A151151kx+byx+42x

AODBODAODBODAOB+4△

ABCD2EFCDDF124ADABDCF42FFHHAHFDAHDFFHADABCDAB2CDAB4ADBD2CD90DF4AFAFAECF2FFHH

AHFDAHDFFHADEH1EF1

m

1m223m

2360312+840%m5010n20223

79.21212

1111

2222

1111

1212eq\o\ac(△,Rt)ABCC1OABOABDBCACF22MAB相OF12

21MOFACOEOE1OFBOFOB122MFABMMBFMFMBMBFMFB

MBFMFBCFMACMABl32BCDBCEB15cm1E2ECD0.8EEsin640.44tan64CDECsinBCM75sin462EHCDHCECDM

ECCEEC+

212ECsinBCM67.5cm212E67.5+322EEHCDEH8064CECE71.1cm605021xy1yx2x3w23wwx+5022250x501060x10x102250

22增大22增大3w40+x+50xx+2000+2450ax30x20x20wABCABC30ACDα018010CEBEC2GAGC3G2

3G1

DE

DBEABC2CGO++180ECBG303ABKDKACOOGOB

OAOAGCAOC60AGCGOBDBBBDADAKDKBKAKBDBKBDDKBDK60DAB

G

y+bxcA10yC0

2221232221212PPABP3QxDAQBQMDMDNCbc2PPG22BGBHHAHPPAPHxAHP3AQBNMNDM+DNx++A0

y+2x32xHABBxBIIx3B3A1

HHOA1OCHHeq\o\ac(△,Rt)AOCsinACO

ABcosBHAGBGGHBAGPABeq\o\ac(△,Rt)ABGAGBBAGBG

BH2HBI+BAGHBIBAGACOeq\o\ac(△,Rt)BIHHBI

cosHBI

BH＝

xHAHy+