浙江省杭州市蕭山區(qū)城區(qū)六校2021年(3月份)中考數(shù)學(xué)質(zhì)檢試卷

2222222222222021年浙江省杭州蕭山區(qū)城區(qū)校中考數(shù)質(zhì)檢試卷（月份）一選題本題有個(gè)題每題3分共分2021|A2021BCx3

DA2xxBx+3xC2xxx+3nAn+Cnn

B4D2+2ABCD⊙BA54

B62

C72

D82ABCDacdAacBbdCcDaACADBACEF

221121212122112121212A3

B

C6

DAC

BDlax+bx+clcx+bxacl

xm00l0y

xyxyxx1A

B

C

D10EFGHGABCDADFHABCDBDEADABC245菱形S矩形A4B

C

D二填題本題6小題每分共24分）+

121

222213614y

xy

3

2

8

9

5

15CD2FDEDFCFEDF+16DBCDEAB1DBC2DBC三解題本題7個(gè)小，分）解應(yīng)出文說、明程推步.17cm1yx23y18

1m2319AD1122ABC2BAD7450cmACsin37tan37sin2120A/B/1A22.5/B3090021BD⊙OACADED2

2212312132212312131D23DBFFAO22mx10+2601x202nyyyny23ABCDFADFOG12tan3

22222222222222222222021年浙江省杭州蕭山區(qū)城區(qū)校中考數(shù)質(zhì)檢試卷（月份）參考答案與試題解析一選題本題有個(gè)題每題3分共分2021|A2021BC2021|x3

DA2xxBx+3xC2xxx+3x32x+3xnAn+Cnn

B4D2+24mn2mn+ABCD⊙BA54

B62

C72

D82

ABCD⊙O108DB72CABCDa1a1+aaCacdAacBbdCcDacdabeDACADBACEF

222221121212222211212121A3

B

C6

DBDCDADBCABCrADBACBDADBCACOAr39DAC

BD解Clax+bx+clcx+bxacl

xm00l0y

xyxyxx1

21222221222212222222122222122221222222A

B

C

Dax+ax++bx+，+c0am

+cm0a++alx0xx

yax0xaxbx+bx+accaxxD10EFGHGABCDADFHABCDBDEADABC245菱形S矩形A4B

C

DEGEMBDBDEGa

EFH2AOD22EFH2AOD22EGEEMBDACBDOEGDOEDEBGABCDADBCADBCAC2BDABCADCEDADBGDEAEABGEEGABFHaFHADeq\o\ac(△,Rt)EMDEMED2S2FHEM矩形△

eq\o\ac(△,Rt)AODADasincosa4AOOD2cosasincos菱形2sin菱形矩形C

4cos二填題本題6小題每分共24分）+

2扇22222扇22223

+2121262136

Srn120

S12

1214yxy

3

2

8

9

5

y2+bxbcax++2

2222yxy815CD2FDEDFCFEDF+ACCFEDFDDHACH

AFABCDDEDCA5CFECFE

AF

ADDHACAHAD

DHHFAFAH

eq\o\ac(△,Rt)DHFDF

DF+

16DBCDEAB1DBC112DBC5ADDABADB90EDBBEDED2BD1DCBDDCBDBEkDADBCDABDACEDB9030EDAEEB

B

2BDCDBDkkAEDB+FDCBED+EDBBEDFDCBCBEDCDF

3BEDCBDCD2BDk

3BE78

2222三解題本題7個(gè)小，分）解應(yīng)出文說、明程推步.17cm1yx23yx224xx23

x8

240xyyy818

1m100n6023mn232000＝n

60%2

32000200019AD1122ABC2BAD7450cmACsin37tan37sin21

2Oeq\o\ac(△,Rt)AODODeq\o\ac(△,Rt)OCABCADCABCABC2ADCBCDACBDBAD37ACDBCDeq\o\ac(△,Rt)AODODADsin3730cm

40cmeq\o\ac(△,Rt)

80OA40+8012020A/B/

1A22.5/B30900Axy2xB424+240030900a1500150021BDOACADAEED1D23DBFFAOABCCD1AB3OBAD

eq\o\ac(△,Rt)ABDBDABOAFOOAFABABCCD

CDABCD2ABCDADBAB+ED

3FAOABAD90eq\o\ac(△,Rt)ABDAB

ADOBAB

OBFBABOFOAF90O

22123121322122121211221231213221221212113122mx10+2601x202nyyyny0mm0m224mmm13mm0m1yx2m0nyyy

n

y

當(dāng)

y

11

23ABCDFADFOG12tan3

2ADGAGF90FAGAGFAFGFαBEF3FF