【附5套中考模拟试卷】湖北省咸宁?019-2020学年中考数学第三次调研试卷含解?- 百度文库 ر

1ݹɶɵõۣ

2ȡSTõRȡEFõGGRMNڵPɵõ ⡿

(1)MN?32?52?34;

2ȡSTõRȡEFõGGRMNڵP

⿼ͼ-ӦͼԳ-̾⣬ȷͼǽĹؼ 201y=2x2+x+32ACB=41㣻3D

?1?ѵA,B뼴ߵĽʽ.

775 832?2?BHACڵHBHijȣACBĶ.

?3?ӳCDxڵGDCEסAOCֻܡCAO=DCE.ֱCDķ̣ߵ

õD.

?a?b?3?0? 1⣬?93a?b?3?0,?2?4??a??2

?b?1ߵıʽΪy??2x2?x?3 2BHACڵH

Aǣ10Cǣ03Bǣ

30 2AC=10AB=

355 OC=3BC=225?3 2BH?AC?OC?ABBAD=BH?10?BH?310 4Rt BCHУBH?33105BHC=90oBC= 24sin?ACB?2 2֡ߡACBǣ?ACB?45? 3ӳCDxڵG Rt AOCУAO=1AC=10

cos?CAO?AO10 ?AC10ߡDCEסAOCֻܡCAO=DCE AG = CG

11AC1010 22cos?GAC???AGAG10AG=1Gǣ40 ߵCǣ03lCD:y??3x?3 47?3x????x?0??y??x?38? ãᣩ. 4??752?y?3?y???y??2x?x?3?32?D?217.6 m

CDкĶBCAC߶ΪAB ⡿

⣺⣬BDC45㣬ADC50㣬ACD90㣬CD40 m

?775?,?. ?832?RtBDCУtanBDC

BCCD40 m

RtADCУtanADC

AB7.6m

ABĸ߶ԼΪ7.6 m 㾦

Ҫ˽ֱεӦãȷӦǺϵǽؼ 22֤

1֪áBDF=BCDٸݡBFD=DFC֤BFDסDFCӶBFDF=DFFCбμã

2֪֤AEGסADCõAEG=ADC=90㣬ӶEGBC̶ɣ1ɵ

EGBF? EDDFBFDFEGDF?? Ӷ ֤. DFCFEDCF1ߡACB=90㣬BCD+ACD=90㣬

CDRtABCĸߣADC=BDC=90㣬A+ACD=90㣬A=BCD EACе㣬

DE=AE=CEA=EDAACD=EDC ߡEDC+BDF=180-BDC=90㣬BDF=BCD ֡ߡBFD=DFC BFDסDFC BFDF=DFFC DF2=BFCF

AC=EDDF 2AE

AEAG? ADAC֡ߡA=A AEGסADC AEG=ADC=90㣬 EGBC

EGBF? EDDFɣ1֪DFDסDFC

BFDF? DFCFEGDF? EDCF

EGCF=EDDF.

23148㣨1֤3

1CDԲܽǶʹֱĶɵýۣ

3 4??PB??PD? 1ȸݵεʵãABE=AEB֤BCG=DACɵ CDԵԲܽȣͬԵԲܽǺԲĽǵĹϵɵýۣ

AG=OF3OOGABG֤COFաOAGOG=CF=xOF=aOA=OC=1x-aݹɶз̵ã1x-a1=x1+a1a=⡿ 1CD ADǡOֱ ACD=90㣬 ACB+BCD=90㣬 ADCG

AFG=G+BAD=90㣬 ߡBAD=BCD ACB=G=48㣻 1AB=AE ABE=AEB

ߡABC=G+BCGAEB=ACB+DAC ɣ1ãG=ACB BCG=DAC

3xʽɵýۣ 4??PB? CDADǡOֱADPC