发布时间 : 星期一 文章2018-2019学年上海市杨浦区初三数学第一学期期末质量调研(参考答案)2019.01更新完毕开始阅读28cf161aa88271fe910ef12d2af90242a995ab69
杨浦区初三数学期末试卷参考答案及评分建议2019.1
一、选择题：（本大题共6题，每题4分，满分24分）
1． C； 2． D； 3． A； 4． B； 5． D； 6． B； 二、填空题：（本大题共12题，每题4分，满分48分）
7． 5； 8．
1:3； 9． 10； 10． 5或3； 11． 4； 12． -2；
352 14．
13． <； <； 15． 3:2； 16． 270 ； 17． y??2x2+4x； 18． 24；
13三、解答题：（本大题共7题，满分78分） 19．解：··························· （3分） （1）∵□ABCD，∴BO=OD，AD//BC，AD=BC. ·
∴ED······································································· （1分） DG. ·
?BCGB ∵点E为边AD的中点，∴.∴DG1. ············· （1分） 11ED?AD?BC?22GB2 ∵BO=OD，∴OG1. ·························································· （1分）
?DG2uuuuurruuurruruuruuuruuruuurrr··············· （1分） （2）∵AB?a，BC?b，∴BD?BA?AD?BA?BC??a?b. · ∵BO=OD，OG·············································· （2分） 1，∴OG1. ·??DG2BD6 ∴uuu··························································· （1分） r1uuur1rr. ·
GO?DB?(a?b)6620．解：（1）∵二次函数y?ax2?bx?c图像过点(1,?2)、(?1,0)和
∴
??a?b?c??2,??a?b?c?0,?3?c??.2?3，
(0,?)2（3分） ∴
123.（2分） 1∴二次函数解析式为?y?x?x??a?2,22??b??1,?3?c??.2?（2）
. ··············································· （1分） 12312y?x?x??(x?1)?2222x y … … -1 0 0 ?3 21 -2 2 ?3 23 0 … … ································································································ （2分）
························································································· （2分） 图略 ·
21．解：（1）作AH⊥BC于H．
在Rt△ACH中，∵
·································· （1分） 2，∴AH2. ·
cosC==2AC2································································· （1分） ∵AC=2，∴CH=1. ·
∴AH=1. ·················································································· （1分） 在Rt△ABH中，∵
····································· （1分） 1，∴AH1. ·tanB==5BH5·················································································· （1分） ∴BH=5. ·
······································································ （1分） ∴BC=BH+CH=6． ·
······················································· （1分） （2）∵BD=CD，BC=6，∴CD=3. ·
··················································· （1分） ∵CH=1，∴DH=2. ∴AD=5. ·AH15sin?ADH=??AD55 ∴∠ADC的正弦值为5．
5在Rt△ADH中，························· （1分,1分） ． ·
22．解：由题意可知∠AEC=30°··········· （3分） ，∠ADC=60°，∠BDC=45°，FG=15. ·设CD=x米，则在Rt△ACD中，由
············· （1分） AC得AC=3x. ·
tan?ADC?DC又Rt△ACE中，由
········································· （1分） EC得EC=3x. ·
cot?AEC?AC····················································································· （1分） ∴3x=15+x. ························································································ （1分） ∴x=7.5. ·
··························································· （1分） ∴AC=7.53.∴AH=7.53+1.5. ·
· （1分） ∵在Rt△BCD中，∠BDC=45°，∴BC=DC=7.5.∴AB=AC﹣BC=7.5(3?1)． ························ （1分） 答：AH的高度是(7.53+1.5)米，AB的高度是7.5(3?1)米. ·
23．证明：··········· （2分） （1）∵∠ACD=∠B，∠BAC=∠CAD，∴△ADC∽△ACB. ·
∵∠ACD=∠BAE，∠ADE=∠CDA，∴△ADE∽△CDA. ············· （2分） ∴△ADE∽△BCA. ····························································· （1分） ∴ADDE. ····································································· （1分）
=BCAC···························· （1分） （2）∵△ADE∽△BCA，∴AEDE，即AEAB. ·
==ABACDEAC
∵△ADE∽△CDA，∴AEDE，即AE··························· （1分） AC. ·
==ACADDEAD ∴AE2ABAC=?DE2ACAD···························································· （2分） AB.
AD ∵点E为CD中点，∴DE=CE. ·················································· （1分）
·········································································· （1分） ∴AE2AB. ·=CE2AD
24．解：（1）作DH⊥y轴，垂足为H，∵D（1,m）（m>0），∴DH= m，HO=1. ∵
tan?COD1，∴OH1，∴m=3. ...................................................................... （1分）
=3DH3∴抛物线y=ax2+bx+c的顶点为D（1,3）. 又∵抛物线y=ax2+bx+c与y轴交于点C（0,2），
2∴ìa=-1,∴抛物线的表达式为y=-x2+2x+2. ....... （1分） a+b+c=3,（分）∴ì???????bíb=2,?=1,?í-??2a????c=2.????c=2.（2）∵将此抛物线向上平移，
..................................... （1分） ∴设平移后的抛物线表达式为y=-x2+2x+2+k(k>0)，.
则它与y轴交点B（0,2+k）.
∵平移后的抛物线与x轴正半轴交于点A，且OA=OB，∴A点的坐标为（2+k,0）. .（1分） ∴0=-(2+k)2+2(2+k)+2+k.∴k=-2,k=1.
12∵k>0，∴k=1.
∴A（3,0），抛物线y=-x2+2x+2向上平移了1个单位. . ...................................... （1分） ∵点A由点E向上平移了1个单位所得，∴E（3,-1）. . .............................................. （1分） （3）由（2）得A（3,0），B（0, 3），∴AB=32. ∵点P是抛物线对称轴上的一点（位于x轴上方），且∠APB=45°,原顶点D（1,3）,
∴设P（1,y），设对称轴与AB的交点为M，与x轴的交点为H，则H（1,0）.
y ∵A（3,0），B（0, 3），∴∠OAB=45°, ∴∠AMH=45°. ∴M（1,2）. ∴BM=2.
B M O P ∵∠BMP=∠AMH, ∴∠BMP=45°.
∵∠APB=45°, ∴∠BMP=∠APB.
∵∠B=∠B，∴△BMP∽△BPA. ......................................... （2分） ∴BPBA.∴BP2=BA?BM=BMBP32?26
H A x
∴BP2=1+(y-3)2=6.∴y=3+1∴P(1,3+5，y2=3-.. ..................................... （1分）
5（舍）5). . ................................................................................................................... （1分）
25．（本题满分14分，第（1）小题4分，第（2）、（3）小题各5分） 解：（1）∵AD//BC，∴DEAEAD.∵E为AB中点，∴AE=BE. ∴AD= BF，DE= EF.
==EFEBBF∵AD=3，AB=6，∴BF=3，BE=3. ∴BF=BE.
························································ （1分） ∵AB⊥BC，∴∠F=45°且EF=32. ·
···················································································· （1分） ∴DF=2EF=62. ····························································· （1分） ∵DF⊥DC，∠F=45°，∴CF=12. ·
··································································· （1分） ∴BC= CF-BF=12-3=9. ·（2）∠DCE的大小确定，
···················································· （1分） 1. 2tan?DCE作CH⊥AD交AD的延长线于点H，∴∠HCD+∠HDC=90°. ∵DF⊥DC，∴∠ADE+∠HDC=90°. ∴∠HCD=∠ADE.
······································· （2分） 又∵AB⊥AD，∴∠A=∠CHD. ∴△AED∽△HDC. ·∴DE························································································ （1分） AD. ·
=DCCH∵AB⊥AD，CH⊥AD，AD//BC，∴CH =AB=6. ∵AD=3，CH=6，∴DE1.即=tan?DCEDC2··········································· （1分） 1. ·
D A 2B F E
（3）当点E在边AB上，设AE=x，
C