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.doc_bak716

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． 52； 8．
9． 10； 10． 53或35； 11． 4； 12． -2；
13． <； 14．
<； 15． 3:2； 16． 270 ； 17． 22+4y x x =-； 18． 2413；
三、解答题：（本大题共7题，满分78分） 19．解：（1）∵□ABCD ，∴BO =OD ，AD //BC ，AD =BC . ·
··························· （3分） ∴ED
DG BC GB
=. ·
······································································· （1分） ∵点E 为边AD 的中点，∴11
22ED AD BC ==.∴12
DG GB =
. ············· （1分） ∵BO =OD ，∴12OG DG =. ·························································· （1分）
（2）∵AB a =uu u r r ，BC b =uu u r r ，∴BD BA AD BA BC a b =+=+=-+u u u r u u r u u u r u u r u u u r r r . ················ （1分） ∵BO =OD ，12OG
DG =，∴16
OG BD =. ··············································· （2分） ∴11()
66GO DB a b ==-u u u r u u u r r r . ·
··························································· （1分） 20．解：（1）∵二次函数2y ax bx c =++图像过点(1,2)-、(1,0)-和
3(0,)
2
-，
∴
2,0,3.
2a b c a b c c ⎧
⎪++=-⎪
-+=⎨⎪⎪=-⎩
（3分） ∴
1,21,3.
2a b c ⎧
=⎪⎪
=-⎨⎪⎪=-⎩
∴二次函数解析式为21322y x x =--.（2分） （2）
22
131(1)2
222y x x x =--=--. ··············································· （1分）
··（2分）
图略 ·
························································································· （2分） 21．解：（1）作AH ⊥BC 于H ．
在Rt △ACH
中，∵
cos C
，∴AH AC . ·
·································· （1分）
∵AC CH =1. ·
································································· （1分） ∴AH =1. ·················································································· （1分） 在Rt △ABH 中，∵
1tan =5B ，∴1=
5
AH BH . ······································ （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
··················································· （1分） 在Rt △ADH
中，sin =AH ADH AD ∠=
． ·
························· （1分,1分） ∴∠ADC
22．解：由题意可知∠AEC =30°，∠ADC =60°，∠BDC =45°，FG =15. ············ （3分） 设CD =x 米，则在Rt △ACD 中，由
tan AC ADC DC
∠=
得AC
. ·
············· （1分） 又Rt △ACE 中，由
cot EC AEC AC
∠=
得EC =3x . ·
········································· （1分） ∴3x =15+x . ······················································································ （1分） ∴x =7.5. ·
······················································································· （1分） ∴AC
=∴AH
=. ·
··························································· （1分） ∵在Rt △BCD 中，∠BDC =45°，∴BC =DC =7.5.∴AB =AC ﹣BC
=1)． ·· （1分） 答：AH 的高度是
()米，AB
的高度是1)米. ·
······················· （1分） 23．证明：（1）∵∠ACD =∠B ，∠BAC =∠CAD ，∴△ADC ∽△ACB . ·
··········· （2分） ∵∠ACD =∠BAE ，∠ADE =∠CDA ，∴△ADE ∽△CDA . ············· （2分） ∴△ADE ∽△BCA . ····························································· （1分） ∴AD DE BC AC =. ····································································· （1分）
（2）∵△ADE ∽△BCA ，∴AE DE AB AC =，即AE AB DE AC
=. ·
···························· （1分）
∵△ADE ∽△CDA ，∴AE DE AC AD =，即AE AC DE AD =. ·
··························· （1分） ∴2
2AE AB AC AB DE AC AD
AD
=?.
···························································· （2分） ∵点E 为CD 中点，∴DE CE =. ·················································· （1分）
∴2
2AE AB CE AD =. ··········································································· （1分）
24．解：（1）作DH ⊥y 轴，垂足为H ，∵D （1,m ）（0m >），∴DH = m ，HO =1. ∵
1tan 3COD
?，∴13
OH DH =
，∴m =3. ...................................................................... （1分）
∴抛物线2y ax bx c =++的顶点为D （1,3）. 又∵抛物线2y ax bx c =++与y 轴交于点C （0,2），
∴3,1,22.a b c b a c ì++=ïïïïï-=íïïïï=ïî（2分）∴1,2,2.a b c ì=-ïïï=íïï=ïïî∴抛物线的表达式为222y x x =-++. ....... （1分） （2）∵将此抛物线向上平移，
∴设平移后的抛物线表达式为222(0)y x x k k =-+++>，.
..................................... （1分） 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） ∴20(2)2(2)2k k k =-+++++.∴122,1k k =-=.
∵0k >，∴1k =.
∴A （3,0），抛物线222y x x =-++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 =∵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）.
∵A （3,0），B （0, 3），∴∠OAB =45°, ∴∠AMH =45°. ∴M （1,2）.
∴BM =
.
∵∠BMP =∠AMH , ∴∠BMP =45°.
∵∠APB =45°, ∴∠BMP =∠APB .
∵∠B =∠B ，∴△BMP ∽△BP A . ......................................... （2
∴BP BA BM BP
=.∴26BP BA BM
=?
∴221(3)6BP y =+-=.∴1233y y =+
=-
.. ..................................... （1分）
∴(1,3P +
. . ................................................................................................................... （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 ，∴DE
AE AD EF EB BF
==.∵E 为AB 中点，∴AE =BE . ∴AD = BF ，DE = EF .
∵AD =3，AB =6，∴BF =3，BE =3. ∴BF =BE .
∵AB ⊥BC ，∴∠F=45°且EF = ·
························································ （1分） ∴DF =2EF =···················································································· （1分） ∵DF ⊥DC ，∠F=45°，∴CF =12. ·
···························································· （1分） ∴BC = 1239CF BF -=-=. ···································································· （1分） （2）∠DCE 的大小确定，
1tan 2
DCE
?. ···················································· （1分） 作CH ⊥AD 交AD 的延长线于点H ，∴∠HCD+∠HDC =90°. ∵DF ⊥DC ，∴∠ADE+∠HDC =90°. ∴∠HCD =∠ADE .
又∵AB ⊥AD ，∴∠A =∠CHD . ∴△AED ∽△HDC . ········································ （2分） ∴DE
AD DC CH
=. ·
························································································ （1分） ∵AB ⊥AD ，CH ⊥AD ，AD //BC ，∴CH =AB =6. ∵AD =3，CH =6，∴12DE
DC =.即1tan 2
DCE ?. ·
··········································· （1分） （3）当点E 在边AB 上，设AE =x ，
x
A B C
D E
F
∵AD //BC ，∴AD AE BF EB =，即36x BF x =-.∴
183x BF x -=. ∵△AEF 的面积为3，∴1
18332x
x x
-鬃=. ∴4x =. ·
······························································································· （1分） ∵AD =3，AB ⊥AD ，∴DE =5. ∵12
DE
DC =，∴DC =10.
∵DF ⊥DC ，∴
1
51025
2
DCE S =创=V . ························································· （1分） 当点E 在边AB 延长线上，设AE =y ，
∵AD //BC ，∴AD AE BF EB =，即36y BF y =-.∴318
y BF y -=. ∵△AEF 的面积为3，∴1
31832y y y
-鬃=.∴8y =. ········································· （1分）
∵AD =3，AB ⊥AD ，∴DE
.
联结CE ，作CH ⊥AD 交AD 的延长线于点H ，同（1）可得12
DE
DC =.
················· （1分） ∴DC
=∵DF ⊥DC
，∴
1
73
2
DCE S =V . ················································ （1分） 综上，当△AEF 的面积为3时，△DCE 的面积为25或73.。