杨浦区2024学年度第一学期期末考试初三数学试卷

由天下分享时间：2025/1/21 11:21:05 加入收藏我要投稿点赞

∴AH=1. ·················································································· （1分）

1AH1在Rt△ABH中，∵tanB=，∴····································· （1分） =. ·

5BH5∴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. ···················································· （1分）在Rt△ADH中，sin?ADH= ∴∠ADC的正弦值为AH15． ·························· （1分,1分） ??AD555． 522．解：由题意可知∠AEC=30°，∠ADC=60°，∠BDC=45°，FG=15. ············ （3分）

AC设CD=x米，则在Rt△ACD中，由 tan?ADC?得AC=3x. ·············· （1分）

DC又Rt△ACE中，由cot?AEC?EC得EC=3x. ·········································· （1分） AC∴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. ························ （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）

ADDE. ····································································· （1分） =BCACAEDEAEAB（2）∵△ADE∽△BCA，∴，即. ····························· （1分） ==ABACDEACAEDEAEAC ∵△ADE∽△CDA，∴，即. ···························· （1分） ==ACADDEADAE2ABACAB=? ∴. ···························································· （2分） 2DEACADAD ∵点E为CD中点，∴DE=CE. ·················································· （1分）

∴

AE2AB=∴. ··········································································· （1分） CE2AD

第 6 页共 8页

24．解：（1）作DH⊥y轴，垂足为H，∵D（1,m）（m>0），∴DH= m，HO=1.

OH11∵tan?COD∴=，∴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），

ìa+b+c=3,?ì?a=-1,????b?∴?（2分）∴íb=2,∴抛物线的表达式为y=-x2+2x+2. ....... （1分） =1,í-??2a??????c=2.?c=2.??（2）∵将此抛物线向上平移，

∴设平移后的抛物线表达式为y=-x2+2x+2+k(k>0)，. ..................................... （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） ∴0=-(2+k)2+2(2+k)+2+k.∴k1=-2,k2=1. ∵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.

∵∠BMP=∠AMH, ∴∠BMP=45°. ∵∠APB=45°, ∴∠BMP=∠APB.

∵∠B=∠B，∴△BMP∽△BPA. ......................................... （2分） ∴

P B M O H A x

BPBA.∴BP2=BA?BM=BMBP32?26

22∴BP=1+(y-3)=6.∴y1=3+5，y2=3-..................................... （1分） 5（舍）..

∴P(1,3+................................................................................................................... （1分） 5). .

25．（本题满分14分，第（1）小题4分，第（2）、（3）小题各5分）

DEAEAD解：（1）∵AD//BC，∴.∵E为AB中点，∴AE=BE. ∴AD= BF，DE= EF. ==EFEBBF∵AD=3，AB=6，∴BF=3，BE=3. ∴BF=BE.

第 7 页共 8页

∵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. ···································································· （1分）（2）∠DCE的大小确定，tan?DCE1. ···················································· （1分） 2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） ∴

DEAD. ························································································· （1分） =DCCHDE1=.即tan?DCEDC21. ············································ （1分） 2A D ∵AB⊥AD，CH⊥AD，AD//BC，∴CH =AB=6. ∵AD=3，CH=6，∴

（3）当点E在边AB上，设AE=x， ∵AD//BC，∴

ADAE18-3x3x，即.∴BF=. ==BFEBxBF6-xB F C

118-3x

∵△AEF的面积为3，∴鬃E x=3.

2x∴x=4. ································································································ （1分）

DE1∵AD=3，AB⊥AD，∴DE=5. ∵=，∴DC=10.

DC21∵DF⊥DC，∴SVDCE=创························································· （1分） 510=25.

2当点E在边AB延长线上，设AE=y，

∵AD//BC，∴

3y3y-18ADAE，即.∴BF=. ==BFy-6yBFEB13y-18∵△AEF的面积为3，∴鬃········································· （1分） y=3.∴y=8.

2y∵AD=3，AB⊥AD，∴DE=73.

联结CE，作CH⊥AD交AD的延长线于点H，同（1）可得∴DC=273 ∵DF⊥DC，∴SVDCE=DE1················· （1分） =.