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平谷区2018～2019学年度第一学期期末初三数学答案及评分参考
一、选择题（本题共16分，每小题2分）
题号 1 2 3 4 5 6 7 8
A C B C A D B D
答案
二、填空题（本题共
16分，每小题2分）
9．x≥3；10． ；11．2π；12．30；
13． 向左平移3个单位长度得到
（向左平移，或平移3个单位长度，只得1分）；
14．答案不唯一，如： ；15．8；
16． （ 或 或 ，只得1分 ）．
三、解答题(本题共68分，第17~22题，每小题5分，第23~26题，每小题6分，第27，
28题，每小题7分)解答应写出文字说明、演算步骤或证明过程.
17．解：= ···············································································4
= ···································································································5
18．（1）如图·····································································································2
C
A B l
D
（2）完成下面的证明．
证明：连结AC，BC，AD，BD．
∵AC=BC，AD=BD，············································································3
∴CD⊥AB（依据：到线段两个端点的距离相等的点在线段垂直平分线上）．··5
19．解：∵正方形ABCD，
A E D
∴AD∥BC，AD=BC．·····································1
∴∠CAE=∠ACB，∠AEB=∠CBE．····················2
∴△AEO∽△CBO．········································3 O
∴ ．··············································4
∵点E是AD中点， B C
∴ ．
1∴ ．·················································5
20． 解：（1） ；···································································1
（2）①当 时，a+2a－3=0．
解得 a=1．
∴二次函数的表达式为 ；············································2
②
·············································································3
∴二次函数的顶点坐标是 ；····················································4
③如图····························································································5
y
4
3
2
1
–3 –2 –1O 1 2 3 4 5x
–1
–2
–3
–4
–5
21．解：由题意可知，∠ACD=45°，∠CBD=30°··························································1
在Rt ACD中，
∵∠ACD=45°，
∴∠C△AD=∠ACD=45°
∴AD=CD=1200．························································································2
在Rt BCD中，∠CBD=30°
∵ ta△n30°= ，
∴BD=1200 ．·······················································································3
∴AB=BD﹣AD=1200（ ﹣1）．·································································4
答：这条江的宽度AB长1200（ ﹣1）米．·····················································5
22．解：（1）由题意可知A（2,2），
∴k=4；···············································1 y
5
（2）由题意可知 AC=2，
4
∴OB=4．
3
∵点B在x轴上，
2
A
1
∴ 或 ．························3 B 1 C B 2
–5 –4 –3 –2 –1 O 1 2 3 4 5x
–1
–2
当A（2,2）， 时，解得 ；···4
–3
–4
当A（2,2）， 时，解得 ．···5
–5
2综上所述， ．
23．（1）证明：∵∠BAC=90°，点D是BC中点，
∴AD=CD．····················································································1
∵AE∥BC，CE∥AD，
∴四边形ADCE是平行四边形．·························································2
∴平行四边形ADCE是菱形．····························································3
（2）解：∵∠BAC=90°，点D是BC中点，∠B=60°，
∴AD=BD=AB=6．··············································································4
∵菱形ADCE，
∴AD=CD=CE=6．
∵DF⊥CE于点F，∠ECD=∠ADB=60°，
∴ ．
∴CF=3．··························································································5
∴EF=3．··························································································6
24．解：（1）连接OD，EF交于点G． C
∵⊙O与AC相切于点D，
∴OD⊥AC于D． D F
∵∠ACB=90°，
∴OD∥BC．·························1
G
∵BE是⊙O的直径，
∴∠EFB=90°． A E O B
∴EF∥AC．·························2
∴OD⊥EF．
∴DE=DF．··························3
（2）在Rt ABC中，∠ACB=90°，BC=3，sinA=
∴AB=5．···························································································4
设⊙O△的半径为r，则AO=5﹣r．
在Rt AOE中， ．
△
∴ ．··························································································5
∴AE= ．···························································································6
25．解：（1）经测量m的值是 5.7 （保留一位小数）．···········································1
（2）如图·······························································································4
3（3）结合函数图象，解决问题：当∠PAC=30°，AD的长度约为 5. 2 cm.·············6
26．解：（1）直接写出点C的坐标 （ 0, 3 ） ；．····················································1
（2）∵抛物线y=ax2+bx+3（a≠0）经过（1,0），
∴ ．·················································································2
（3）①当t=3时，D（3,3）．
解得抛物线的表达式为 ．···········································3
②∵3
