1、2022 年福州市初中畢業(yè)班適應(yīng)性試卷數(shù)學(xué)試題答案及評分標(biāo)準(zhǔn)評分說明：1本解答給出了一種或幾種解法供參考，如果考生的解法與本解答不同，可根據(jù)試題的主要考查內(nèi)容比照評分參考制定相應(yīng)的評分細那么2對于計算題，當(dāng)考生的解答在某一步出現(xiàn)錯誤時，如果后繼局部的解答未改變該題的內(nèi)容和難度，可視影響的程度決定后繼局部的給分，但不得超過該局部正確解容許給分?jǐn)?shù)的一半；如果后繼局部的解答有較嚴(yán)重的錯誤，就不再給分3解答右端所注分?jǐn)?shù)，表示考生正確做到這一步應(yīng)得的累加分?jǐn)?shù)4只給整數(shù)分?jǐn)?shù)選擇題和填空題不給中間分一、選擇題：每題 4 分，總分值 40 分1b2d3a4c5c6c7d8b9c10d二、填空題：每題 4 分，

2、總分值 24 分11 - 1213 231231415 315答案不唯一，如“0 -4 ，而 021616 2 19三、解答題：此題共 9 小題，共 86 分解容許寫出文字說明、證明過程和演算步驟ìx + y = 1，î17解： í4x + y = -8 - ，得3x = -9 ，························

3、··········································· 3 分x = -3 ·····&

4、#183;·················································&

5、#183;··········· 5 分將 x = -3 代入，得 y = 4 ··································&

6、#183;······················· 7 分ìx = -3î這個方程組的解是í y = 4···················&

7、#183;····································· 8 分18證明：acbd，cab = ebd········&#

8、183;·················································&#

9、183;······· 3 分a在cab 和ebd 中eìÐcab = Ðebd，íïÐabc = Ðd，îï ac = be，dcbcabebd， ························&

10、#183;········································ 6 分ab = bd·······&#

11、183;·················································&#

12、183;·················· 6 分19解：原式= 1 - x - 2 × 3(x + 1)·························&

13、#183;································ 3 分x + 1 (x + 2)(x - 2)= 1 - 3 x + 2= x + 2 - 3 ·········

14、··················································

15、··········· 4 分x + 2x + 2= x - 1 ， ···································

16、········································· 5 分x + 23當(dāng) x =- 2 時，原式= 3 - 2 - 13 - 2 + 2= 3 - 33

17、83;················································· 6

18、分= 1 -3 ················································

19、3;···· 8 分20解：1依題意得60 - 200 ´ 0.12 = 36 ·······································&

20、#183;········· 2 分小張離開效勞區(qū)時，油箱內(nèi)還有 36 升汽油 ··························· 3 分2依題意得 s = 36 ······

21、··················································

22、········ 5 分a360，當(dāng) 0.08a0.1 時，s 隨 a 的增大而減小，······························ 6 分360s450····&#

23、183;·················································&#

24、183;············ 8 分211答案不唯一，如：aod 和aob，abd 和abc 等 ·················· 2 分2ghepfq···········

25、3;·········· 4 分如圖，efh 為所求作的三角形 ····································&#

26、183;············ 5 分證明：作 gpef 于點 p，hqef 交 ef 延長線于點 q由尺規(guī)作圖可得：gh = ef，fh = eg，四邊形 efhg 是平行四邊形，ghef， ·······················

27、3;············································· 6 分gp = hq ··

28、3;·················································

29、3;················ 7 分sefg = 1 ef·gp，sefh = 1 ef·hq，22sefg = sefh， ························

30、;······································· 8 分efh 和efg 的面積相等22解：1由表 2 數(shù)據(jù)可得，使用了節(jié)水龍頭后，50 天日用水量小于 0.3 的頻數(shù)為 1 +

31、 5 + 13 = 19， ························· 1 分50 天的日用水量少于 0.3 m3 的頻率為19 ，·················

32、83;··········· 2 分50由頻率估計概率得該家庭使用節(jié)水龍頭后，日用水量少于 0.3 m3 的概率約為19 ······························

33、3;···· 3 分502該家庭未使用節(jié)水龍頭 50 天日用水量的平均數(shù)為：0.05 ´1 + 0.15 ´ 3 + 0.25 ´ 2 + 0.35 ´ 4 + 0.45 ´ 9 + 0.55 ´ 26 + 0.65 ´ 550= 0.48； ····················

34、83;·················································

35、83;······· 6 分該家庭使用節(jié)水龍頭 50 天日用水量的平均數(shù)為：0.05 ´1 + 0.15 ´ 5 + 0.25 ´13 + 0.35 ´10 + 0.45 ´16 + 0.55 ´ 5 = 0.35；····· 8 分50估計使用節(jié)水龍頭后，一年可節(jié)省水：(0.48 - 0.35)×365 = 47.45 m3 ······

36、3;····································· 10 分231證明：ef = bd， e»f = b»d ， e»f - b»f = b»d - b

37、»f ，即 b»e = d»f ，············································

38、3;·················· 1 分bde = dbf，bfdc····························

39、3;···································· 2 分 d»f = d»f ，dbf = def，bde = fed ······

40、················································· 3 分bd

41、 = bc，c = bde，fed = c，efbc ·············································

42、83;··················· 4 分2解：連接 dfab 為直徑，bc 為切線，abbc，abc = 90°，efbc，bgf = abc = 90°，abef， ·················

43、83;·················································

44、83; 5 分fg = eg = 1 ef， b»f = b»e ，2bdf = bdeaeh = 4，hf = 2，ef = fh + he = 6，dhg = fg - hf = 1 ef - hf = 12 b»e = b»e ，hebfe = bde = dbf，fgbh = fh = 2bc在 rtbgh 中，cosbhg = hg = 1 ，bh2bhg = 60°， ··············

45、;················································ 6 分由1得fed

46、= bde = 30°，bdf = 30°，dfe = 180° - bdf - bde - def = 90°，de 為直徑····································&

47、#183;···························· 7 分在 rtdef 中 ，de = ef = 4 3 ，cos 30°3圓的半徑為2 ·············

48、;············································ 8 分 b»e = b»e ，bde = 30&

49、#176;， b»e 所對的圓心角為 60°， ············································&#

50、183;·· 9 分 b»e 的長= 60p ´ 2 3 = 2 3p ······································ 10 分1803241證明：c

51、dab，cd = ab，四邊形 abcd 是平行四邊形，········································· 1 分abc = 90°，平行四邊形 a

52、bcd 是矩形············································ 2 分ab = bc，矩形 abcd 是正方形·

53、;················································· 3 分由

54、得bad = 90°，ab = ad 由旋轉(zhuǎn)得baebcd，ae = cd，bae = bcd = 90°，ae = bc，eab = cba = 90°are = brc，arebrc，································

55、;························· 4 分ar = brbfce，cfg = 90°，fcb + fbc = 90°fbc + fba = 90°，eagdfrfcb = fba， ··········

56、·················· 5 分bccbrbag，ag = br， ····························&#

57、183;····································· 6 分ag = 1 ab = 1 ad，22g 是 ad 的中點·······

58、;··················································

59、; 7 分2點 g 仍然是 ad 的中點 ··············································

60、83;············· 8 分證明如下：延長 cd，bg 交于點 m，延長 ea 交 cm 于點 nabcd，abc = 90°，bcd = 90°，bag = mdg，abg = dmg 由旋轉(zhuǎn)得baebcd，bae = bcd = 90°，cd = ae，ban = 90°，四邊形 abcn 是矩形ab = bc，矩形 abcn 是正方形， ······

61、······································· 9 分bc = cn = an，cne = 90°，eandfgcen + ecn = 90°mcfg = 9

62、0°，ecn + bmc = 90°，bmc = cen，bmccen， ············ 10 分cm = ne，cm - cd = ne - ae，即 dm = an， ····················· 11 分ab = dm，

63、bcabgdmg，ga = gd，g 是 ad 中點 ················ 12 分251解：將 a -1 ，0代入 y = ax2 + bx - 3a a > 0 得a - b - 3a = 0 ， ····· 1 分 b = -2a ，···········

64、··················································

65、············ 2 分 y = ax2 - 2ax - 3a 當(dāng) y = 0 時， ax2 - 2ax - 3a = 0 ， ····························

66、3;·················· 3 分解得 x1 = -1 ，x2 = 3a -1 ，0，b3，0··························

67、············································· 4 分2點 p 為第四象限內(nèi)拋物線上一個動點，設(shè) pm，n

68、其中 m0，n0，且 n = am2 - 2am - 3a = a(m2 - 2m - 3) = a(m + 1)(m - 3) 根據(jù)勾股定理得 ap2 = n2 + (m + 1)2，bp2 = n2 + (3 - m)2，ab2 = 16 在 rtabp 中，apb = 90°，ap2 + bp2 = ab2，即 n2 + (m + 1)2 + n2 + (3 - m)2 = 16，················

69、3;···························· 5 分整理得 n2 = - m2 + 2m + 3， n2 = - n ，an0， n = - 1 0. ············

70、3;·················································

71、3;·········· 6 分a又拋物線的頂點縱坐標(biāo)是-4a ， -4a - 1 0， ··································&

72、#183;································ 7 分aa 1 2a3， 1 a3··································