2022年全国各地中考数学压轴题参考答案及评分标准（二）.docx

本文格式为Word版，下载可任意编辑2022年全国各地中考数学压轴题参考答案及评分标准（二） 2022年全国各地中考数学压轴题 参考答案及评分标准（二） 101．解：（1）当t＝4－3时，四边形FBCG为正方形． ················································· 1分 当0＜t≤4时，四边形AEGD为平行四边形． ············································ 2分 （2）由题意知点D、C的坐标分别为（1，3），（5，3） ······························· 4分 ∵抛物线经过原点O（0，0），∴设抛物线的解析式为y＝ax＋bx（a≠0）． 2 将D、C两点坐标代入得 ?3a＝－???a＋b＝ 3?5 ·，解得?································································· 6分 ???b＝63?25a＋5b＝ 3?5?∴抛物线的解析式为y＝－（3）存在． 3263x＋x ··················································· 7分 55 ∵点Q在抛物线上，∴点Q(x，－过点Q作QM⊥x轴于M，如图． 3263x＋x) 55 y Q D G C ∵∠DAB＝60°，CD＝4，∴AB＝5． ∴S△ABQ＝ 1AB·QM 2O (A) E F M B 32631＝×5×|－x＋x| 55212 ＝|－3x＋63x| ····································································· 8分 2 x ∵EG的延长线与抛物线交于x轴的上方 32632x＋x＞0，∴－3x＋63x＞0． 5512 ∴S△ABQ＝(－3x＋63x) 2∴－ ∵S梯形ABCD＝ 1(CD＋AB)·BC 231(4＋5)×2× 229＝························································································ 9分 3 ·2＝ 若S△ABQ＝S梯形ABCD，那么 2 192 (－3x＋63x)＝3． 22 整理得x－6x＋9＝0，解得x＝3． ···························································· 10分 把x＝3代入抛物线的解析式，得y＝－ 36392 ×3＋×3＝3，即MQ555 ＝ 93． 593MQ95∵∠QEM＝60°，∴EM＝＝＝． ······································ 11分 o5tan603∴t＝3－ 96＝（秒）． 55∴存在这样的时刻t，使得△ABQ的面积与梯形ABCD的面积相等，此时t ＝6秒． 5····································································································· 12分 102．解：（1）A（0，2），B（4，0） ····················································································· 2分 设直线AB的解析式为y＝kx＋b，那么有 1??b＝ 2?k ＝－，解得?2 ?4k＋b＝ 0???b＝ 2y A D ∴直线AB的解析式为y＝－ 1x＋2． ······································ 3分 2O E C B x 图① （2）ⅰ）①当点E在原点和x轴正半轴上时，重叠片面为△CDE，如图① ∴S＝S△CDE＝ 11CE·CD＝BC·CD 22 ＝ 1112 (4－x)(－x＋2)＝x－2x＋4 2241BO＝2，∴2≤x＜4． ···················· 4分 2∵当点E与原点O重合时，CE＝ ②当点E在x轴的负半轴上时，设DE与y轴交于点F，那么重叠片面为梯形CDFO，如 图② ∵△FEO∽△ABO，∴ OA11OF＝＝，∴OF＝OE． 22OEOBF E O y A D B x 1又∵OE＝4－2x，∴OF＝(4－2x)＝2－x 2∴S＝S梯形CDFO＝ 1(OF＋CD)·OC 2 C 1132 x＋2x． ················ 5分 ＝[2－x＋(－x＋2)]· x＝－ 224图② ∵当点C与原点O重合时，点C的坐标为（0，0） ∴0＜x＜2． ······························································································ 6分 ?32－x＋2x（0＜x＜2） ??4综合①②得S＝? ··················································· 7分 1?x2－2x＋4（2≤x＜4） ??4 2121x－2x＋4＝(x－4)，∴抛物线的对称轴为x44 ⅱ）①∵当2≤x＜4时，S＝ ＝4 ∵抛物线开口向上，∴当2≤x＜4时，S随x的增大而减小 ∴当x＝2时，S的最大值＝②当0＜x＜2时，S＝－ 轴为x＝ 4 3 21····································· 8分 (2－4)＝1． · 4323424x＋2x＝－(x－)＋，∴抛物线的对称4433∵抛物线开口向下，∴当x＝综合①②，当x＝ 44时，S有最大值为． ····················· 9分 3344时，S有最大值为． ······································ 10分 3335，0）和（，0）． ····································· 14分 22ⅲ）存在，点C的坐标为（ 附：详解：①当Rt△ADE以点A为直角顶点时，作AE⊥AB交x轴负半轴于点E， 如图③ ∵Rt△AOE∽Rt△BOA，∴∵OA＝2，∴OE＝1 ∵OE＋2OC＝4，∴OC＝∴点C的坐标为（ 13×(4－1)＝ 22E O OEOA1＝＝ 2OAOBy A D C B x 3，0） 2y A 图③ ②当Rt△ADE以点E为直角顶点时，如图④ ∵∠AED＝90°，∴∠AEO＋∠DEC＝90° ∵∠DEC＝∠DBC，∴∠AEO＋∠DBC＝90° ∵∠BAO＋∠DBC＝90°，∴∠AEO＝∠BAO ∴Rt△AOE∽Rt△BOA，∴∴OC＝1＋ OEOA1＝＝，∴OE＝1 2OAOBO E D C B x 图④ 15×(4－1)＝ 225，0） 2∴点C的坐标为（ 综上所述，存在这样的点C，使得△ADE为直角三角形，点C的坐标为： C1（ 103．解：（1）把x＝0代入y＝－2x＋4，得y＝4． ∴C（0，4）． ··································································································· 1分 35，0）和C2（，0） 22 ∵矩形OABC，∴BC＝OA＝3，AB＝OC＝4． ∴B（－3，4）． ································································································ 2分 （2）∵二次函数y＝－ 12 x＋bx＋c的图象经过B、C两点 2 3?1?2?－?(－3)－3b＋c ＝ 4?b ＝?∴?2 解得?··············································· 4分 2 ·???c ＝ 4?c ＝ 4∴二次函数的解析式为y＝－ 123x－x＋4． ················································· 6分 22 （3）证明：连结AC，在Rt△AOC中，AC＝OA2＋OC2＝32＋42＝5． ∵y＝－2x＋4，当y＝0时，x＝2，∴D（2，0）． ∵AD＝OA＋OD＝3＋2＝5，∴AC＝AD． ∵P是CD的中点，∴AP⊥CD． ··································································· 9分 （4）存在． ············································································································· 10分 方法一：假设四边形APCM为矩形，过点M作MN⊥x轴于N． 在Rt△COD中，∵CD＝OC2＋OD2＝42＋22＝25． 1∴MA＝PC＝CD＝5 2∵MA∥CD，∴∠MAN＝∠CDO． 又∵∠MNA＝∠COD＝90°，∴△MNA∽△COD． ····································· 12分 ∴ MNNAMA． ＝＝ COODCD525y B 525C P D O x ∴MN＝4?＝2，NA＝2?＝1． M N A ∵NO＝NA＋OA＝1＋3＝4 ∴M（－4，2）． ················································· 13分 把x＝－4代入y＝－ 123132 x－x＋4，得y＝－×(－4)－×(－4)＋4＝2． 2222 ∴点M在抛物线上． ∴存在这样的点M，使以A、P、C、M为顶点的四边形为矩形． ············ 14分 方法二（原参考答案中没有，本人添加，仅供参考）： 过点作AM∥CD交抛物线于点M，连结CM． ∵MA∥CD，直线CD的斜率为－2． ∴设直线MA的解析式为y＝－2x＋m，把A（－3，0）代入，得m＝－6． ∴直线MA的解析式为y＝－2x－6 ?y ＝－2x－6?x ＝－4?联立? 解得 ?123y ＝ 2y ＝－x－x＋4??22? ∴M（－4，2）． ∴MA＝(－4＋3)2＋22＝5 在Rt△COD中，∵CD＝OC2＋OD2＝42＋22＝25． 1∴PC＝CD＝5 2∴MA＝PC． ∵MA∥CD，∴四边形APCM为平行四边形． 又∵A。