中考数学专题复习动态图形面积问题

专题复习动态图形面积问题1．（09山东聊城）如图，已知正方形ABCD的边长与Rt△PQR的直角边PQ的长均为4厘米，QR＝8厘米，AB与QR在同一条直线l上．开始时点Q与点B重合，让△PQR以1厘米/秒速度在直线l上向左匀速运动，直至点R与点A重合为止，t秒时△PQR与正方形ABCD重叠部分的面积记为S平方厘米．（1）当t＝3秒时，求S的值．（2）求S与t之间的函数关系式，并写出自变量t的取值范围．（3）写出t为何值时，重叠部分的面积S有最大值，最大值是多少？2．（09吉林长春）如图，直线y＝－43x＋6分别与x轴、y轴交于A、B两点；直线y＝45x与AB交于点C，与过点A且平行于y轴的直线交于点D．点E从点A出发，以每秒1个单位的速度沿x轴向左运动．过点E作x轴的垂线，分别交直线AB、OD于P、Q两点，以PQ为边向右作正方形PQMN，设正方形PQMN与△ACD重叠部分（阴影部分）的面积为S（平方单位），点E的运动时间为t（秒）．（1）求点C的坐标；（2）当0＜t＜5时，求S与t之间的函数关系式；（3）求（2）中S的最大值；（4）当t＞0时，直接写出点（4，29）在正方形PQMN内部时t的取值范围．AEPCQMNDOBxyABQPRCDl3．（09山西省）如图，已知直线l1：y＝32x＋38与直线l2：y＝－2x＋16相交于点C，l1、l2分别交x轴于A、B两点．矩形DEFG的顶点D、E分别在直线l1、l2上，顶点F、G都在x轴上，且点G与点B重合．（1）求△ABC的面积；（2）求矩形DEFG的边DE与EF的长；（3）若矩形DEFG从原地出发，沿x轴的反方向以每秒1个单位长度的速度平移，设移动时间为t（0≤t≤12）秒，矩形DEFG与△ABC重叠部分的面积为S，求S关于t的函数关系式，并写出相应的t的取值范围；（4）S是否存在最大值？若存在，请直接写出最大值及相应的t值，若不存在，请说明理由．4．（09湖南邵阳）如图，直线l的解析式为y＝－x＋4，它与x轴、y轴分别相交于A、B两点．平行于直线l的直线m从原点O出发，沿x轴的正方向以每秒1个单位长度的速度运动，它与x轴、y轴分别相交于M、N两点，运动时间为t秒（0＜t≤4）．（1）求A、B两点的坐标；（2）用含t的代数式表示△MON的面积S1；（3）以MN为对角线作矩形OMPN，记△MPN和△OAB重合部分的面积为S2，①当2＜t≤4时，试探究S2与t之间的函数关系式；②在直线m的运动过程中，当t为何值时，S2为△OAB的面积的165？ABOyxFED(G)l1l2CxNOyABPMmlxNOyABPMmlFPE答案部分：1．解：（1）当t＝3秒时，如图①所示．设PR与BC交于点M，则QB＝3，BR＝QR－QB＝5．∵Rt△MBR∽Rt△PQR，∴BRBM＝QRQP，即5BM＝84．∴BM＝25．·······································2分∴S＝21(QP＋BM)·QB＝21×(4＋25)×3＝439(平方厘米)．·································3分（2）①当0≤t≤4时，如图①所示，则QB＝t，BR＝8－t．由（1）知BRBM＝QRQP，即tBM8＝84，∴BM＝28t．∴S＝21(QP＋BM)·QB＝21×(4＋28t)·t＝－41t2＋4t．······················································5分②当4＜t≤8时，如图②所示．设PR分别与DA、CB交于点M、N，则QB＝t，BR＝8－t，QA＝t－4，AR＝QR－QA＝8－(t－4)＝12－t．∵Rt△MAR∽Rt△PQR，∴ARAM＝QRQP，即tAM12＝84，∴AM＝212t．∵Rt△NBR∽Rt△PQR，∴BRBN＝QRQP，即tBN8＝84，∴BN＝28t．∴S＝21(AM＋BN)·AB＝21×(212t＋28t)×4＝20－2t．···································8分③当8＜t≤12时，如图③所示．设PR交DA于点M，则QB＝t，RB＝t－8，AR＝AB－RB＝4－(t－8)＝12－t．∵Rt△MAR∽Rt△PQR，∴ARAM＝QRQP．即tAM12＝84，∴AM＝212t．∴S＝21AM·AR＝21×212t×(12－t)＝41t2－6t＋36．·······································10分综上所述，S＝3664122044122ttttt（3）当t＝4时，重叠部分的面积S有最大值，最大值是12平方厘米．··················12分ABQPRCDlM图①ABQPRCDlM图②NABCDlM图③QPR2．解：（1）由题意，得xxyy45643＝＋＝－解得4153＝＝yx∴点C的坐标为（3，415）．·············································································1分（2）根据题意，得AE＝t，OE＝8－t．∴点Q的纵坐标为45(8－t)，点P的纵坐标为43t．∴PQ＝45(8－t)－43t＝10－2t当MN在AD上时，10－2t＝t，∴t＝310．···························································3分当0＜t≤310时，S＝t(10－2t)，即S＝－2t2＋10t当310≤t＜5时，S＝(10－2t)2，即S＝4t2－40t＋100．···········································5分（3）当0＜t≤310时，S＝－2t2＋10t＝－2(t－25)2＋225当t＝25时，S最大值＝225；当310≤t＜5时，S＝4t2－40t＋100＝4(t－5)2，S随t的增大而减小∴t＝310时，S最大值＝4(310－5)2＝9100∵225＞9100，∴S的最大值为225．···································································7分（4）4＜t＜522．·························································································10分3．解：（1）将y＝0代入y＝32x＋38，得x＝－4，∴点A的坐标为（－4，0），∴OA＝4．将y＝0代入y＝－2x＋16，得x＝8，∴点B的坐标为（8，0），∴OB＝8．∴AB＝OA＋OB＝4＋8＝12········································································1分联立1623832＋＝＋＝－xxyy，解得65＝＝yx，∴点C的坐标为（5，6）．··························2分∴S△ABC＝AB21·yC＝21×12×6＝36···························································3分（2）∵点D在直线l1上且xD＝xB＝8，∴yD＝32×8＋38＝8．∴点D的坐标为（8，8）．········································································4分又∵点E在直线l2上且yE＝yD＝8，∴－2xE＋16＝8，∴xE＝4．∴点E的坐标为（4，8）．·········································································5分∴DE＝8－4＝4，EF＝8．·········································································6分（3）过点C作CM⊥AB于M．∵C（5，6），∴OM＝5，CM＝6，∴MB＝OB－OM＝8－5＝3，∴AM＝AB－MB＝12－3＝9，AB－FG＝12－4＝8，AG＝AB－GB＝12－t，AF＝AG－FG＝12－t－4＝8－t．①当0≤t＜3时，如图1，矩形DEFG与△ABC重叠部分为五边形CHFGR（t＝0时，为四边形CHFG）．∵Rt△RGB∽Rt△CMB，∴GBRG＝MBCM，即tRG＝36，∴RG＝2t．∵Rt△AFH∽Rt△AMC，∴AFHF＝AMCM．即tHF－8＝96，∴HF＝32(8－t)．∴S＝S△ABC－S△BRG－S△AFH＝36－21×t×2t－21(8－t)×32(8－t)＝－34t2＋316t＋344即S＝－34t2＋316t＋344（0≤t＜3）．·····················································8分②当3≤t＜8时，如图2，矩形DEFG与△ABC重叠部分为梯形HFGR．∵Rt△AGR∽Rt△AMC．∴AGRG＝AMCM，即tRG－12＝96，∴RG＝32(12－t)．＝21(HF＋RG)·FG＝21[32(8－t)＋32(12－t)]×4＝－38t＋380即S＝－38t＋380（3≤t＜8）．························10分③当8≤t＜12时，如图3，矩形DEFG与△ABC重叠部分为Rt△AGR．∴S＝SRt△AGR＝21AG·RG＝21(12－t)×32(12－t)＝31t2－8t＋48ABOyxFEDl1l2CGMHR图1ABOyxFEDl1l2CGMR图3ABOyxFEDl1l2CGMHR图2即S＝31t2－8t＋48（8≤t＜12）．····················12分（4）存在，当t＝2时，S最大＝20．························14分4．解：（1）当x＝0时，y＝4；当y＝0时，x＝4．∴A(4，0)，B(0，4)；····································2分（2）∵MN∥AB，∴ONOM＝OBOA＝1．∴OM＝ON＝t．∴S1＝21OM·ON＝21t2；································4分（3）①当2＜t≤4时，易知点P在△OAB的外面，且点P的坐标为(t，t)．设PN、PM与直线l分别相交于E、F两点，如图2．则F点的坐标满足4ttxy－＝＝即F(t，4－t)．同理可得E(4－t，t)，则PF＝PE＝|t－(4－t)|＝2t－4．·······································6分∴S2＝S△PNM－S△PEF＝S△MON－S△PEF＝21t2－21PE·PF＝21t2－21(2t－4)2＝－23t2＋8t－8；8分②S△OAB＝21×4×4＝8当＜t≤2时，S2＝21t2．由题意，得21t2＝165×8＝25．解得t1＝－5＜0，t1＝5＞2，均不合题意，舍去；·········································10分当2＜t≤4时，S2＝－23t2＋8t－8＝25．解得t3＝37，t4＝3．综上所述，当t＝37或t＝3时，S2为△OAB的面积的165．····································12分xNOyABPMml图1xNOyABPMmlFPE图2