如图,在矩形 ABCD 中, AB =3, BC =4.动点 P 从点 A 出发沿 AC 向终点 C 运动,同时动点 Q 从点 B 出
如图,在矩形ABCD中,AB=3,BC=4.动点P从点A出发沿AC向终点C运动,同时动点Q从点B出发沿BA向点A运动,到达A点后立刻以原来的速度沿AB返回.点P、Q运动速...
如图,在矩形 ABCD 中, AB =3, BC =4.动点 P 从点 A 出发沿 AC 向终点 C 运动,同时动点 Q 从点 B 出发沿 BA 向点 A 运动,到达 A 点后立刻以原来的速度沿 AB 返回.点 P 、 Q 运动速度均为每秒1个单位长度,当点 P 到达点 C 时停止运动,点 Q 也同时停止.连接 PQ ,设运动时间为 t ( t >0)秒. (1)求线段 AC 的长度;(2)当点 Q 从点 B 向点 A 运动时(未到达 A 点),求△ APQ 的面积 S 关于 t 的函数关系式,并写出 t 的取值范围;(3)伴随着 P 、 Q 两点的运动,线段 PQ 的垂直平分线为 l :①当 l 经过点 A 时,射线 QP 交 AD 于点 E ,求 AE 的长;②当 l 经过点 B 时,求 t 的值.
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2014-10-07
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试题分析:(1)在矩形 ABCD 中, (2)过点 P 作 PH ⊥ AB 于点 H , AP=t , AQ = 3- t , 由△ AHP ∽△ ABC ,得 ![](https://iknow-pic.cdn.bcebos.com/b3fb43166d224f4a9cbd3ca10af790529822d158?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,∴ PH= , ![](https://iknow-pic.cdn.bcebos.com/472309f790529822b3177671d4ca7bcb0a46d458?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) , ![](https://iknow-pic.cdn.bcebos.com/d01373f082025aafd70e3179f8edab64034f1a47?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) . (3) ①如图②,线段 PQ 的垂直平分线为 l 经过点 A ,则 AP=AQ , 即3 -t=t ,∴ t= 1.5,∴ AP=AQ= 1.5, 延长 QP 交 AD 于点 E ,过点 Q 作 QO ∥ AD 交 AC 于点 O , 则 ![](https://iknow-pic.cdn.bcebos.com/83025aafa40f4bfb8eecd1e6004f78f0f7361847?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) , ![](https://iknow-pic.cdn.bcebos.com/730e0cf3d7ca7bcbcf5f889dbd096b63f624a858?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,∴ PO=AO -AP= 1. 由△ APE ∽△ OPQ ,得 ![](https://iknow-pic.cdn.bcebos.com/4afbfbedab64034f00f18db4acc379310a551d47?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) . ②(ⅰ)如图③,当点 Q 从 B 向 A 运动时 l 经过点 B , BQ = CP = AP = t ,∠ QBP =∠ QAP ∵∠ QBP +∠ PBC =90°,∠ QAP +∠ PCB =90° ∴∠ PBC =∠ PCB CP = BP = AP = t ∴ CP = AP = AC = ![](https://iknow-pic.cdn.bcebos.com/eac4b74543a982264dfd64958982b9014a90eb40?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ×5=2.5 ∴ t =2.5. (ⅱ)如图④,当点 Q 从 A 向 B 运动时 l 经过点 B , BP = BQ =3-( t- 3)=6- t , AP = t , PC = 5-t , 过点 P 作 PG ⊥ CB 于点 G 由△ PGC ∽△ ABC , 得 ![](https://iknow-pic.cdn.bcebos.com/d52a2834349b033b83acb36816ce36d3d539bd58?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) , BG =4- ![](https://iknow-pic.cdn.bcebos.com/14ce36d3d539b600e4ea4005ea50352ac65cb758?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) = 由勾股定理得 ![](https://iknow-pic.cdn.bcebos.com/f703738da9773912381acad3fb198618367ae240?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,即 ![](https://iknow-pic.cdn.bcebos.com/5243fbf2b21193135867b00266380cd791238d58?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) ,解得 ![](https://iknow-pic.cdn.bcebos.com/faf2b2119313b07e23991fba0fd7912397dd8c58?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto) . 点评:本题考查矩形,相似三角形,要求考生掌握矩形的性质,相似三角形的判定方法,会判定两个三角形相似 |
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