题霸题霸学习平台
← 返回公开题库
八年级数学解答题一般
题目
如图11,AB=12AB=12,ACABAC\bot AB,BDABBD\bot AB,AC=BD=8AC=BD=8.点PP在线段ABAB上以每秒22个单位的速度由点AA向点BB运动,同时,点QQ在线段BDBD上由BB点向点DD运动.它们的运动时间为t(s)t\left(s\right).

(1)(1)若点QQ的运动速度与点PP的运动速度相等,当t=2t=2时,ACP\triangle ACPBPQ\triangle BPQ是否全等,请说明理由,并判断此时线段PCPC和线段PQPQ的位置关系;
(2)(2)如图22,将图11中的"ACABAC\bot AB,BDABBD\bot AB"改为"CAB=DBA=60\angle CAB=\angle DBA=60^{\circ}",其他条件不变.设点QQ的运动速度为每秒xx个单位,是否存在实数xx,使得ACP\triangle ACPBPQ\triangle BPQ全等?若存在,求出相应的xx,tt的值;若不存在,请说明理由.
知识点:三角形章节:第13章 三角形 / 13.1 三角形的概念

答案与解析

答案

(1)结论:ACP\triangle ACPBPQ\triangle BPQ全等.
理由如下:当t=2t=2时,AP=BQ=2×2=4AP=BQ=2\times 2=4
BP=ABAP=124=8BP=AB-AP=12-4=8
BP=AC\therefore BP=AC
A=B=90\because \angle A=\angle B=90^{\circ}
ACP\triangle ACPBPQ\triangle BPQ中,
{AP=BQA=BCA=PB\left\{\begin{array}{l}{AP=BQ}\\{∠A=∠B}\\{CA=PB}\end{array}\right.
ACP\therefore \triangle ACPBPQ(SAS)\triangle BPQ\left(SAS\right)
结论:PCPQPC\bot PQ.
证明:ACP\because \triangle ACPBPQ\triangle BPQ
ACP=BPQ\therefore \angle ACP=\angle BPQ
APC+BPQ=APC+ACP=90\therefore \angle APC+\angle BPQ=\angle APC+\angle ACP=90^{\circ}.
CPQ=90\therefore \angle CPQ=90^{\circ}
即线段PCPC与线段PQPQ垂直.

(2)(2)①若ACP\triangle ACPBPQ\triangle BPQ
AC=BPAC=BPAP=BQAP=BQ
{8=122t2t=tx\therefore \left\{\begin{array}{l}{8=12-2t}\\{2t=tx}\end{array}\right.
解得
{t=2x=2\left\{\begin{array}{l}{t=2}\\{x=2}\end{array}\right.
②若ACP\triangle ACPBQP\triangle BQP
AC=BQAC=BQAP=BPAP=BP
{8=xt2t=122t\left\{\begin{array}{l}{8=xt}\\{2t=12-2t}\end{array}\right.
解得
{t=3x=83\left\{\begin{array}{l}{t=3}\\{x=\frac{8}{3}}\end{array}\right.
综上所述,当{t=2x=2\left\{\begin{array}{l}{t=2}\\{x=2}\end{array}\right.{t=3x=83\left\{\begin{array}{l}{t=3}\\{x=\frac{8}{3}}\end{array}\right.时,
使得ACP\triangle ACPBPQ\triangle BPQ全等.

解析

(1)结论:ACP\triangle ACPBPQ\triangle BPQ全等.
理由如下:当t=2t=2时,AP=BQ=2×2=4AP=BQ=2\times 2=4
BP=ABAP=124=8BP=AB-AP=12-4=8
BP=AC\therefore BP=AC
A=B=90\because \angle A=\angle B=90^{\circ}
ACP\triangle ACPBPQ\triangle BPQ中,
{AP=BQA=BCA=PB\left\{\begin{array}{l}{AP=BQ}\\{∠A=∠B}\\{CA=PB}\end{array}\right.
ACP\therefore \triangle ACPBPQ(SAS)\triangle BPQ\left(SAS\right)
结论:PCPQPC\bot PQ.
证明:ACP\because \triangle ACPBPQ\triangle BPQ
ACP=BPQ\therefore \angle ACP=\angle BPQ
APC+BPQ=APC+ACP=90\therefore \angle APC+\angle BPQ=\angle APC+\angle ACP=90^{\circ}.
CPQ=90\therefore \angle CPQ=90^{\circ}
即线段PCPC与线段PQPQ垂直.

(2)(2)①若ACP\triangle ACPBPQ\triangle BPQ
AC=BPAC=BPAP=BQAP=BQ
{8=122t2t=tx\therefore \left\{\begin{array}{l}{8=12-2t}\\{2t=tx}\end{array}\right.
解得
{t=2x=2\left\{\begin{array}{l}{t=2}\\{x=2}\end{array}\right.
②若ACP\triangle ACPBQP\triangle BQP
AC=BQAC=BQAP=BPAP=BP
{8=xt2t=122t\left\{\begin{array}{l}{8=xt}\\{2t=12-2t}\end{array}\right.
解得
{t=3x=83\left\{\begin{array}{l}{t=3}\\{x=\frac{8}{3}}\end{array}\right.
综上所述,当{t=2x=2\left\{\begin{array}{l}{t=2}\\{x=2}\end{array}\right.{t=3x=83\left\{\begin{array}{l}{t=3}\\{x=\frac{8}{3}}\end{array}\right.时,
使得ACP\triangle ACPBPQ\triangle BPQ全等.

自由组卷

围绕这道题再组一份练习 →

完整试卷

浏览同年级试卷结构 →