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九年级数学解答题一般
题目
如图,在四边形ABCDABCD,AB,ABCD,ADC=90CD,\angle ADC=90^{\circ},ACBCAC\bot BC,DC=9cmDC=9cm,AD=12cmAD=12cm.点PPAA点出发,沿ABAB向点BB匀速运动,同时点QQBB点出发,沿BCBC向点CC匀速运动,运动速度均为5cm/s5cm/s,当其中一点到达终点时,两点都停止运动.设运动时间为t(s)(0<t<4)t\left(s\right)\left(0 \lt t \lt 4\right).
(1)(1)求线段ABAB的长度;
(2)t(2)t为何值时,以BBPPQQ为顶点的三角形与ADC\triangle ADC相似?
(3)(3)是否存在某一时刻tt,使得四边形DPQCDPQC的面积等于144cm2144cm^{2}?若存在,求出此时tt的值;若不存在,说明理由.
(4)(4)是否存在某一时刻tt,使DPPQDP\bot PQ?若存在,直接写出此时tt的值;若不存在,说明理由.
知识点:勾股定理、矩形的判定章节:第1章 特殊的平行四边形 / 1.2 矩形的性质与判定

答案与解析

答案

(1)在RtACDRt\triangle ACD中,由勾股定理得,AC=92+122=15AC=\sqrt{{9}^{2}+1{2}^{2}}=15
ACBC\because AC\bot BC
ACB=90\therefore \angle ACB=90^{\circ}
ACB=ADC\therefore \angle ACB=\angle ADC
DC\because DCABAB
ACD=CAB\therefore \angle ACD=\angle CAB
ACB\therefore \triangle ACBCDA\triangle CDA
ACCD=ABAC\therefore \frac{AC}{CD}=\frac{AB}{AC}
159=AB15\therefore \frac{15}{9}=\frac{AB}{15}
AB=25\therefore AB=25
(2)(2)由题意知,BP=255tBP=25-5tBQ=5tBQ=5t
由(1)知,B=DAC\angle B=\angle DAC
BPQ\triangle BPQADC\triangle ADC时,
BPAD=BQAC\therefore \frac{BP}{AD}=\frac{BQ}{AC}
255t12=5t15\therefore \frac{25-5t}{12}=\frac{5t}{15}
解得t=259t=\frac{25}{9}
BPQ\triangle BPQACD\triangle ACD时,
BPAC=BQAD\therefore \frac{BP}{AC}=\frac{BQ}{AD}
255t15=5t12\therefore \frac{25-5t}{15}=\frac{5t}{12}
解得t=209t=\frac{20}{9}
综上:t=259t=\frac{25}{9}209\frac{20}{9}时,以BBPPQQ为顶点的三角形与ADC\triangle ADC相似
(3)(3)QHABQH\bot ABHH

BQ=5t\because BQ=5t
QH=3tQH=3t
\therefore四边形DPQCDPQC的面积=S梯形ABCDSADPSBPQ=S_{梯形ABCD}-S_{\triangle ADP}-S_{\triangle BPQ}
=12(9+25)×1212×12×5t12×(255t)×3t=\frac{1}{2}(9+25)×12-\frac{1}{2}×12×5t-\frac{1}{2}×(25-5t)\times 3t
152t2135t2+204=144\therefore \frac{15}{2}{t}^{2}-\frac{135t}{2}+204=144
解得t1=1t_{1}=1t2=8(舍去)t_{2}=8(舍去)
\thereforet=1t=1时,四边形DPQCDPQC的面积等于144cm2144cm^{2}
(4)(4)DPPQDP\bot PQ时,
DPA+QPH=90\therefore \angle DPA+\angle QPH=90^{\circ}
APD+ADP=90\because \angle APD+\angle ADP=90^{\circ}
ADP=QPH\therefore \angle ADP=\angle QPH
DAP=QHP\because \angle DAP=\angle QHP
DAP\therefore \triangle DAPPHQ\triangle PHQ
ADPH=APHQ\therefore \frac{AD}{PH}=\frac{AP}{HQ}
12259t=5t3t\therefore \frac{12}{25-9t}=\frac{5t}{3t}
解得t=8945t=\frac{89}{45}
t=8945\therefore t=\frac{89}{45}时,DPPQDP\bot PQ.

解析

(1)在RtACDRt\triangle ACD中,由勾股定理得,AC=92+122=15AC=\sqrt{{9}^{2}+1{2}^{2}}=15
ACBC\because AC\bot BC
ACB=90\therefore \angle ACB=90^{\circ}
ACB=ADC\therefore \angle ACB=\angle ADC
DC\because DCABAB
ACD=CAB\therefore \angle ACD=\angle CAB
ACB\therefore \triangle ACBCDA\triangle CDA
ACCD=ABAC\therefore \frac{AC}{CD}=\frac{AB}{AC}
159=AB15\therefore \frac{15}{9}=\frac{AB}{15}
AB=25\therefore AB=25
(2)(2)由题意知,BP=255tBP=25-5tBQ=5tBQ=5t
由(1)知,B=DAC\angle B=\angle DAC
BPQ\triangle BPQADC\triangle ADC时,
BPAD=BQAC\therefore \frac{BP}{AD}=\frac{BQ}{AC}
255t12=5t15\therefore \frac{25-5t}{12}=\frac{5t}{15}
解得t=259t=\frac{25}{9}
BPQ\triangle BPQACD\triangle ACD时,
BPAC=BQAD\therefore \frac{BP}{AC}=\frac{BQ}{AD}
255t15=5t12\therefore \frac{25-5t}{15}=\frac{5t}{12}
解得t=209t=\frac{20}{9}
综上:t=259t=\frac{25}{9}209\frac{20}{9}时,以BBPPQQ为顶点的三角形与ADC\triangle ADC相似
(3)(3)QHABQH\bot ABHH

BQ=5t\because BQ=5t
QH=3tQH=3t
\therefore四边形DPQCDPQC的面积=S梯形ABCDSADPSBPQ=S_{梯形ABCD}-S_{\triangle ADP}-S_{\triangle BPQ}
=12(9+25)×1212×12×5t12×(255t)×3t=\frac{1}{2}(9+25)×12-\frac{1}{2}×12×5t-\frac{1}{2}×(25-5t)\times 3t
152t2135t2+204=144\therefore \frac{15}{2}{t}^{2}-\frac{135t}{2}+204=144
解得t1=1t_{1}=1t2=8(舍去)t_{2}=8(舍去)
\thereforet=1t=1时,四边形DPQCDPQC的面积等于144cm2144cm^{2}
(4)(4)DPPQDP\bot PQ时,
DPA+QPH=90\therefore \angle DPA+\angle QPH=90^{\circ}
APD+ADP=90\because \angle APD+\angle ADP=90^{\circ}
ADP=QPH\therefore \angle ADP=\angle QPH
DAP=QHP\because \angle DAP=\angle QHP
DAP\therefore \triangle DAPPHQ\triangle PHQ
ADPH=APHQ\therefore \frac{AD}{PH}=\frac{AP}{HQ}
12259t=5t3t\therefore \frac{12}{25-9t}=\frac{5t}{3t}
解得t=8945t=\frac{89}{45}
t=8945\therefore t=\frac{89}{45}时,DPPQDP\bot PQ.

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