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八年级数学解答题一般
题目
如图,在RtABCRt\triangle ABC中,AB=10AB=10,BCACBC\bot AC,PP为线段ACAC上一点,点QQ,PP关于直线BCBC对称,QDABQD\bot AB于点DD,DQDQBCBC交于点EE,连结DPDP,设AP=mAP=m,
(1)(1)BC=8BC=8,求ACAC的长,并用含mm的代数式表示PQPQ的长;
(2)(2)在(1)的条件下,若AP=PDAP=PD,求mm的值;
(3)(3)连结PEPE,若A=60\angle A=60^{\circ},PCE\triangle PCEPDE\triangle PDE的面积之比为1:31:3,求mm的值.
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)在RtABCRt\triangle ABC中,ACB=90\angle ACB=90^{\circ}AB=10AB=10BC=8BC=8
AC=AB2BC2=10282=6\therefore AC=\sqrt{A{B}^{2}-B{C}^{2}}=\sqrt{1{0}^{2}-{8}^{2}}=6
P\because PQQ关于BCBC对称,
PC=CQ=6m\therefore PC=CQ=6-m
PQ=2PC=122m\therefore PQ=2PC=12-2m
(2)(2)AP=PDAP=PD时,A=PDA\angle A=\angle PDA
QDAB\because QD\bot AB
ADQ=90\therefore \angle ADQ=90^{\circ}
PDQ+ADP=90\therefore \angle PDQ+\angle ADP=90^{\circ}Q+A=90\angle Q+\angle A=90^{\circ}
Q=PDQ\therefore \angle Q=\angle PDQ
PD=PQ\therefore PD=PQ
PA=PQ\therefore PA=PQ
m=122m\therefore m=12-2m
m=4\therefore m=4
(3)CP=CQ(3)\because CP=CQ
SPEC=SECQ\therefore S_{\triangle PEC}=S_{\triangle ECQ}
SPDE=3SPEC\because S_{\triangle PDE}=3S_{\triangle PEC}
SPDE\therefore S_{\triangle PDE:}SPEQS_{\triangle PEQ}=3:2=3:2
DE:QE=3:2\therefore DE:QE=3:2
DE=3xDE=3xQE=2xQE=2x
A=60\because \angle A=60^{\circ}ACB=90\angle ACB=90^{\circ}
B=9060=30\therefore \angle B=90^{\circ}-60^{\circ}=30^{\circ}
BE=6x\therefore BE=6x
ADQ=90\because \angle ADQ=90^{\circ}
Q=9060=30\therefore \angle Q=90^{\circ}-60^{\circ}=30^{\circ}
EC=12EQ=x\therefore EC=\frac{1}{2}EQ=x
BC=AB32=53\because BC=AB\cdot \frac{\sqrt{3}}{2}=5\sqrt{3}
6x+x=53\therefore 6x+x=5\sqrt{3}
x=537\therefore x=\frac{5\sqrt{3}}{7}
DQ=5x=2573\therefore DQ=5x=\frac{25}{7}\sqrt{3}CQ=PC=EQ32=157CQ=PC=EQ\cdot \frac{\sqrt{3}}{2}=\frac{15}{7}
m=AP=ACPC=5157=207\therefore m=AP=AC-PC=5-\frac{15}{7}=\frac{20}{7}.

解析

(1)在RtABCRt\triangle ABC中,ACB=90\angle ACB=90^{\circ}AB=10AB=10BC=8BC=8
AC=AB2BC2=10282=6\therefore AC=\sqrt{A{B}^{2}-B{C}^{2}}=\sqrt{1{0}^{2}-{8}^{2}}=6
P\because PQQ关于BCBC对称,
PC=CQ=6m\therefore PC=CQ=6-m
PQ=2PC=122m\therefore PQ=2PC=12-2m
(2)(2)AP=PDAP=PD时,A=PDA\angle A=\angle PDA
QDAB\because QD\bot AB
ADQ=90\therefore \angle ADQ=90^{\circ}
PDQ+ADP=90\therefore \angle PDQ+\angle ADP=90^{\circ}Q+A=90\angle Q+\angle A=90^{\circ}
Q=PDQ\therefore \angle Q=\angle PDQ
PD=PQ\therefore PD=PQ
PA=PQ\therefore PA=PQ
m=122m\therefore m=12-2m
m=4\therefore m=4
(3)CP=CQ(3)\because CP=CQ
SPEC=SECQ\therefore S_{\triangle PEC}=S_{\triangle ECQ}
SPDE=3SPEC\because S_{\triangle PDE}=3S_{\triangle PEC}
SPDE\therefore S_{\triangle PDE:}SPEQS_{\triangle PEQ}=3:2=3:2
DE:QE=3:2\therefore DE:QE=3:2
DE=3xDE=3xQE=2xQE=2x
A=60\because \angle A=60^{\circ}ACB=90\angle ACB=90^{\circ}
B=9060=30\therefore \angle B=90^{\circ}-60^{\circ}=30^{\circ}
BE=6x\therefore BE=6x
ADQ=90\because \angle ADQ=90^{\circ}
Q=9060=30\therefore \angle Q=90^{\circ}-60^{\circ}=30^{\circ}
EC=12EQ=x\therefore EC=\frac{1}{2}EQ=x
BC=AB32=53\because BC=AB\cdot \frac{\sqrt{3}}{2}=5\sqrt{3}
6x+x=53\therefore 6x+x=5\sqrt{3}
x=537\therefore x=\frac{5\sqrt{3}}{7}
DQ=5x=2573\therefore DQ=5x=\frac{25}{7}\sqrt{3}CQ=PC=EQ32=157CQ=PC=EQ\cdot \frac{\sqrt{3}}{2}=\frac{15}{7}
m=AP=ACPC=5157=207\therefore m=AP=AC-PC=5-\frac{15}{7}=\frac{20}{7}.

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