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八年级数学解答题一般
题目
如图,在平面直角坐标系中,A(4,0)A\left(4,0\right),B(0,3)B\left(0,3\right),以线段ABAB为直角边在第一象限内作等腰直角三角形ABCABC,AB=ACAB=AC,BAC=90\angle BAC=90^{\circ},则点CC坐标为____.
知识点:点的坐标、全等三角形的性质、全等三角形的判定、等腰三角形的性质章节:第13章 三角形 / 13.1 三角形的概念

答案与解析

答案

CDxCD\bot x轴于点DD,则CDA=AOB=90\angle CDA=\angle AOB=90^{\circ}
BAC=90\because \angle BAC=90^{\circ}
DAC=OBA=90OAB\therefore \angle DAC=\angle OBA=90^{\circ}-\angle OAB
CDA\triangle CDAAOB\triangle AOB中,
{CDA=AOBDAC=OBAAC=BA\left\{\begin{array}{l}{∠CDA=∠AOB}\\{∠DAC=∠OBA}\\{AC=BA}\end{array}\right.
CDA\therefore \triangle CDAAOB(AAS)\triangle AOB\left(AAS\right)
A(4,0)\because A\left(4,0\right)B(0,3)B\left(0,3\right)
DC=OA=4\therefore DC=OA=4DA=OB=3DA=OB=3
OD=OA+DA=4+3=7\therefore OD=OA+DA=4+3=7
C(7,4)\therefore C\left(7,4\right)
故答案为:(7,4)\left(7,4\right).

解析

CDxCD\bot x轴于点DD,则CDA=AOB=90\angle CDA=\angle AOB=90^{\circ}
BAC=90\because \angle BAC=90^{\circ}
DAC=OBA=90OAB\therefore \angle DAC=\angle OBA=90^{\circ}-\angle OAB
CDA\triangle CDAAOB\triangle AOB中,
{CDA=AOBDAC=OBAAC=BA\left\{\begin{array}{l}{∠CDA=∠AOB}\\{∠DAC=∠OBA}\\{AC=BA}\end{array}\right.
CDA\therefore \triangle CDAAOB(AAS)\triangle AOB\left(AAS\right)
A(4,0)\because A\left(4,0\right)B(0,3)B\left(0,3\right)
DC=OA=4\therefore DC=OA=4DA=OB=3DA=OB=3
OD=OA+DA=4+3=7\therefore OD=OA+DA=4+3=7
C(7,4)\therefore C\left(7,4\right)
故答案为:(7,4)\left(7,4\right).

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