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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,点A(0,5)A\left(0,5\right),B(3,0)B\left(3,0\right),点CC在第二象限,且AB=ACAB=AC,过点CCCDyCD\bot y轴于点DD,AD=OBAD=OB.求点CC的坐标.
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

CDy\because CD\bot y轴,
ADC=90\therefore \angle ADC=90^{\circ}
AOB=90\because \angle AOB=90^{\circ}
RtACDRt\triangle ACDRtBAORt\triangle BAO中,
{AC=BAAD=BO\left\{\begin{array}{c}AC=BA\\ AD=BO\end{array}\right.
RtACD\therefore Rt\triangle ACDRtBAO(HL)Rt\triangle BAO\left(HL\right)
CD=AO\therefore CD=AO
A(0,5)\because A\left(0,5\right)B(3,0)B\left(3,0\right)
AO=5\therefore AO=5BO=3BO=3
CD=AO=5\therefore CD=AO=5
AD=OB=3\therefore AD=OB=3
OD=AOAD=53=2\therefore OD=AO-AD=5-3=2
\becauseCC在第二象限,
C(5,2)\therefore C\left(-5,2\right).

解析

CDy\because CD\bot y轴,
ADC=90\therefore \angle ADC=90^{\circ}
AOB=90\because \angle AOB=90^{\circ}
RtACDRt\triangle ACDRtBAORt\triangle BAO中,
{AC=BAAD=BO\left\{\begin{array}{c}AC=BA\\ AD=BO\end{array}\right.
RtACD\therefore Rt\triangle ACDRtBAO(HL)Rt\triangle BAO\left(HL\right)
CD=AO\therefore CD=AO
A(0,5)\because A\left(0,5\right)B(3,0)B\left(3,0\right)
AO=5\therefore AO=5BO=3BO=3
CD=AO=5\therefore CD=AO=5
AD=OB=3\therefore AD=OB=3
OD=AOAD=53=2\therefore OD=AO-AD=5-3=2
\becauseCC在第二象限,
C(5,2)\therefore C\left(-5,2\right).

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