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,在ABC\triangle ABC中,C=90\angle C=90^{\circ},ABC=60\angle ABC=60^{\circ}.
(1)(1)ACAC上找一点DD,使得点DDAABB的距离相等;(尺规作图,保留作图痕迹,不写作法和证明)
(2)(2)在(1)的条件下,若AC=6AC=6,求点DDABAB的距离.
知识点:展开图折叠成几何体、角平分线、线段垂直平分线的性质章节:未标注

答案与解析

答案

(1)如图,点DD即为所求.

(2)(2)设线段ABAB的垂直平分线交ABAB于点EE,则点DDABAB的距离即为DEDE的长.
C=90\because \angle C=90^{\circ}ABC=60\angle ABC=60^{\circ}
A=30\therefore \angle A=30^{\circ}.
\becauseDDAABB的距离相等,即AD=BDAD=BD
A=ABD=30\therefore \angle A=\angle ABD=30^{\circ}
CBD=ABCABD=30\therefore \angle CBD=\angle ABC-\angle ABD=30^{\circ}
ABD=CBD\therefore \angle ABD=\angle CBD
BD\therefore BDABC\angle ABC的平分线,
C=90\because \angle C=90^{\circ}DEABDE\bot AB
DE=CD\therefore DE=CD.
RtADERt\triangle ADE中,A=30\angle A=30^{\circ}
AD=2DE\therefore AD=2DE
AD=2CD\therefore AD=2CD
AC=AD+CD=3CD=6\therefore AC=AD+CD=3CD=6
CD=DE=2\therefore CD=DE=2
\thereforeDDABAB的距离为22.

解析

(1)如图,点DD即为所求.

(2)(2)设线段ABAB的垂直平分线交ABAB于点EE,则点DDABAB的距离即为DEDE的长.
C=90\because \angle C=90^{\circ}ABC=60\angle ABC=60^{\circ}
A=30\therefore \angle A=30^{\circ}.
\becauseDDAABB的距离相等,即AD=BDAD=BD
A=ABD=30\therefore \angle A=\angle ABD=30^{\circ}
CBD=ABCABD=30\therefore \angle CBD=\angle ABC-\angle ABD=30^{\circ}
ABD=CBD\therefore \angle ABD=\angle CBD
BD\therefore BDABC\angle ABC的平分线,
C=90\because \angle C=90^{\circ}DEABDE\bot AB
DE=CD\therefore DE=CD.
RtADERt\triangle ADE中,A=30\angle A=30^{\circ}
AD=2DE\therefore AD=2DE
AD=2CD\therefore AD=2CD
AC=AD+CD=3CD=6\therefore AC=AD+CD=3CD=6
CD=DE=2\therefore CD=DE=2
\thereforeDDABAB的距离为22.

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 →