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八年级数学填空题一般
题目
如图,直线l1l_{1},l2l_{2},l3l_{3}分别过正方形ABCDABCD的三个顶点AA,DD,CC,且相互平行,若l1l_{1},l2l_{2}的距离为11,l2l_{2},l3l_{3}的距离为22,则正方形的边长为______.
知识点:平行线之间的距离、全等三角形的判定、勾股定理、正方形的性质章节:未标注

答案与解析

答案

如图,过点DDEFl1EF\bot l_{1}l1l_{1}于点EE,交l3l_{3}于点FF

l1\because l_{1}l2l_{2}l3l_{3}
EFl2\therefore EF\bot l_{2}EFl3EF\bot l_{3}
AED=ADC=CFD=90\therefore \angle AED=\angle ADC=\angle CFD=90^{\circ}DE=1DE=1DF=2DF=2
ADE+DAE=90\therefore \angle ADE+\angle DAE=90^{\circ}ADE+CDF=90\angle ADE+\angle CDF=90^{\circ}
DAE=CDF\therefore \angle DAE=\angle CDF
\because四边形ABCDABCD是正方形,
AD=CD\therefore AD=CD
ADE\therefore \triangle ADEDCF\triangle DCF
AE=DF=2\therefore AE=DF=2DE=CF=1DE=CF=1AD=DE2+AE2=5AD=\sqrt{D{E^2}+A{E^2}}=\sqrt{5}
即正方形的边长为5\sqrt{5}.
故答案为:5\sqrt{5}.

解析

如图,过点DDEFl1EF\bot l_{1}l1l_{1}于点EE,交l3l_{3}于点FF

l1\because l_{1}l2l_{2}l3l_{3}
EFl2\therefore EF\bot l_{2}EFl3EF\bot l_{3}
AED=ADC=CFD=90\therefore \angle AED=\angle ADC=\angle CFD=90^{\circ}DE=1DE=1DF=2DF=2
ADE+DAE=90\therefore \angle ADE+\angle DAE=90^{\circ}ADE+CDF=90\angle ADE+\angle CDF=90^{\circ}
DAE=CDF\therefore \angle DAE=\angle CDF
\because四边形ABCDABCD是正方形,
AD=CD\therefore AD=CD
ADE\therefore \triangle ADEDCF\triangle DCF
AE=DF=2\therefore AE=DF=2DE=CF=1DE=CF=1AD=DE2+AE2=5AD=\sqrt{D{E^2}+A{E^2}}=\sqrt{5}
即正方形的边长为5\sqrt{5}.
故答案为:5\sqrt{5}.

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