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八年级数学填空题一般
题目
如图,点PPAOB\angle AOB内一点,分别作出PP点关于OAOAOBOB的对称点P1P_{1},P2P_{2},连接P1P2P_{1}P_{2}OAOAMM,交OBOBNN,P1P2=20P_{1}P_{2}=20,则PMN\triangle PMN的周长为______.
知识点:轴对称的性质章节:第二章 轴对称 / 2.1 轴对称及其性质

答案与解析

答案

\becausePP关于OAOAOBOB的对称点P1P_{1}P2P_{2}
PM=P1M\therefore PM=P_{1}MPN=P2NPN=P_{2}N
PMN\therefore \triangle PMN的周长=PM+MN+PN=P1M+MN+P2N=P1P2=PM+MN+PN=P_{1}M+MN+P_{2}N=P_{1}P_{2}
P1P2=20\because P_{1}P_{2}=20
PMN\therefore \triangle PMN的周长=20=20.
故答案为:2020.

解析

\becausePP关于OAOAOBOB的对称点P1P_{1}P2P_{2}
PM=P1M\therefore PM=P_{1}MPN=P2NPN=P_{2}N
PMN\therefore \triangle PMN的周长=PM+MN+PN=P1M+MN+P2N=P1P2=PM+MN+PN=P_{1}M+MN+P_{2}N=P_{1}P_{2}
P1P2=20\because P_{1}P_{2}=20
PMN\therefore \triangle PMN的周长=20=20.
故答案为:2020.

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