如下图,p是矩形ABCD边AD上任一点,pm⊥bd与m,pn⊥AC与n,若AB=3,AD=4.则pm+pn=()
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/02 18:32:11
![如下图,p是矩形ABCD边AD上任一点,pm⊥bd与m,pn⊥AC与n,若AB=3,AD=4.则pm+pn=()](/uploads/image/z/12646910-38-0.jpg?t=%E5%A6%82%E4%B8%8B%E5%9B%BE%2Cp%E6%98%AF%E7%9F%A9%E5%BD%A2ABCD%E8%BE%B9AD%E4%B8%8A%E4%BB%BB%E4%B8%80%E7%82%B9%2Cpm%E2%8A%A5bd%E4%B8%8Em%2Cpn%E2%8A%A5AC%E4%B8%8En%2C%E8%8B%A5AB%3D3%2CAD%3D4.%E5%88%99pm%2Bpn%3D%EF%BC%88%EF%BC%89)
如下图,p是矩形ABCD边AD上任一点,pm⊥bd与m,pn⊥AC与n,若AB=3,AD=4.则pm+pn=()
如下图,p是矩形ABCD边AD上任一点,pm⊥bd与m,pn⊥AC与n,若AB=3,AD=4.则pm+pn=()
如下图,p是矩形ABCD边AD上任一点,pm⊥bd与m,pn⊥AC与n,若AB=3,AD=4.则pm+pn=()
假设AB,CD交点O
AC=√(AD^2+CD^2)=5
AO=DO=1/2AC
S△AOD=1/4S□ABCD=1/4*AB*AD=3
又S△AOD
=SAOP+S△DOP
=1/2AO*PN+1/2DO*PM
=1/2*1/2AC(PN+PM)
=5/4(PM+PN)
所以:
5/4(PM+PN)=3
PM+PN=2.4
试问PM+PN与AB有何关系?说明理由 因为PM CD PN AB 所以PN/AB=PC/BC=(BC-BP)/BC 1 PM/DC=BP/BC 2 又因为AB=CD 所以1+2得 (PN+PM )
相似三角形原理,PN/CD=AP/AC,则PN=AP·CD/AC。
同理可得:PM=PD·BA/BD
由题意可知:AB=3,AD=4,AC,BD是矩形的对角线,则AC=BD=5
PM+PN=AP·CD/AC+PD·BA/BD=0.6AP+0.6PD=0.6(AP+PD)=0.6AD=2.4