Home/ Intrebari/Q 71898
Urmator
In Process

AniDeȘcoală.ro Latest Intrebari

denidan
Pe: 2011-02-26T16:03:02+02:00 In: In:

geometrie VII

ABCD paralelogram
M apartine lui AC
MC = 2AM ,
MN || AD
MP || DC
AD = 3 cm
DC = 6 cm

a) MN = ?
b) DEM DM perpendicular pe NP
c ) Daca Aria lui ABCD = 9 radical din 3 ,calculati Aria lui MNPC =?

Similare

2 raspunsuri

  1. yuxdar

    \rm{MC=2AM (1)\\MC=2AM \Rightarrow AC= 3AM (2)\\MN || AD \Rightarrow \Delta MNC \sim \Delta ADC \Rightarrow \frac{MN}{AD}=\frac{MC}{AC} (3)\\(1), (2), (3) \Rightarrow \frac{MN}{AD}= \frac{2AM}{3AM}\Rightarrow \frac{MN}{3}=\frac{2}{3}\Rightarrow MN=2cm.}

  2. denidan

    yuxdar wrote: \rm{MC=2AM (1)\\MC=2AM \Rightarrow AC= 3AM (2)\\MN || AD \Rightarrow \Delta MNC \sim \Delta ADC \Rightarrow \frac{MN}{AD}=\frac{MC}{AC} (3)\\(1), (2), (3) \Rightarrow \frac{MN}{AD}= \frac{2AM}{3AM}\Rightarrow \frac{MN}{3}=\frac{2}{3}\Rightarrow MN=2cm.}

    Iti multumesc ,mult de tot..pfuu privind rezolvarea parca nu mai pare atat de incalcita..pana la punctul asta..in leg cu puctul c) cred ca e usor de aflat aria paralelogramului MNPC ,e ceva legat de un raport,cred
    ..daca am spus o prostie imi cer scuze..ma tot gandesc 🙁

Raspunde

Raspunde