Problema 1
intr un triunghi echilateral, h=L√3/2 si Aria = L²√3/4
a) L²=25√3*4/√3 = > L²=25*4 => L=10cm
L=10 => h=5√3cm
b) L²=48√3*4/√3 = > L²=48*4 => L=8√3cm
L=8√3 => h=12cm
c) L²=225√3*4/√3*2 = > L²=225*2 => L=15√2cm
L=15√2 => h=15√6/2cm
d) L²=27√3*4/√3 => L²=27*4 =>L=6√3cm
L=6√3cm => h=6√3*√3/2=9cm
Problema 2
Fie DD' perpendiculara pe AC, cu D'∈AC, cf ipoteza DD'=8cm
fie BB' perpendiculara pe AC, cu B''∈AC
BB' || DD'
CD=DB
rezulta triunghiurile CDD' si CBB' sunt asemenea
rezulta BB'=2DD'=16cm
Atunci Aria ABC=BB'*AC/2=16*12/2=96cm²
Fie DE perpendiculara pe AB cu E ∈ AB
Aria DAB =Aria ABC - Aria DAC =>
DE*AB/2=BB'*AC/2-DD'*AC/2
DE*16=16*12-8*12
DE*16=8*12 => DE=6cm
