Question

Un trunchi de piramida trunghiulara regulata are razele cercurilor circumscrise bazelor de 4 cm si 2 cm , iar inaltimea
radical din 3 . aflati aria laterala si volumul

Ajutor! Multumesc

Answer 1

R=l=4
r=l=2
h=√3
Pb=3*2=6
PB=4*2=8
h=l²√3/2
h=4²√3/2
h=4√3
h-inaltimea bazei mari-bl
OO'-inaltimea trunchiului
OL-å3
OL=1/3*4√3
OL=4√3/3
inΔSLO dr.  cu  O=90
SL²=SO²+OL²

SL²=(2√3/3)²+(4√3/3)²
SL²=12/9+48/9
SL=√60/9
SL=2√15/3
Al=Pb+PB)*ap/2
PB=3*4=12
Pb=3*2=6
Al=(6?12)*2√15/3/2
Al=18*2√15/3/2
Al=12√15/2=6√15
V=h/3(AB+Ab+√AB*Ab)
V=√3/3(4√3+√3+√4√3*√3)
V=√3/3(5√3+2√3)
V=√3/3*7√3
V=21/3=7