Question

M-am tot gandit la aceasta problema dar nu mi-a dat deloc, va rog mult ajutati-ma!!!!!!!

Answer 1

La calculele am folosit rotunjurile ...

Answer 2

A'M'_I_B'C'
AM_I_BC
O'O inaltimea trunchiului de piramida
AM=L√3/2=4√3/2=2√3 cm
A'M'=l√3/2=3√3/2 cm
OM=L√3/6=4√3/6=2√3/3 cm
O'M'=l√3/6=3√3/6=√3/2 cm
OA=L√3/3=4√3/3 cm
O'A'=l√3/3=3√3/3=OA=√3 cm
A'N_I_AM
NO=A'O'=√3
AN=AO-NO=4√3/3-√3=(4√3-3√3)/3=√3/3
√[(AA')²-AN²]=√(√39/3)²-(√3/3)²=√(39/9-3/9)=√36/9=√(√39/3)²-(√3/3)²=√4=2 cm

V=h/3[Abazei mari+Abazei mici+√(Abazei mari+Abazei mici)]
Abazei mari=L²√3/4=16√3/4=4√3
Abazei mici=l²√3/4=9√3/4=2,25√3
V=2/3[4√3+2,25√3+√(4√3 ×9√3/4)]=
=2/3(4√3+2,25√3+3√3)=
=2/3 ×9,25√3=18,5√3/3= 18,5/√3=18,5 /1,7=10,88 cm³

greutatea bomboanei
10,88 cm³ × 1,3 g/cm³=14,15 g ≈ 14 g