Question

Am nevoie urgenta de rezolvarea problemelor astea ..

Answer 1

1)a)Ducem MM' perpendiularar pe AB si ABCD=dreptunghi de aici rezulta ca MM' este paralel si egal cu BC deci are lungimea 27 m.
Aria ABCD=L•l=27•36=972 m^2 si Aria ABM=l•h/2=
=27•36/2=486 m^2
Aria fig.nehas.=Aria ABCD-Aria ABM=972-486=486 m^2
b)In triunghiul DMA dreptunghic DM=DC/2=36/2=18 m
=>AM^2=AD^2+DM^2
AM=√(27^2+18^2)√1053=9√13 m
P∆AMB=AM+MB+AB=9√13+9√13+36
c)Ducem AN perpendiculara pe MB
A∆AMB=MM'•AB/2=AN•MB/2=>MM'•AB=AN•MB
36•27=9√13•AN
AN=972/9√13=108/√13=108√13/13
sin unghi AMB=sin unghi AMN=AN/MA=(108√13/13)/9√13=12/13
Deci sin unghi AMB=12/13
2)a)V=Ab•h=36pi•15=113•15=1695cm^3=1695ml
b)A totala=Al+2•Ab
A totala=2pi•R•G+piR^2
=2•pi•6•15+pi•6^2
=180pi+36pi=216pi
c)V paralelipiped=L•l•h=30•12•20=7200cm^3
7200/1695=4 cutii