Question

URGENTTTTTTTT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

a)
AB=BC=6
MB=AB/2=6/2=3
A(MBC)=3×6/2=9 cm²


b)
AM=MB
AO=OD'   ⇒ MO este linie mijlocie in triunghiul AMD' ⇒ MO II BD' ⇒
MO II (DBB')

c)
OM II BD'
m∡(OM, BC')=m(∡D'BC')
BD'=√(6²+6²+6²)=6√3
sin∡(OM, BC')=sin(∡D'BC')=D'C'/BD'=6/6√3=1/√3=√3/3