Răspuns :
ABCD romb
AC = 16 cm
BD = 12 cm
AC = 16/2 = 8
BD = 12/2 = 6
AB = 8² + 6² = 64 + 36 = √100 = 10 cm ( teorema lui Pitagora )
sau ( 16/2 )² + ( 12/2 )² = 8² + 6² = 64 + 36 = √100 = 10.
AC = 16 cm
BD = 12 cm
AC = 16/2 = 8
BD = 12/2 = 6
AB = 8² + 6² = 64 + 36 = √100 = 10 cm ( teorema lui Pitagora )
sau ( 16/2 )² + ( 12/2 )² = 8² + 6² = 64 + 36 = √100 = 10.
AC= 16, BD= 12
AC n BD = {O}
BO=OD = [tex] \frac{BD}{2} [/tex] = 6cm
AO=OC= [tex] \frac{AC}{2} [/tex] = 8 cm
In ΔAOB, m(BOA)=90° → aplic Teorema lui Pitagora
AC_|_BD
AO²+OB²=AB²
8²+6²=AB²
100=AB² → AB= √100
AB= 10cm
AC n BD = {O}
BO=OD = [tex] \frac{BD}{2} [/tex] = 6cm
AO=OC= [tex] \frac{AC}{2} [/tex] = 8 cm
In ΔAOB, m(BOA)=90° → aplic Teorema lui Pitagora
AC_|_BD
AO²+OB²=AB²
8²+6²=AB²
100=AB² → AB= √100
AB= 10cm