Question

ABCD romb
AC=16
BD=12
-------------
AB=?

Answer 1

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. 

Answer 2

AC= 16, BD= 12 
AC  n BD = {O}
BO=OD = $\frac{BD}{2}$ = 6cm
AO=OC= $\frac{AC}{2}$ = 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