============================================================
95)
Parallelogramma:
base $b= 3,5+26 = 29,5\,cm;$
altezza $h= \sqrt{lo^2-plo^2} = \sqrt{12,5^2-3,5^2} = 12\,cm$ $(teorema\, di\, Pitagora);$
area $A= b·h = 29,5×12 = 354\,cm^2.$
Home
Profilo
Menu