notam ABCD trapezul
ducem CM perpendiculara pe AB
DC/AB=2/3 DC = 2AB/3
AB=DC+12 AB=2AB/3+12 3AB=2AB+36 AB=36 ⇒ DC=2×36/3=24
teorema inaltimii
CM =√AM×MB=√24×12=12√2
teorema catetei
AC=√AB×AM=√36×24=12√6 (diagonala mica)
BC=√AB×MB=√36×12=12√3
teorema lui Pitagora
DB=√AD²+AB²=√[(12√2)²+36²]=√(288+1296) =√1584=12√11 (diagonala mare)
P=24+36+12√3+12√2=12(5+√3+√2)
A=(24+36)×12√2/2=360√2