fie CM inaltimea trapezului
in triunghiul dreptunghic CMB
sin B=CM/BC
√2/2=CM/16
CM=8√2
MB=8√2
aplicam teorema inaltimii
CM²=AM·MB
(8√2)²=AM·8√2
AM=128/8√2=8√2
AB -baza mare=16√2
DC-baza mica=8√2
DA=8√2
P=16√2+16+8√2+8√2=32√2+16=16(2√2+1)
DB²=(8√2)²+(16√2)²=128+512=640
DB=2√185
AC=16