se duc diagonalele trapez
notam DO=x
OB=16-x
Aplicam teorema asemanarii in triunghiurile DOC BOA deoarece DC//AB
[tex] \frac{DC}{AB} [/tex]=[tex] \frac{DO}{OB} [/tex]
[tex] \frac{6}{18} [/tex]=[tex] \frac{x}{16-x} [/tex]
[tex] \frac{1}{3} [/tex]=[tex] \frac{x}{16-x} [/tex]
3x=16-x
3x+x=16
4x=16
x=16/4
x=4 deci DO=4 BO=16-4=12