Question

question no. 5 and 6 ​

Answer 1

ans 6=in triangle AOB and triangleCOD angleCDO=angleOBA(as AB parallel to CD),similarly angleOCD=angle OAB

hence triangleAOB similar to triangle COD

then by CPST,

DO/BO=CD/AB,

DO/15=6/20 =cross multiplication we get =20×OD =90 then OD=4.5