MathematicsHard202×since 2002Q5855Let OA→=2a⃗,OB→=6a⃗+5b⃗\overrightarrow{O A}=2 \vec{a}, \overrightarrow{O B}=6 \vec{a}+5 \vec{b}OA=2a,OB=6a+5b and OC→=3b⃗\overrightarrow{O C}=3 \vec{b}OC=3b, where OOO is the origin. If the area of the parallelogram with adjacent sides OA→\overrightarrow{O A}OA and OC→\overrightarrow{O C}OC is 15 sq. units, then the area (in sq. units) of the quadrilateral OABCO A B COABC is equal to:A32B38C35D40Check answerSkip