Question

ABCD is a parallelogram.M is the mid point of AB.MC and BD intersect at O.If area of MOB is 80 sq. m,then find area of triangle DOC?

Answer 1

Since OMB OS a triangle area would be 1/2*a/2*h=80
From this u get a equation a*h=320
(a/2 since MB is a base and M is amid point)
Since ODC is a triangle of base CD
1\2*a*h
In before equation we got a*h=320
Substitute in this equation
Therefore you will get area of trial as 169

Answer 2

consider triangle MCB,triangle DBC
they lie in the same lls so, their height is the same ,but base is half
therefore,1/2ar(DCB)=ar(MCB)
since,ar(DCB)=ar(DOC)+ar(BOC)
         ar(MCB)=ar(MOB)+ar(BOC)
therefore,1/2[ar(DOC)+ar(BOC)]=ar(MOB)+ar(BOC)..........ar(BOC)-gets cancelled
       so, 1/2ar(DOC)+ar(MOB)
       so,ar(DOC)=2ar(MOB)
       so,ar(DOC)=2(80m^2)
       so,ar(DOC)=160m^2