The diasonals of a quadrilateral ABCD meet at O.It is known that |ABO| = 6cm sq ,|BOC| = 4cm sq and |AOD| = 9 cm2.FIND |COD| and show that BC is...

Let ABCD be quadrilateral with O as the intersection of
the diagonals AC and BD.

Now draw a perpendicular from the
vertex A  to meet the diagonal BD at X.

Then  areaof
traiangle AOD/ area of triangle B = 9 sq cm/6 sq cm {(1/2)DO*AX}/{(1/2)OB*AX} = OD/DB.
Therefore = OD/DB = 9/6 = 3/2....(1).

Draw a perpendicular
from the vertex B to meet the diagonal AC at Y.

Then area
of AOB/area of BOC = 6 sq cm/4 sq cm = {(1/2)AO*YB}/{(1/2)OC*YB} = AO/OC. Therefore
AO/OC = 6/4 =3/2...(2).

From (1) and (2),
OD/DC = AO/OC. Therefore by Thale's theorem  (for the  intercepts) the lines AD and BC
are ||

Now AD||BC. Therefore triangles ADB
and ADC having the common base AD, and their vertexes B and C being on the line BC || to
AD must have the same area = 9sq cm+6 sq cm = 15 sq
cm.

Therefore area of triangle ACD =
15cm.

Therefore the area of triangle COD = area of ACD -
area of AOD = 15sq cm - 9 sq cm = 6sq cm.

So
area of COD = 6sq cm.