Question

Please explain this question. 4(b)

Answer 1

Given :ABCD is a parallelogram

Construction :Draw a line GH parallel to AB and EF parallel to AD

To prove:(i)-------
(ii)-------
Proof : In a parallelogram ABCD,
AB parallel to GH ( CONSTRUCTION ) -(1)eq
Thus, AD parallel to BC ( AG Parallel to BH ) -(2)eq

From eq-1 and 2 ----
ABGH is a parallelogram.

Now,
In triangle AOB, and parallelogram ABGH are lying on the same base and between the same parallel lines AB and GH.

# Ar ( Triangle AOB) = 1/2 Ar (ABGH) --3 eq
Also,
In triangle COD and parallelogram CDGH are lying on the same base and between the same parallel CD and GH.
#Ar ( Triangle COD ) = 1/2 Ar ( CDGH ) --4 eq

Adding equations 3 and 4 ----
---#Ar (Triangle AOB ) + Ar (Triangle COD) = 1/2 Ar (ABGH) + Ar (CDGH)
---#Ar (Triangle AOB) + Ar ( Triangle COD) = 1/2 Ar (ABCD).

SIMILARLY, By construction AEDF is a parallelogram

similarly, area of triangle AOD =1/2 AREA OF AEFD ------5 eq
Also,
area of triangle BOC = 1/2 AREA OF BCFE -----6 eq

Adding equation 5 and 6 ------
---#area of triangle AOD + area of triangle BOC = 1/2 (AREA OF (AEFD) + AREA OF ( BCFE ) )
--#AR of triangle AOD + AR of triangle BOC = 1/2 AREA OF (ABCD)