posted by DEEPI .
ABCD IS A PARALLELOGRAM.P IS A POINT ON SIDE CD.AREA OF TRIANGLE ADP=48 SQUARE CENTIMETRE AND AREA OF TRIANGLE BCP=30 SQUARE CENTIMETRE FIND AREA OF PARALLELOGRAM ABCD
Drop an altitude PQ to AB. So, the height of ABCD is PQ.
The area of ABCD is AB*PQ
The area of triangle APB is AB*PQ/2
So, 48+30 = 78 is the rest of the area of ABCD. It is also half that area.
Area ABCD = 156
by basic geometry:
CP:PD = 30:48 = 5:8
let CP=5x, and PD = 8x
let the height between AB and CD = h
so area of BCP + area of ADP = 5xh +8xh = 13xh = 78
area of ABC = 13xh = 78
area of whole thing = 78+30+48 = 156