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
To find the area of parallelogram ABCD, we can use the following formula:
Area of parallelogram = base × height
In this case, we are given the areas of triangles ADP and BCP, which share the same base (CD). Therefore, we can find the total area of parallelogram ABCD by summing up the areas of both triangles.
Area of parallelogram ABCD = Area of triangle ADP + Area of triangle BCP
Let's substitute the given values:
Area of parallelogram ABCD = 48 cm² + 30 cm² = 78 cm²
Therefore, the area of parallelogram ABCD is 78 square centimeters.