the diagonal of a parallelogram are 18 cm and 30cm respectively.one side of a parallelogram is 12cm. find the area of a parallelogram?

Refer to the figure below.
ABCD is a parallelogram with AB = 12 cm.
The diagonals AC (18 cm) and BD (30 cm) meet at M.

The two diagonals of parallelogram bisect each other.
Hence, AM = 18/2 = 9 cm
and BM = 30/2 = 15 cm

Use Heron's formula to calculate the area of ΔABM:
s = (12 + 15 + 9)/2 = 18
Area of ΔABM = √[18 × (18 - 12) × (18 - 15) × (18 - 9)] = 54

The two diagonals divide the parallelogram into 4 triangles with equal areas.
Area of the parallelogram = 54 × 4 = 216 (square units)

Refer to the figure :

In any parallelogram, the two diagonals ( in our case AC and BD ) bisect each other 

(say at O) such that areas of the four triangles are equal to each other.  Hence --

Area (A)  of whole of the parallelogram = 4 * (Area of any of the triangles)

=>  A  =  4 * ( Area of triangle AOB )  ........................ (1)

For the area of Triangle AOB we will use HERON's formula. According to this 

formula -- If sum of all the three sides = P then ,

P/2 is known as Semi-perimeter and denoted as ' s '

Then area of the triangle --

= √[ s ( s - a ) ( s - b ) ( s - c ) ]

Where a, b and c are measurement of the three sides.

In our case  s  = (1/2) ( 9 + 12 + 15 ) = 18 cm

Hence Area of triangle AOB = √ [ 18 ( 18 - 9 ) ( 18 - 12 ) ( 18 - 15 ) ] cm²

=> √[ 18 * 9 * 6 * 3 ]  =  54 cm²

Hence Area of the Parallelogram  =  4 * 54  =   216 cm²  ............... Answer  

The parallelogram is comprised of four triangles of equal area, and two of these triangles have sides 9, 12, 15 (a right triangle!), so the area of the parallelogram is 4×½×9×12 = 216 cm².