Question 3904
 Let AB be a side of the pentagon and O be the center.

  then angle AOB = 360/5 = 72 degrees.
 
 And the area of AOB = 1/2 AOsin angle AOB
                     = 1/2 r^2 sin AOB = 1/2 * 144 sin 72'
                     = 72 sin 72 = 68.48 cm^2

 So the area of the pentagon = 5 area of triangle AOB = 342.38 cm^2

 Kenny