Area of triangle abc = 12.8 square units area of triangle acd = 18.4 square units area of triangle aed = 14.8 square units a polygon abcde is shown. the vertices a and d are joined with a dashed line, and the vertices a and c are joined with a dashed line. detailed show you work : )

Question is not complete.
Assuming the side lengths of a REGULAR pentagon is required.

Sum of areas of the pentagon = 12.8+14.8+18.4=55
Area of triangle formed by the centre and one of the sides = 55/5=11 sq. units.
For a pentagon, the interior angle is (5-2)*180/5=108 degrees.
Therefore base angle of each isosceles triangle (base = side) 
=108/2=54 degrees.

Let length of each pentagon side = c
Area of triangle, 
(c/2)(c/2)tan(54 degrees)/2=11
c^2tan(54)/8 = 11 sq. units
c^2=11*8/tan(54)=63.94 sq. units
c=8.00 units = side length of pentagon

You must find the surface of the block therefore 98187 dived by 5628 = 583m squared