What is the area of a pentagon with sides of equal length, measuring 5v+3?

Assuming we have a regular pentagon,

form central triangles, there will be 5 equal ones
Each one is an isosceles triangle with a central angle of 72°
and two equal angles of 54°
we need the height, h

tan 54 = h/((1/2)(5v+3))
h = (1/2)(5v+3) tan 54

area of one of them
= (1/2) base x height
= ...
total area is 5 times that

you do the algebra