geometry
posted by cindy .
Find the area of a regular pentagon with a side of 6 ft. Give the answer to the nearest tenth.
ok can't remember how to do this

A regular pentagon can be drawn as 5 conguent isosceles triangles with a central angle of 72° each, and equal base angles of 54°.
So just find the area of one of these by first finding the altitude of one of them
tan 54° = h/3
h = 3tan54°
area of one of them = (1/2)base x height
= (1/2)(6)(3tan54)
= 9tan 54°
so the whole pentagon = 45tan 54°
= appr. 61.9 square ft 
You forgot to give us the lenght of the side of the line that connects from the middle of the pentagon to a side of the pentagon.

It's there; you just need to recognize it:
h = 3tan54°