how do you find the area of a pentagon when you are only given the radius

Melody is indeed right; http://www.icoachmath.com/math_dictionary/pentagon.html

In addition to that it could also be an inscribed circle touching each edge instead of the vertices.

You treat each side as a triangle . The length of two of the sides are the radius. The angle between those two sides will be 72°. To calculate the length of the last side, you take half of the angle, so you get a triangle with a right angle. This allows you to use the sine function for the last side. But because you divided the angle by two, your result will be the third side's length divided by two.

Summarized:

Side 1 = Side 2 = r

Side 3 = 2*(sin(36°)*r)

The term "radius" implies that we want to find the area of a pentagon inscribed in a circle. We have 5 triangles. Using trigonometry, each triangle has an area = (1/2)*(r^2)*sin(72º).

And since we have 5 triangles, the total area = (5/2)(r^2)*sin(72º)

Also, if we know the length of a side, the total area is given by 1.7204*(s^2) where s is the side length.

I am thinking about both of your answers.

Since the question talks about a radius it is fair to assume that the pentagon is inscribed in a circle.  That is, the 5 vertices must lie on the circumference of the circle.  However,  why have you both assumed that it is a regular pentagon.  I don't think that all the angles have to be 72 degrees.  BUT if it is not a regular pentagon then there is not enough information to get an answer.  On that basis, maybe your second assumption are not so unreasonable.

Would anyone like to comment?

Thanks reinout, I hadn't thought of that possibility.  

Does everyone deserve brownie points?  What do you think reinout?

definitely