Distance Halfway Around Track Between Inner and Outer Edges

Question

Here is a problem from an old Algebra book that has me a little stumped. I am not clear exactly what they are asking for or how to solve it.

(Problem 60.) Show that T (the area of the Track) is the product of w and the distance around the track halfway between the inner and outer edges.

The answer to problem 59 which is related to the area of the track is T = w[2s + PI(2r +w)]

I am not sure if that is needed to solve problem 60 or not.

I have included images from the book here.

Dolciani; Wooton; Sorgenfrey; Brown. Algebra: Structure and Method (Book 1). Houghton Mifflin Company, 1979, p. 163, ISBN: 0-395-26637-8.

Answer 1

I failed to recognize that what the question was asking about a perimeter--specifically the perimeter between the inner and outer circle.

Outer perimeter = 2 * PI * (r+w) + 2s

Inner perimeter = 2 * PI * r + 2s

Perimeter halfway between inner and outer circle = 2PI(r+1/2w)

So...

P = Perimeter

P=2 * PI*(r+1/2w)

P=2 * PI*r + PI * w + PI * s

P=2s + PI (2r+w)

Since the answer to number 59 was

T= w[2s + PI (2r+w)]

we can say that the area of the track, T, is w times the perimeter of the track at a distance of halfway between the inner and outer edges of track T.