Question 8895
Let's say that we have a wheel whose radius is r. We first have to find the circumference of the circle, which is {{{ C = 2*pi*r }}}. This is, as you know, the length of the string you would wrap around the outermost edge of the wheel ONCE AROUND. If you stretched out this string and laid it on the ground, it will be {{{2*pi*r}}} long.


Now, if the thickness of the wheel is 1/2 the radius, the thickness {{{ T = (1/2)r}}}. If the wheel was rolled out 1 rev, the track's length will be {{{2*pi*r}}} and its thickness will be {{{(1/2)r}}}. The track would then be a rectangle whose dimensions are {{{2*pi*r}}} and  {{{(1/2)r}}}. Since it has those dimensions, it would then make sense to find the area of the track which would be length * width. The length is {{{2*pi*r}}} and the width (AKA, the tire's thickness) is {{{(1/2)r}}}. So:


{{{ A = (2*pi*r)*(1/2)r }}} <------ Start here


{{{ A = pi*r^2 }}} <----- Area of the circle is equal to the area of the track that the wheel would leave behind if it was rolled exactly 1 rev. (That's if the wheel's thickness were 1/2 the radius.