<PRE><B>Question 195460</B>
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So what you have is something that is sort of the shape of an allergy capsule -- straight on the sides and round on the ends.  If the inner radius of the semicircle at the end is &lt;b&gt;&lt;i&gt;r&lt;/i&gt;&lt;/b&gt;, then the width of the soccer field, which is the same as the diameter of the semicircle has to be &lt;b&gt;2&lt;i&gt;r&lt;/i&gt;&lt;/b&gt;.   Since the soccer field is twice as long as it is wide, the length of the soccer field has to be &lt;b&gt;4&lt;i&gt;r&lt;/i&gt;&lt;/b&gt;.


First, let&#39;s calculate the area of just the soccer field and the semicircle parts at the ends inside of the track.  What we actually have is a circle, if you count both semicircular ends, plus a rectangle.  The circle is radius &lt;b&gt;&lt;i&gt;r&lt;/i&gt;&lt;/b&gt;, and the rectangle is &lt;b&gt;2&lt;i&gt;r&lt;/i&gt;&lt;/b&gt; by &lt;b&gt;4&lt;i&gt;r&lt;/i&gt;&lt;/b&gt;.


So the area of the circle is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_{ic} = \pi r^2]


And the area of the rectangle part is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_{sf} = 2r\cdot4r = 8r^2]


Add these together to get the whole area of the inside of the track:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_{Ti} = \pi r^2 + 8r^2 = (\pi + 8)r^2]


Now let&#39;s look at the area of the region bounded by the outside edge of the track:


Again, we have two semicircles, but this time the radius is &lt;b&gt;&lt;i&gt;r&lt;/i&gt; + 4&lt;/b&gt;, and the area of both semicircles is then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_{oc} = \pi (r + 4)^2 = \pi(r^2 + 8r + 16) = \pi r^2 + 8\pi r + 16\pi ]


As for the rectangle part, we have increased the width by 4m on each side, but not changed the length, so the new width is &lt;b&gt;2&lt;i&gt;r&lt;/i&gt; + 8&lt;/b&gt; and the length is still &lt;b&gt;4&lt;i&gt;r&lt;/i&gt;&lt;/b&gt; and the area is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_{sf+} = (2r + 8)\cdot4r = 8r^2 + 32r]


And again, let&#39;s add these two areas to get the total area of the outside figure:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_{To} = \pi r^2 + 8\pi r + 16\pi + 8r^2 + 32r = (\pi + 8)r^2 + (8\pi + 32)r + 16\pi].


Now, if you subtract the area of the shape bounded by the inside of the track from the area bounded by the outside of the track, you will have the area of just the track:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_{To} - A_{Ti} = \{(\pi + 8)r^2 + (8\pi + 32)r + 16\pi\} - \{(\pi + 8)r^2\} = (8\pi + 32)r + 16\pi ]


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1a.  For the straight part of the track, it doesn&#39;t matter how far away from the inner edge you are, you are still going to run &lt;b&gt;4&lt;i&gt;r&lt;/i&gt;&lt;/b&gt; meters on each side, for a total of &lt;b&gt;8&lt;i&gt;r&lt;/i&gt;&lt;/b&gt; meters.  But when you go around the two semicircles (both together making a whole circle), you will run the circumference of a circle with radius &lt;b&gt;&lt;i&gt;r&lt;/i&gt; + a&lt;/b&gt; where &lt;b&gt;&lt;i&gt;a&lt;/i&gt;&lt;/b&gt; is your distance from the inner edge. So,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\pi(r + 0.5)]


And then the total distance that you will run is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8r + 2\pi(r + 0.5)]


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1b.  Your friend, running 0.5 meter in from the outside edge of a 4 meter wide track must be 3.5 meters from the inside edge, so his/her distance must be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 8r + 2\pi(r + 3.5)]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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