Question 195460: Please help to resolve this problem. It has 3 parts.
1. A running track 4 m wide goes around a soccer field that is twice as long as it is wide. At each end of the soccer field the track is a semi-circle with inner radius is r. Find the formula for the area of the track in terms of (pi) and r.
1a. Suppose that you plan to run once around the track described in problem above, if you stay 0.5 m from the inner edge of the track, how far will you run? (Hint: The circumfrence of a circle is 2(pi)r. Your anwser will be in terms of pi and r).
1b. Suppose that a friend stays 0.5 m from the outer edge of the track. How much farther does your friend run than you do?
Thanks a lot for your help.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
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 r, then the width of the soccer field, which is the same as the diameter of the semicircle has to be 2r. Since the soccer field is twice as long as it is wide, the length of the soccer field has to be 4r.
First, let'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 r, and the rectangle is 2r by 4r.
So the area of the circle is:
And the area of the rectangle part is:
Add these together to get the whole area of the inside of the track:
r^2)
Now let'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 r + 4, and the area of both semicircles is then:
^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 2r + 8 and the length is still 4r and the area is:
\cdot4r = 8r^2 + 32r)
And again, let's add these two areas to get the total area of the outside figure:
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:
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't matter how far away from the inner edge you are, you are still going to run 4r meters on each side, for a total of 8r 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 r + a where a is your distance from the inner edge. So,
)
And then the total distance that you will run is:
)
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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:
)
John
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