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Question:

# A dog is on a 20 ft leash which is attached to a silo with radius 10 ft. What area can the dog roam?

Serious answers only...show all work.

## Answer:

A= Pi R^2 A = 3.14x10^2 A = 3.14x10 A = 31.4^2 A = 985.96 FtSq
in all seriousness...I would also need to know how high(in feet) the lease is attached to the silo.
Area = Pi times r times r Radius = 20 ft + 10 ft = 30 feet Total Area = Pi * 30 * 30 = 2,827 sq ft Now subtract the area of the silo Pi * 10 * 10 = 314 The dog has 2,827 - 314 = 2,513 sq ft in which to roam.
So this quesiton is tough because you need to factor in how the silo blocks the dog's access to certain areas, noting that the leash can't go through the silo but has to go around. Because of this, all previous answers (as of 4:07 EST) are wrong. Note that the furthest along the length of the silo that the dog can reach is when the leash is taut against the silo, which is precisely (20ft)/(10ft) = 2 radians along the edge in either direction. We may construct a semicircle in which the leash is pulled completely away from the building, which has an area of: A = (π*20^2)/2 = 200π Calculating the rest of the area requires some calculus: Now take the point where the leash is attached to be the top of our circular silo. Say that the leash is stretched taut against the wall, covering an angle equal to q (in radians). Then the length of the rope that is stretched against the wall is simply 10q. Now, we can state that the leash can be stretched tangentially from the last point that the leash is taut, to a length of (20 - 10q). Well this problem is more difficult than I anticipated, so I am going to take an intuitive leap. The boundaries of an area equivalent to the two which I am trying to find should be given by q = 0 r = 10 + 20 - 10q = 30 - 10q r = 10 in order to find the area of such a region, take the following integral: A = (1/2) (0...2)∫[(30 - 10q)^2 - 10^2]dq of course, there are two such areas on each side, so the total area should be: A = (0...2)∫[(30 - 10q)^2 - 10^2]dq the integral itself is not difficult, and evaluating it you find A = 2000/3 Thus, the TOTAL area in which the dog can roam should be given by: A = 200π + 2000/3 which is about 1295 square ft Anyway, there's an answer, I hope it's the right one. I invite anyone to refute my solution, or provide a more lucid line of reasoning to arrive at the same one. In the mean time, I'll double check my work.

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