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A goat is tied to an external corner of a rectangular shed measuring 4 m by 6 m. If the goat’s rope is 8 m long, what is the total area, in square meters, in which the goat can graze?

Well, it seems like the goat can turn a full circle of radius 8 m, and a rectangular shed's diagonal is less than 8m (actually √52), and so shouldn't it be just 6 x 4 = 24 sq metre? The answer says it is 53 pi, and I have no clue why it is so or why my way of solving doesn't work.

Updated: Oh, and the only area given is that of the shed's. How can I know the full area in which the goat can actually graze on?

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    $\begingroup$ The goat is outside of the shed, so it wraps around the shed. $\endgroup$ – Cameron Williams Sep 26 '16 at 14:22
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    $\begingroup$ "What is the total area, in square meters, in which the goat can graze?" Infinite. Reasoning: the very first thing the goat is going to do is get bored, figure out that the reason it can't wander freely is because of the stupid rope and collar, and will then either A) slip its head out of the collar, or B) chew through the rope. We raise goats.They have the problem-solving abilities of a 7-year-old human. Seriously - you don't confine a goat - you just convince it that it will have more fun/be better fed/have life easier "inside" than "outside". If a goat wants out, it will find a way. $\endgroup$ – Bob Jarvis - Reinstate Monica Sep 26 '16 at 16:51
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    $\begingroup$ 6 x 4 that's the area of the shed. The goat grazes outside of the shed, not inside $\endgroup$ – njzk2 Sep 26 '16 at 16:59
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    $\begingroup$ How long is the neck of the goat? It adds to the circles :) unless the rope is tied to its mouth, in which case eating in general might become problematic. $\endgroup$ – Konerak Sep 27 '16 at 11:42
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    $\begingroup$ @AmaniKilumanga As a simplifying assumption, we posit that the goat lives on an infinite, flat plane. ;-) $\endgroup$ – Kevin Sep 27 '16 at 15:16
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Parts of three different circles. enter image description here

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    $\begingroup$ Man, with that goat you have on the picture, you are definitely the winner of this post :D $\endgroup$ – Futurologist Sep 26 '16 at 15:58
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    $\begingroup$ You have a magically appearing $m^2$ in an equation. $\endgroup$ – Carsten S Sep 26 '16 at 16:06
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    $\begingroup$ Don't know what @CarstenS means by magically appearing m2. The units are clearly in meters, so the area would be in squared meters. Perhaps he/she means that the m is missing and should be included in your previous terms. And +1 for the goat. $\endgroup$ – AgapwIesu Sep 26 '16 at 19:20
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    $\begingroup$ @AgapwIesu This is math.SE, so I'm thinking it's because meters weren't in the rest of the equation in this answer (even if they are in the question) $\endgroup$ – Izkata Sep 26 '16 at 21:42
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    $\begingroup$ Wow nice answer. Clear case of 'a picture says more than a thousand words'. I was subtracting the shed area from the full cirkel area but your picture shows very clearly why that is wrong. Great answer, +1! $\endgroup$ – Stijn de Witt Sep 27 '16 at 7:30
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Your answer is perfectly correct. Your issue is that you make an esoteric conclusion little would think of:

You don't have to be outside the shed to have a leash connected to the outside of the shed.

Your answer is correct under this interpretation. Of course, then I would counter and say that no area is available for grazing. Then I'd say the answer is 0. Of course, we will then assume there is a bed of sod in the shed, and that the floor is empty.

Of course, the intended answer is for the goat to be outside the shed, but meh. Your mileage may vary.

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    $\begingroup$ IMO this is a borderline NAA, which could be a lot better as a comment. Also, the direct implication of the statement "external" is that the goat is outside. While the matter of exact wording and problem formulation can be nit-picked, the intention of the problem's author is quite obvious here. $\endgroup$ – user81774 Sep 30 '16 at 16:19
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$$\frac{3}{4}\pi 8^2+\frac{1}{4}\pi (8-6)^2+\frac{1}{4}\pi (8-4)^2=53\pi$$

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    $\begingroup$ An equation without a result and zero explanation isn't going to help anyone. $\endgroup$ – JPhi1618 Sep 26 '16 at 17:54

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