# A goat tied to a corner of a rectangle

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?

• The goat is outside of the shed, so it wraps around the shed. Sep 26, 2016 at 14:22
• "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. Sep 26, 2016 at 16:51
• 6 x 4 that's the area of the shed. The goat grazes outside of the shed, not inside Sep 26, 2016 at 16:59
• 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. Sep 27, 2016 at 11:42
• @AmaniKilumanga As a simplifying assumption, we posit that the goat lives on an infinite, flat plane. ;-) Sep 27, 2016 at 15:16

• You have a magically appearing $m^2$ in an equation. Sep 26, 2016 at 16:06
$$\frac{3}{4}\pi 8^2+\frac{1}{4}\pi (8-6)^2+\frac{1}{4}\pi (8-4)^2=53\pi$$