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Goat Tied by a 10-Meter Rope

Date: 11/28/2001 at 06:24:24
From: Sadia
Subject: Calculus and integration

1) A goat is tied to the corner of a 5-by-4-meter shed by a 10-meter 
piece of rope. What is the area grazed by the goat?

2) The shed is now a circle with radius r, and the rope is 2r. What is 
the area grazed now?

I first tried using the formula for the area of a sector, but this 
created further questions. My college friends say the answer lies in 
the formula for the equation of a circle, but I have not studied 
calculus and the books I have don't even seem to be in English.

The second part is about integration, but again I have no previous 
knowledge of calculus at all. I am not after an answer but guidance, 
as I can easily get the answer from a friend, but I want to know how 
it's done for my own benefit.

Date: 11/28/2001 at 15:36:38
From: Doctor Rob
Subject: Re: Calculus and integration

Thanks for writing to Ask Dr. Math, Sadia.

No calculus is required, just some analytic geometry.

Draw the following diagram:

               ,-'           `-.
            ,-'                 `-.
          ,'                       `.
        ,'                           `.
      ,'                               `.
     /                                   \
    /                                     \
   |                                       |
   |    5    B    5    A                   |
 E o---------o---------o                   +
   |        /|         |                   |
    \     5/ |4        |4                  |
     \    / C|    5    |                  /
      \  /_.-o-----__--oD                /
       `o'---''''''    |               ,'
       G \      6      |             ,'
          \            |6          ,'
           '._         |        ,-'
               `-._    |    _,-'

ABCD is the rectangular shed, and the rope is tied at A.

The area the goat can graze consists of the 3/4 of the circle of 
radius 10 with center A whose boundary is EBADFE, plus a sector of a 
circle of radius 5 and center B whose boundary is EBGE, plus a sector 
of a circle of radius 6 and center D whose boundary is FDGF, plus the 
two triangles BCG and DCG. The tricky part is to determine the length 
of the segment CG.

Let d be the length of CG. By using the Law of Cosines on the two 
triangles, one gets

   (d^2+4^2-5^2)/(2*4*d) = cos(<BCG),
   (d^2+5^2-6^2)/(2*5*d) = cos(<DCG)
                         = cos(3*Pi/2-<BCG)
                         = -sin(<BCG).

Now square both expressions and add then, and, since
sin^2(<BCG) + cos^2(<BCG) = 1, you get

   (d^2+4^2-5^2)^2/(2*4*d)^2 + (d^2+5^2-6^2)^2/(2*5*d)^2 = 1.

When fractions are cleared and the expressions are simplified, this 

  41*d^4 - 2402*d^2 + 3961 = 0.

Solving this, and rejecting extraneous roots, one finds that
d = sqrt([1201-800*sqrt(2)]/41), or approximately 1.303177637 meters.

Using Heron's Formula for the area of a triangle when you know all 
three sides gives the areas of the two triangles, which are
(160*sqrt[2]-150)/41 and (250*sqrt[2]-260)/41 square meters.

Using the Law of Cosines to compute <GDC and <GBC gives

   <GDF = Pi/2 - <GDC,
        = arcsin([65+40*sqrt(2)]/123) = 1.418084117... radians,
   <GBE = Pi/2 - <GBC,
        = arcsin([12+20*sqrt(2)]/41) = 1.383671627... radians.

Of course the angle <EAF at the center of the 3/4 circle with radius 
10 is

   <EAF = 3*Pi/2 radians.

Now the sector area formula K = (1/2)*r^2*theta, where r is the radius 
and theta the central angle, will give you the areas of the three 
circle sectors. Add these to the areas of the two triangles, and 
you'll have the total graze area.

Filling in the details and the rest I'll leave to you.

A problem rather simple to state ended up using some fairly 
complicated math, and the answer turned out to be quite messy!

Feel free to write again if I can help further.

- Doctor Rob, The Math Forum   
Associated Topics:
College Euclidean Geometry
High School Euclidean/Plane Geometry

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