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Another Grazing Cow


Date: Wed, 7 Jun 1995 12:53:17 -0500
Subject: A math riddle
From: Ken Ryland

A friend came by my office at work yesterday touting a math riddle that he
claimed was insoluble (at least until the present). Perhaps you would like
a crack at it.

A man has a barn that is 20 ft by 10 ft. He tethers a cow to one corner of
the outside of the barn using a 50-ft rope. What is the total area that the
cow is capable of grazing?

  -------------------------
  |                       |
  |                       |
  |                       |
  |                       |
  |                       |
  |                       |10 ft
  -------------------------

             20 ft         \
                            \ 50 ft
                             \
                              \
                               \
                                cow
(you get the idea.)

The smart alec who posed this question is Jack McVey. His 
e-mail address is mcvey@vax882.rac.ray.com if you would like 
to send him a copy of your response. 

Hope to hear from you soon.

Yours truly,

Ken Ryland


Subject: Re: A math riddle
From: Dr. Ken
Date: Thu, 8 Jun 1995 15:02:24 -0400 (EDT)

Hello there!

Well, it seems to me that this problem is far from insoluble.  
In fact, I think that if I get you started a little bit and 
if you know how to do some integration, you'll be able to 
solve it yourself!  Also, I made a nice sketch of the problem 
using The Geometer's Sketchpad:

   

Basically, the shape of the grazing area is circular with a 
little chunk out of it.  In my drawing, the barn looks like 
this:

                                             COW
                                            /
                                           /
                                          /
                                         /
                                        /
                                       /
                                      /
                                     /
                                    /
                                   /
                                  /
                                 /
                                /
                               /
                              /
                             /
                            /
                           /
                          /
             ____________/
            |            |
            |            |
            |            |   
            |            |
            |            |
            |            |
            |            |
            |            |
            |            |
            |            |
            |            |
            |____________|

With this setup, the chunk that's missing from the grazing 
circle is in the lower left-hand corner.  The right half of 
the grazing area, then, is just the area of the semicircle, 
which is 1250.  To find the area of the remaining grazing 
area (including the barn, which you subtract later), you 
use integration.  You see how the grazing area in the lower 
left-hand corner is the intersection of two circles?  Well, 
you need to find that intersection point, and then integrate 
to find the area of the grazing area to the left of the 
intersection point, then the area to the right of the 
intersection point.

I'm trying not to do too much for you here so that you can 
grab all the glory with your friend.  If you want more help, 
please write back.   

Ken "Dr. Math"
    
Associated Topics:
High School Calculus
High School Conic Sections/Circles
High School Euclidean/Plane Geometry
High School Geometry

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