Grazing Area


Tonya DeGeorge


The following investigation is an attempt to answer the following question:

In order to answer this question, we need to assume that the goat can move freely on this line as the stake is placed along the line segment from the midpoint of the side of the shed to the edge of the silo.  In order to imagine this, refer to the picture below:



Before computing the area in which the goat can graze, letŐs imagine that the goat can move freely and the silo and shed were not obstacles.  For instance, suppose the silo and shed were not there but the goat could only move along the 92 ft (as shown above).  Since the goatŐs tether is only 76.7 feet long, that would mean that the goat could only move 76.7 ft left and right of the line segment shown above.  The diagram below is a graphical representation (not drawn to scale):

So the area can be broken up in three pieces: the top semicircle (with a radius of 76.7 ft), the rectangle in the middle (with a width of 92 ft and a length of 153.4 ft), and the bottom semicircle (also with a radius of 76.7 ft).  This would represent the maximum possible area the goat can graze.  LetŐs compute this:

Area of rectangle:

A = l x w

A = 153.4 x 92

A = 14,112.80 square feet


Area of top semicircle + area of bottom semicircle = area of one complete circle


Area of circle:


Total Area = area of rectangle + area of circle

Total Area = 14,112.80 + 18,472.27 = 32,585.07 square feet


However, this area would represent the area the goat could graze if it did not have any restrictions.  Since the shed and the silo are present, the goat would have to travel around the shed and silo in order to get to the other side example of shed shown below).  The purple curves represent the path of the goat as it tries to move around the shed.




To account for this, we need to look at the two ŇendsÓ of the field.  LetŐs begin by look at the end of the field with the shed.  By using the construction in GSP, we can see that there is a point of intersection from where the goat will move from the right and from the left.  To compute this area, letŐs first compute area A1 and A2, which are both bounded by the side of the shed.



First note that the length on either side of the shed is 66.7 ft long (the original radius was 76.7 ft but the shed takes up 10 ft of the radius).  Hence, A1 and A2 are each a quarter of a circle with radius 66.7 ft.  To find the combined area, we could just find the area of half a circle of radius 66.7 ft:






To find the area of the small piece above the square shed, we could approximate the area by finding the area of rectangle ABCD and subtracting the area of rectangle BCGF.  The length of AB is approximately 46.7 ft (since 66.7 ft – 20 = 46.7 ft).  And AD is 20 ft.  Therefore, the area of rectangle ABCD is approximately 934 square feet (since 46.7 x 20 = 934 square feet).


In addition, the length of FG is also 20 ft.  To find the area of rectangle BCGF, we need to find the length of BF.  To find the length of this, we can set up a proportion:


So the area of rectangle BCGF:

A = 8.57 x 20

A = 171.30 square feet


Now to find the area of rectangle AFGD:

Area of rectangle AFGD = area of rectangle ABCD – area of rectangle BCGF

Area of rectangle AFGD = 934 – 171.30

Area of rectangle AFGD = 762.69 square feet


So, the total area around the shed is (top ŇendÓ):

Total area around shed = areas of ¼ circles + area of rectangle AFGD

Total area around shed = 6,984.76 + 762.69

Total area around shed = 7,747.45 square feet


Now letŐs look at the bottom ŇendÓ of the field.  Below is a diagram of what we get when the goat moves around the silo:


To approximate the area of the bottom end of the field, we could first find the area of the semicircle and then subtract the area of the silo.  From previous work, we know that the area of the semicircle is approximately 18,472.27 / 2 = 9,236.14 square feet.  Now we need to subtract the area of the silo.  Since the silo is 20 ft in diameter, which means the radius is 10 ft.


Hence, the area of the bottom end of the field is:

Area of bottom end of the field = area of semicircle – area of silo

Area of bottom end of field = 9,236.14 – 314

Area of bottom end of field = 8,922.14 square feet


However, this does not take into account the area the goat cannot reach.  In order to address this, I thought it would be best to construct a triangle, as shown below:



We know that the length of QS and QR are both 76.7 ft long, since that is the length of the tether the goat is tied to.  The ratio of QT to QU is approximately .23.


So the length of UT is approximately equal to 17.64 ft (since 76.7 x .23 = 17.64 ft).  And therefore, QU must be 59.06 ft (since 76.7 – 17.64 = 59.06 ft).  From here, we can set up another proportion to find the length of SR:



From here, we know the following:

Using the Pythagorean theorem, we can find the length of side UR:


Thus, SR = 2(UR)

SR= 2(14.56)

SR = 29.12 feet


Area of triangle RST:


Hence, the area of the bottom end of the field:

Area of bottom of field = area of (semicircle – silo) – area of triangle RST

Area of bottom of field = 8,922.14 – 256.8

Area of bottom of field = 8,665.34 square feet


Hence, the total area the goat can graze is:

Total Area = area of top end of field + area of middle rectangle + area of bottom end of field

Total Area = 7,747.45 + 14,112.80 + 8,665.34

Total Area = 30,525.59 square feet