GRAZING!
By Brandie
Thrasher
Farmer Jones
has a goat that is tethered to a stake, in between the barn and silo. Mr. Jones
wants to know how much area his goat will be able to graze given the
measurements shown.
(The barn is measured
at 20x20ft, the silo is 20ft in diameter, the tether the goat is tied to is
76.7ft long, and the center of the distance from the barn to the silo is 92ft.)
A good way to
approach this problem is to picture the total area, without the buildings
The field
would have a width of (76.7 * 2) ft and a length of 92ft. The arcs would have a
radius of 76.7ft. This would be the measurement, with the two buildings (the
barn and the silo) not there. With this example, we can calculate the total
area as being the area of the rectangle plus the area of the two semi-circles
(arcs).
Since the area
of a semi-circle is (pr2)/2
and we have two halves, we can just use the area of a circle, being pr2
Thus, we have
A
= (length * width) + (pr2)
A = (92 * 153.4) + (p*(76.7)2)
A
= 14112.8 + 18481.64
A
= 32594.44ftsq
This looks
good, but the goat will not actually get to graze the entire area, because the
barn and silo will be in the way.
The area that
can possibly be grazed over the barn looks similar to this:
We can
calculate this area usind the area of a semi-circle (because we have two
quarters of a circle), where the radius ir reduced from the original value by
10, thus it is 66.7.
A = (pr2)/2
A = (p (66.7)2)/2
A = 6988.30ftsq
We can also
find the area of the triangle formed by the intersection of the arcs and the
base of the barn. Using proportions, we can find the measured length of the
height and find the equal proportion of measurement in regards to the barn.
20ft/1.64 = x/3.07
20 * 3.07 = 1.64x
61.4 = 1.64x
x = 37.49
We can now
calculate the area of the triangle
A = ½ b * h
A = ½ (20 * 37.49)
A = 374.9ftsq
Using the
hight of our iscosoles triangle a using proportions to find the base, we can
find the area of the two smaller triangles. Since these two triangles are
similar we will find the area of one and multiply it by 2.
A = ½ b * h
A = ½ (10.12 * 37.49)
A = 189.7 * 2
A = 379.4ftsq
Together the
area aound the barn would be:
A = 6988.30 + 374.9 + 379.4
A = 7747.6ftsq
Now lets take a
look at the area that could be grazed around the silo:
The thinner
green line represents thepossible grazed area, in which they appear to be two
sectors of a circle. We can calculate the area of a circle (in degrees) as:
A = (q/360)*pr2
Since our line
separating the two sectors is perpindicular to the bare of the silo/rectangular
field, we can say that q = 90¡, and for now we will incluse the area containing the silo, and
subtract that area from the end.
Thus the area
of the two sctors will be:
A = 2[(90/360)*p(76.7)2]
A = 2[(0.25)*5882.89]
A = 2[1470.72]
A = 2941.445ftsq
We must now
take out the area consumed by the silo:
A of
silo = pr2
A = p(10)2
A = 314.16
The total
possible area grazed around the silo:
A = 2941.45 – 314.16
A = 2627.29ftsq
Now that we
have found the possible grazing area around the barn and silo, we can take
those valus and ad them to the area of the rectangular pasture to find or total
possible area grazed (based on the drawings and drawing scales used, so this is
just an approximation):
A = (Area of rectangular field) + (Area grazed
around barn) + (Area grazed around silo)
A = (14112.8) + (7747.6) + (2627.29)
A = 24487.69ftsq
