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