Final Assignment
Grazing Area
by Emily Bradley
Farmer Jones had a goat on a tether. He tied the end of the tether not attached
to the goat to a stake in a field. Over what area could the goat graze? Of
course you need to know something about the length of the tether and about
the field.
There are two structures in the field:
-- a shed that is 20 ft long and 20 feet wide (square)
-- a silo that is 20 ft in diameter
The center of the shed and the center of the silo are on a line and the distance
apart is 92 feet. The distance from center to center, if you wanted to use this
data, is 112 feet.
The tether for the goat is 76.7 feet long. The stake to which the tether is tied
is somewhere along the line of centers between the shed and the silo.
Explore the area over which the goat can graze as the stake is moved along this
line segment from the midpoint of the side of the shed to the edge of the silo.
Grazing Area
| Total | The following image represents the field. Let us calculate the area of each part of the field, to find total area before finding the optimum placement.
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| Part 1 | The purple section is 1/4 the area of rectangle with radius 66.7'. A = 1/4 pi 66.7^2 = 3494.2 sq ft The blue section is a right triangle with side length 10, hypotenuse length 46.7. Applying Pythagorean theorem, 10^2 + b^2 = 46.7^2 b = 45.617'. Using A = 1/2 b h = 1/2 * 10 * 45.617 = 228.1 sq ft We are interested in finding the angle at the base of this triangle. cos (theta) = adjacent/hypotenuse = 10/46.7 theta = 77.6 degrees, therefore the complement angle in the green section is 12.4 degrees The area of the green section is A = (12.4/360) pi * r^2 = .034 * pi * 46.7^2 = 235.3 sq ft So the area of the entire shaded region is 3494.2 + 228.1 + 253.3 = 3975.6 sq ft We double this to include the area of the opposite side, so the entire grazable area of part 1 = 7951.2 sq ft
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| Part 2 | Finding the area of this middle region is simple because it is a rectangle. A = lw = 92 * 153.4 = 14112.8 sq ft
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| Part 3 | The following grazable area was found by tracing the endpoint of the line tangent to the circle. The area can be found by the radii and corresponding circles that estimate this area The pink area is of a 45 degree section of a circle with radius 76.7 ft. A = (45/360) * pi * 76.7^2 = .125 * pi * 5882.9 = 2310.2 sq ft The blue area is of a 54.5 degree section of a circle with radius 61 ft. A = (54.4/360) * pi * 61^2 = .151 * pi * 3721 = 1765.2 sq ft The red section is that of a triangle with side length 10, hypotenuse length 61, minus the area of 1/4 the circle with radius 10. First solving for the length of b in 10^2 + b^2 = 61^2, b = 60.175 ft. Solving A = 1/2 * b * h = 1/2 * 10 * 60.175 = 300.9sq ft The area of the quarter circle is, A = (1/4) pi * 10^2 = 25pi = 78.6 sq ft 300.9 - 78.6 = 222.3 sq ft for the red region 2310.2 + 1765.2 + 222.3 = 4297.7 sq ft
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| Therefore the total grazable area is 7951.2 + 14112.8 + 4297.7 = 26361.7 sq ft |
