**Old Farmer Jones had a Goat**

by

Patty Wagner

Farmer Jones has a field with a silo and a shed 92 feet apart. The silo is 20 feet in diameter and the shed is 20 feet wide by 20 feet long. He ties a clothesline between the silo and shed and hooks a tether to it. The tether is 76.7 feet long. Farmer Jones attaches his prized goat to the tether and hopes the grazing area afforded the goat by this set-up is sufficient to keep it well-fed. He knows that the rule of thumb is two goats per acre of land, so he knows his goat should have at least a half acre of grazing area, which is 21,780 square feet. However, he's not sure how to go about figuring the grazing area his set-up gives his goat, so he asked for my help. This is how I found the approximate grazing area.

The diagram at left approximates the grazing area of the goat. (The diagram is not to scale.) Notice that I could divide the area into three sections. The area of the center section is fairly easy to compute:

Area_{center}= 2(76.7 feet)(92 feet)

Area_{center}= 14,113 ft^{2}This is not enough area to sustain the goat, so I needed to look at the ends of the field. Beginning with the shed end, I could see that the grazing area on one side of the shed could be viewed as three different areas.

Area 1 = 1/4πr^{2}= 1/4π(66.7)^{2}

Area 1 ≈ 3,494 ft^{2}To find Area 3, I noted that it is a right triangle. So

h^{2}= 46.7^{2}- 10^{2}≈ 45.6 ft

Area 3 ≈ 1/2(10 ft)(45.6 ft) ≈ 228 ft^{2}To find area 2, I recognized that it is an arc section of a circle with radius 46.7 ft. The angle in section A3 can be easily computed since it is a right triangle.

cosθ ≈ 10/46.7

θ ≈ 77.6°

The angle, α, in the arc section is 90° - 77.6° ≈ 12.4° ≈ .07π

Area 2 = 1/2r^{2}α ≈ 1/2(46.7)(.07π) ≈ 240 ft^{2}The goat's grazing area on this side of the field is

Area_{shed}≈ 2(3,494 ft^{2}+ 228 ft^{2}+ 240 ft^{2}) ≈ 7,924 ft^{2}The area of the center plus the area around the shed was a little more than the 1/2 acre the goat needed but it was cutting it awfully close, so I turned my attention to the silo end of the field. On this side my task was more difficult. The best I could do was estimate the grazing area using what I knew would be the area if the silo was not there and subtracting an approximation of the grazing area lost to the silo.

I used Geometer's Sketchpad® to help myself find some dimensions; click here for a link to the program. I used the "trace" feature to mark out the actual outline of the grazing area. When I created similar ratios between silo and tether length, I could see that the ratio of AB to AE = .21

To get an approximate value of the amount of grazing area lost from the position of the silo, I considered a chord that passes through point B that is parallel to the tangent through E. The area of the segment created by the chord will approximate the lost grazing area.

AB = (.21)(76.7) = 16.1 ft

EB = 60.6 ftThe chord length = 2√(r

^{2}- d^{2}) = 2√(76.7^{2}- 60.6^{2}) ≈ 94 ftThe arc section = 1/2πr

^{2}θtan 1/2θ = 47/60.6

θ ≈ 76° ≈ .42π

So the arc section ≈ 1/2(76.7)

^{2}(.42π) ≈ 3,881 ft^{2}The lost grazing area is approximately equal to the area of the arc section minus the area of triangle EKJ.

area of ΔEKJ = 1/2(94 ft)(60.6 ft) = 2,848 ft

^{2}so the approximate lost grazing area is3,881 ft

^{2}- 2,848 ft^{2}≈1,033 ft^{2}If the silo was not there, the goat could have grazed on A = 1/2πr

^{2}= 1/2π76.7^{2}=9,241 ft^{2}The silo itself results in a loss of π20

^{2}≈ 1,257 ft^{2}and the silo causes an additional loss of ≈ 1,033 ft^{2}. So on this end, the goat can graze approximately

Area_{silo}≈ 9,241 ft^{2}- 1,257 ft^{2}- 1,033 ft^{2}≈ 7,000 ft^{2}

So altogether, I estimate the goat has approximately

Area_{center}+ Area_{shed}+ Area_{silo}

≈ 14,113 ft^{2}+ 7,924 ft^{2}+ 7,000 ft^{2}

≈ 29,037 square feetor 2/3 of an acre of grazing area. This should be plenty to keep Farmer Jones's goat fat and happy!

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