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Alice plans to put her dog on 20foot leash in her fence in backyard. She will temporary attach the leash to the back of her rectangle shaped home which is centered in the lot.
A) Alice is considering two places to attach the leash to the back wall of her house these two places are labeled X and Y in the diagram. Point X is at the corner of the house. Point X is at the corner of the house. Point Y is 20 feet from corner X. The dimensions of her house are 50 feet by 100 feet. How many square feet of yard space are available to the dog if the dog is tied to the leash is at attached at point X?
B) Which of the two positions X or Y provides the greater area for the dog to walk? Show all calculations to explain your answer.
C) Alice buy a new 35 foot leash and decides to attach it to a corner of the fence in the backyard at point Z. Will the placement of the new 35 foot leash provide more square feet of yard space for the dog than the placement of the 20 foot leash at either point X or Point Y? The dog is unable to jump the fence.
the corner of the house makes 3/4 of a circle that the jackalope can play in... the area of any section of a cirlce is:
You can determine the area of a circle by this formula because:
(2pi)(r^2)/2 = pi(r^2)
the radians measure of 3/4 of a circle is:
(3pi/2)(20^2)/2 which gives us:
3(20)(20)/2 = 60(10) = 600/2 = 300
300pi ft^2 to romp around in
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if the jackalope is tied up at the point Y, it only has the area of half a circle to graze upon;
(pi)(20)(20)/2 = 200pi ft^2
the part of the circle formed by the inside corner of the fence is 1/4 of a circle...but the jackalope has 35 ft of radius to explore and conquer with.
(pi/2)(35)(35)/2 = pi(17.5)(17.5) = 306.5pi ft^2
while this may seem most opportune for the little jackalope, it tends to be rather clostraphobic