Posted 22 Feb 2013 You put your dog on a leash. Your leash is 80 feet in length. The dog is tethered to part of the exterior of your house (the tie is not part of the 80 feet). Your home's dimensions are 60ft by 40ft. How much yard does the dog have access to? 0 Share this post Link to post Share on other sites

0 Posted 25 Apr 2013 (edited) Can you explain how you found it to be 4100? I found it to be 3425 Edited 25 Apr 2013 by BMAD 0 Share this post Link to post Share on other sites

0 Posted 25 Apr 2013 Chain the dog in the middle of a 100-foot wall. In the front yard he has an 80-foot-radius semicircle of access. When he gets to the ends of the wall, he has 30 feet of chain left. As he walks into the back yard he gains access to a 30-foot-radius semicircle on each side. Area [three semicircles] = pi/2 [80^{2} + 30^{2} + 30^{2}] = pi/2 [6400 + 900 + 900] = pi [3200 + 900] = 4100 pi ft^{2} General area formula (x is distance from end of wall to the chain point): 0<x<20:.. A = pi/2 [80^{2} + (80-x)^{2}] 20<x<50: .A = pi/2 [80^{2} + (80-x)^{2} + (x-20)^{2}] 0 Share this post Link to post Share on other sites

0 Posted 25 Apr 2013 (edited) I see what I did wrong. I have it being quarter circles at the end of radius 30. Chain the dog in the middle of a 100-foot wall.In the front yard he has an 80-foot-radius semicircle of access.When he gets to the ends of the wall, he has 30 feet of chain left.As he walks into the back yard he gains access to a 30-foot-radius semicircle on each side. Area [three semicircles] = pi/2 [80^{2} + 30^{2} + 30^{2}] = pi/2 [6400 + 900 + 900] = pi [3200 + 900] = 4100 pi ft^{2} General area formula (x is distance from end of wall to the chain point): 0<x<20:.. A = pi/2 [80^{2} + (80-x)^{2}] 20<x<50: .A = pi/2 [80^{2} + (80-x)^{2} + (x-20)^{2}] Edited 25 Apr 2013 by BMAD 0 Share this post Link to post Share on other sites

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You put your dog on a leash. Your leash is 80 feet in length. The dog is tethered to part of the exterior of your house (the tie is not part of the 80 feet). Your home's dimensions are 60ft by 40ft. How much yard does the dog have access to?

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