SOLUTION: Terence’s dog has a dog house that is 3’ wide and 5’ long. He is building a fence so that way the dog can stay within 4’ of the doghouse on each side. What

Algebra ->  Length-and-distance -> SOLUTION: Terence’s dog has a dog house that is 3’ wide and 5’ long. He is building a fence so that way the dog can stay within 4’ of the doghouse on each side. What       Log On


   



Question 91982: Terence’s dog has a dog house that is 3’ wide and 5’ long. He is building a fence so that way the dog can stay within 4’ of the doghouse on each side. What would be the area inside the fence?
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
adding 4' to BOTH ends of a 3' width and 5' length gives an area that is 11' by 13' ... or a=11(13) ... a=143sq'