SOLUTION: bob wants to build a kennel that is twice as long as it is wide. if he has 30 feet of fence to build the kennel what are the dimensions ? what is the are enclosed? i know this

Algebra ->  Surface-area -> SOLUTION: bob wants to build a kennel that is twice as long as it is wide. if he has 30 feet of fence to build the kennel what are the dimensions ? what is the are enclosed? i know this       Log On


   

Question 1067144: bob wants to build a kennel that is twice as long as it is wide. if he has 30 feet of fence to build the kennel what are the dimensions ? what is the are enclosed?
i know this is simple but i've been a rut today studying and I could really use some help.
thank you!
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +w+ = the width
+2w+ = the length
I assume the kennel is a retangle
The perimeter of a rectangle is:
+2l+%2B+2w+=+2%2A%282w%29+%2B+2w+
+2l+%2B+2w+=+6w+
You are given that:
+6w+=+30+
+w+=+5+
and
+2w+=+10+
and
+5%2A10+=+50+
the area is 50 ft2