SOLUTION: 1). the perimeter of a rectangle is 50 feet. the length is 11 feet more than width. what are the dimensions of the rectangle?

Algebra ->  Linear-equations -> SOLUTION: 1). the perimeter of a rectangle is 50 feet. the length is 11 feet more than width. what are the dimensions of the rectangle?      Log On


   

Question 85911: 1). the perimeter of a rectangle is 50 feet. the length is 11 feet more than width. what are the dimensions of the rectangle?
Answer by praseenakos@yahoo.com(507) About Me  (Show Source):
You can put this solution on YOUR website!
Question:

the perimeter of a rectangle is 50 feet. the length is 11 feet more than width. what are the dimensions of the rectangle?

Answer:


Assume that width = x feet.


Then length = x +11 feet( given that length is 11 feet more than width)



Perimeter = 2( length + width)



Given, perimeter = 50 feet.


==> 2( length + width) = 50


==> 2( x + 11 + x ) = 50


==> 2( 2x + 11) = 50



==> 2*2x + 2*11 = 50


==> 4x + 22 = 50



Subtract 22 from both sides.....


==> 4x + 22 -22 = 50 - 22


==> 4x = 28


Divide both sides of the expression by 4



==> 4x%2F4+=+28%2F4



===> x = 7


That is width = 7 feet.


So, length = x + 11 = 7 + 11 = 18 feet.


You can check yopur answer by plugging the values of length and width in the given formula, perimeter = 2(length + width)



Hope you found the explanation useful.

Regards.


Praseena.