Question 85911
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/4 = 28/4}}}




===> 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.