Question #e9dfa

1 Answer
May 3, 2017

Dimensions are : width #sqrt(13)#, length #5sqrt(13)#

Explanation:

Original rectangle:
let width be #w#, and length be #5w#

New rectangle:
new width #w-1#, new length #5w+5#
So, new area :#(5w+5)*(w-1) = 60#
#5(w+1)(w-1)=60#
#5(w^2 -1)=60#
#w^2-1 = 12#
#w^2 = 13#
#w = \pm sqrt(13)#
since width cannot be negative, #w = sqrt(13)# and the length is #5w = 5sqrt(13)#