Question 29496
the perimeter of a rectangle is 32m and the area of the rectangle is 55m squared.  Find the dimensions of the rectangle

Let the length of therectangle be L meters
and let the width be B meter.
Given that the perimeter of a rectangle is 32m 
That is 2(L+B)= 32 ----(1)
which on dividing by 2 gives (L+B)=16 ----(2)
Given that the area of the rectangle is 55m 
That is LXB = 55----(3)
By formula (L-B)^2 = (L+B)^2-4LB
=(16)^2-4X(55) [using (2) and (3)]
=256-220
=36
(L-B)^2 = 36 
taking the positive sqroot 
(since length >width by convention, therefore L>B implying (L-B) >0 }
Therefore L-B = 6  ----(4)
We have    L+B= 16----(2)
Adding  (4) and (2)
(L+L) = (6+16)
2L= 22
L=22/2 = 11
L=11 in (2) gives B = 5
Therefore Length of the rectangle = 11m
and width of the rectangle = 5 ms
Verification: Area = length X width = 11X5=55 sq meters which is correct
Therefore our values are correct.