If the perimeter of a rectangle is 46 cm and the area is 132 square cm, then what are the dimensions of the rectangle?

Let the length be l and the width be w
Perimeter will be:
P=2(l+w)=46
thus
l=23-wi

the area will be:
A=l*w=132
hence
l=132/wii
combining i and ii we get:
23-w=132/w
this will be written as:
w(23-w)=132
23w-w^2=132
hence:
w^2-23w+132=0
solving the quadratic above we obtain:
(w-11)(w-12)=0
hence:
width,w=11 or 12
thus
length=132/11 or 132/11
=12 or 11

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