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-w ... i

the area will be:

A=l*w=132

hence

l=132/w ... ii

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