The perimeter of a rectangle is 36ft, and the area of the rectangle is 72ft^2 . How do you find the dimensions?

1 Answer
Apr 18, 2016

You must write a system of equations to represent the problem.

Explanation:

The formula for perimeter of a rectangle is #p = 2L + 2W#. The formula for area is #A = L xx W#

Thus, #L xx W = 72, 2L + 2W = 36#

#W = 72/L -> 2L + 2(72/L) = 36#

#2L + 144/L = 36#

#(2L^2)/L + 144/L = (36L)/L#

We can now eliminate the denominators since all fractions are equal.

#2L^2 + 144 = 36L#

#2L^2 - 36L + 144 = 0#

This is a trinomial of the form #y = ax^2 + bx + c, a != 1# Therefore, this can be factored by finding two numbers that multiply to #a xx c# and that add to b, and following the process shown below. These two numbers are #-12# and #-24#

#2L^2 - 12L - 24L + 144 = 0#

#2L(L - 6) - 24(L - 6) = 0#

#(2L - 24)(L - 6) = 0#

#L = 12 and 6#

Since the length can be the width and vice-versa, the sides of the rectangle measure 12 and 6.

Hopefully this helps!