Question

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

Answer 1

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!