The perimeter of a rectangle is 24 inches. What is the number of inches in the maximum diagonal for this rectangle?

Guest Apr 6, 2021

#1**+1 **

perimeter = 2 x length + 2 x width or 2(length + width)

and since the perimeter is 24 inches:

24 = 2 x length + 2 x width

12 = length + width

diagonal length = \(\sqrt{l^2 + w^2}\), when l = length and w = width, according to pythagoras' theorem.

since they are both squared, we want a big number (example: 2^{2} + 3^{2} < 1^{2} + 4^{2})

this means to maximize the diagonal, the length is 11 and width is 1. (11 + 1 = 12)

\(\sqrt{11^2 + 1^2} = \sqrt{122} ≈ 11.06\)

hope this helps; tell me if I did anything wrong :)

Logarhythm Apr 6, 2021