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

 Apr 6, 2021
 #1
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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: 22 + 32 < 12 + 42)

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 :)

 Apr 6, 2021
 #2
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You didn't stae that the sides were integer lengths....

   we could build a rectangle with sided   .009   and 11.991     ( make one side as small as possible....approaching zero)

      as the small side approaches zero , the diagonal approaches a limit  of 12 units

 Apr 6, 2021

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