SOLUTION: How do I find the length and width of a rectangle with a perimeter of 82 inches and a diagonal of 29 inches?

Question 239058: How do I find the length and width of a rectangle with a perimeter of 82 inches and a diagonal of 29 inches?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
P = 2L + 2W = 82
D = 29

Diagonal forms right triangle with sides of the rectangle.

This means that:

L^2 + w^2 = D^2 which means that:

L^2 + W^2 = 29^2

Since 2L + 2W = 82, this means that:

2L = 82-2W which means that:

L = 41 - W

L^2 + W^2 = 29^2 becomes:

(41-W)^2 + W^2 = 29^2 which becomes:

41^2 - 82W + W^2 + W^2 = 29^2 which becomes:

41^2 - 82W + W^2 + W^2 - 29^2 = 0

combine like terms and simplify to get:

2W^2 - 82W + 840 = 0

divide both side by 2 to get:

W^2 - 41W + 420 = 0

This factors out to:

(W-20) * (W - 21) = 0

This makes W = 20 or W = 21

If W = 20, then L = 21

If W = 21, then L = 20

Your answers are:

L = 20 and W = 21
or:
L = 21 and L = 20

To confirm, plug these values into the original equations and see if the equations are true.

Assume L = 20 and W = 21.

L^2 + W^2 = 29^2 becomes 20^2 + 21^2 = 29^2 which becomes 841 = 841 which is true.

Making L = 21 and W = 20 will yield the same answer.

2L + 2W = 82 becomes:

40 + 42 = 82 which is also true.

The values shown look good.

Answers are:

L = 20 and W = 21
or:
L = 21 and W = 20

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