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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