SOLUTION: Please help. I am still stumped. A rectangle has a perimeter of 82 inches with a diagonal of 29 inches. What is the length and width of the rectangle?

Question 316155: Please help. I am still stumped. A rectangle has a perimeter of 82 inches with a diagonal of 29 inches. What is the length and width of the rectangle?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A rectangle has a perimeter of 82 inches with a diagonal of 29 inches.
What is the length and width of the rectangle?
:
Write the perimeter equation
2L + 2W = 82
simplify, divide by 2
L + W = 41
L = (41-W); we can use this form for substitution
:
Write the pythag equation; a^2 + b^2 = c^2, where a=L, b=W, c=29
L^2 + W^2 = 29^2
L^2 + W^2 = 841
:
Substitute (41-W) for L
(41-W)^2 + W^2 = 841
:
FOIL (41-W)(41-W)
(1681 - 41W - 41W + W^2) + W^2 = 841
1681 - 82W + W^2 + w^2 = 841
:
Arrange as a quadratic equation on the left:
2W^2 - 82W + 1681 - 841 = 0
2W^2 - 82W + 840 = 0
:
Simplify, divide by 2
W^2 - 41W + 420 = 0
:
Factor
(W-20)(W-21) = 0
:
Two solutions
W = 20
and
W = 21
:
Find the Length when W=20
L = 41-W
L = 41-20
L = 21
and
Find the Length when W=21
L = 41-W
L = 41-21
L = 20
:
See if that checks out, find the diagonal (hypotenuse)
d =
d = 29 is the diagonal
:
YOu can confirm the solutions in the perimeter equation
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