SOLUTION: the length of a rectangle exceeds twice it width by 1 inch.
the perimeter of the rectangle is 32 inches. find the length and the width
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the perimeter of the rectangle is 32 inches. find the length and the width
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Question 500874: the length of a rectangle exceeds twice it width by 1 inch.
the perimeter of the rectangle is 32 inches. find the length and the width Answer by geetha_rama(94) (Show Source):
You can put this solution on YOUR website! Let L be the Length of the rectangle and W be the width
L = 2W+1 - > equation 1(Length exceeds twice the wide by 1 inch)
2(L+W) = 32 (Perimter)
L+w=16
Substituting equation 1,
2W+1+W = 16
3W = 15
=> W = 5
=> L = 11
Length of the rectangle = 11 inches
Width of the rectangle = 5 inches
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