SOLUTION: The length of a picture is 1 inch less than twice its width. The frame around the picture has a uniform width of 2 inches and an area of 96 square inches. What are the dimensions
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Question 183801: The length of a picture is 1 inch less than twice its width. The frame around the picture has a uniform width of 2 inches and an area of 96 square inches. What are the dimensions of the picture?
Here's what I tried.
L = 2w-1
W = 2
A = LXW = 96
L = 2(2) -1
L = 4-1
L = 3
doesn't make sense though
Thanks for the help!
You can put this solution on YOUR website! The length of a picture is 1 inch less than twice its width. The frame around the picture has a uniform width of 2 inches and an area of 96 square inches. What are the dimensions of the picture?
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Draw the picture of a rectangle inside a rectangle.
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Label the length and the width of the inner rectangle.
Let width = w
Then length= 2w - 1
Area of the inner rectangle = w(2w-1) = 2w^2-w sq. in.
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Label the length and width of the outer rectangle.
Width is w + 4 inches
Length is 2w-1+4 = 2w +3 inches
Area of the outer rectangle = (w+4)(2w+3) = 2w^2+7w+12 sq. in.
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Area of the picture = outer rectangle area - inner rectangle area
96 = 2w^2+7w+12 -(2w^2-w)
8w + 12 = 96
2w + 3 = 24
2w = 21
w = 21/2 inches (width of the picture)
L = 2w-1 = 21-1 = 20 (length of picture)
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Cheers,
Stan H.
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