# The length of a picture frame is 3 in. greater than the width. The perimeter is less than 52 in. How do you find the dimensions of the frame?

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Conclusion: Width is less than 13 inches Length is less than 16 inches

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Let's denote the width of the picture frame as ( W ) inches. Since the length of the frame is 3 inches greater than the width, we can represent the length as ( W + 3 ) inches.

The perimeter of a rectangle (picture frame) is given by the formula ( P = 2L + 2W ), where ( P ) is the perimeter, ( L ) is the length, and ( W ) is the width.

Given that the perimeter is less than 52 inches, we have the inequality:

( 2(W + 3) + 2W < 52 )

Simplifying and solving for ( W ), we get:

( 2W + 6 + 2W < 52 )

( 4W + 6 < 52 )

( 4W < 46 )

( W < \frac{46}{4} )

( W < 11.5 )

So, the width of the frame must be less than 11.5 inches. Since the width cannot be negative or zero, we consider ( W ) to be an integer. The possible values for ( W ) would be 1, 2, 3, ..., 11.

Next, we can find the corresponding lengths for each possible width ( W ) by using ( L = W + 3 ).

For example, if ( W = 1 ), then ( L = 1 + 3 = 4 ). This would give us a perimeter of ( 2(4) + 2(1) = 8 + 2 = 10 ) inches.

Similarly, we can calculate the perimeters for ( W = 2, 3, \ldots, 11 ) to see which combinations result in a perimeter less than 52 inches. The dimensions of the frame that satisfy the given conditions would be the ones where the perimeter is less than 52 inches.

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