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2 March, 14:54

A square and rectangle have the same perimeters. The lengtj of a side of the square is 4x-1. The length of the rectangle is 2x+1 and the width is x+2 write and solve an equation to find x

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  1. 2 March, 15:20
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    Well, the perimeter of a square is 4 times the side. The side of THIS square is (4x - 1), so the perimeter is 4 x (4x-1) = 16x - 4. Hold that thought while we go over there and work on the rectangle.

    The perimeter of a rectangle is (2 lengths) plus (2 widths). The length of THIS rectangle is (2x + 1), and 2 lengths = 2 x (2x+1) = 4x + 2. The rectangle's width is (x+2). Two widths add up to 2 x (x+2) = 2x + 4.

    The total perimeter of the rectangle is (4x+2) + (2x+4) = 6x + 6.

    Now we have the expressions for the perimeters of both figures.

    AND we know that their perimeters are EQUAL ... THAT's the very powerful piece of information that's going to bust this whole problem wide open.

    (perimeter of the square) = (perimeter of the rectangle)

    The equation that the question asked for: 16x - 4 = 6x + 6

    That's the answer to the question that was posted. But we can't just leave it there. We've got to find out what 'x' actually is, and then find the perimeters of the square and the rectangle.

    Add 4 to each side: 16x = 6x + 10

    Subtract 6x from each side: 10x = 10

    Divide each side by 10 : x = 1

    Side of the square: (4x-1) = 3

    Perimeter of the square: 4 x 3 = 12

    Length of the rectangle: (2x+1) = 3

    Width of the rectangle: (x+2) = 3

    Perimeter of the rectangle = (2 lengths) + (2 widths) = 12

    Do you notice anything strange and exciting? The "rectangle" is actually another square, exactly the same shape and size as the "square"!
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