The perimeter of a rectangle is 40 cm. The length is 14 cm. Let x = width of the rectangle. Ravi says he can find the width usin
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If we let x and y represent length and width, respectively, then we can write equations according to the problem statement.
.. x = y +2
.. xy = 3(2(x +y)) -1
This can be solved a variety of ways. I find a graphing calculator provides an easy solution: (x, y) = (13, 11).
The length of the rectangle is 13 inches.
The width of the rectangle is 11 inches.
______
Just so you're aware, the problem statement is nonsensical. You cannot compare perimeter (inches) to area (square inches). You can compare their numerical values, but the units are different, so there is no direct comparison.
.. x = y +2
.. xy = 3(2(x +y)) -1
This can be solved a variety of ways. I find a graphing calculator provides an easy solution: (x, y) = (13, 11).
The length of the rectangle is 13 inches.
The width of the rectangle is 11 inches.
______
Just so you're aware, the problem statement is nonsensical. You cannot compare perimeter (inches) to area (square inches). You can compare their numerical values, but the units are different, so there is no direct comparison.