the area of a rectangle is 45 ft^2 , and the length of the rectangle is 1 ft more than double the width. Find the dimensions of the rectangle

Answer 1

Suppose the width of the rectangle is w. Then its length would be l = 2w + 1.

Area = w * l = w * (2w + 1) = 45.

So 2w² + w = 45 and 2w² + w - 45 = 0.

2w² + w - 45 = (2w - 9)(w + 5) = 0; w = 9/2 or -5 (ridiculous).

For w = 9/2 , l = 2(9/2) + 1 = 10. The length of the rectangle is 10 feet and its width is 9/2 feet.

Answer 2

Width 4.5 ft and the Lenght is 10ft.

Answer 3

Let x be the width, and y be the length.

We know that y = 2x+1 (the length is 1ft more than double the width).

We know that x * y = 45 (length times width = area)

This gives us a system of equations. If we substitute 2x+1 in the place of y in the second equation, we get x * (2x+1) = 45.

Doing the multiplication on the left side gives us 2x^2 + x = 45.

Subtracting 45 from both sides yields 2x^2 + x - 45 = 0.

This is a polynomial equation and can be solved using any of the various methods (factoring, quadratic formula, etc). The solutions are x = 0, x = 9/2. Since x represents the width, it cannot be 0, so it must be 9/2.

Plugging that back in to the original y = 2x+1, we get y = 2(9/2)+1. Solving that shows that y = 10.

So the width is 9/2 in, and the length is 10 in. If you plug these back into the original question, you will see that they satisfy the requirements.