The length of a rectangle is 11 ft more than twice the width, and the area of the rectangle is 63 ft2. Find the dimensions of the rectangle.Length : ftWidth : ft

Let's denote the width of the rectangle as x (in feet). According to the problem, the length of the rectangle is 11 ft more than twice the width, so we can express the length as 2x + 11 ft.

The area of a rectangle is given by the formula: Area = Length * Width. In this case, the area is 63 square feet. So, we can set up the equation:

63 = (2x + 11) * x

Solving this equation will give us the values of x (width) and 2x + 11 (length).

Step 1: Expand the equation: 63 = 2x^2 + 11x

Step 2: Rearrange the equation to equal zero: 0 = 2x^2 + 11x - 63

Step 3: Divide all terms by 2 to simplify: 0 = x^2 + 5.5x - 31.5

Step 4: Solve the quadratic equation for x. You can use the quadratic formula, which is x = [-b ± sqrt(b^2 - 4ac)] / (2a). In this case, a = 1, b = 5.5, and c = -31.5.

Solving this gives two possible solutions for x: x = 4.5 and x = -7.

Since the width of a rectangle cannot be negative, we discard x = -7. So, the width of the rectangle is 4.5 ft.

Step 5: Substitute x = 4.5 into the length equation (2x + 11): Length = 2*4.5 + 11 = 20 ft.

So, the dimensions of the rectangle are: Length: 20 ft Width: 4.5 ft