Best Answer

Let the length be (2x+2) and the width be x:

length*width = area

(2x+2)*x = 40 sq ft

2x2+2x = 40

2x2+2x-40 = 0

Solving the above quadratic equation works out as:

x = -5 or x = 4 it must be the latter because dimensions can't be negative.

Therefore: width = 4 feet and length = 10 feet

Check: 10*4 = 40 sq ft

Q: Find the width of a rectangle when its length is greater by twice and 2 feet more than its width and has an area of 40 square feet?

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A rectangle with one side that is double the length of the other side can be seen as two squares beside each other, that are half the area of the original rectangle, in this case 18 square inches / 2 = 9 square inches. The side of a square is the square root of the area, in this case the square root of 9 square inches, which is 3 inches. Now think about the rectangle again. What we just counted was the side of a square that made up one half of the rectangle you wanted to calculate the area of. So the shorter side of the rectangle is the same as our square, and the longer side is double the length. So the width of the rectangle is 3 inches, and the length is 2 * 3 inches = 6 inches. Or: Area = length x width (A=lw) Area = 18 (in2) Width = w Length = twice width (2w) 18 = (2w)w 18 = 2w2 9 = w2 3 = w The width is 3 in., the length (2w) is 6 in.