Question

The length of a rectangle is I yd less than twice the width, and the area of the rectangle is 21 yd'. Find the dimensions of the rectangle.Length:

Answer 1

length of the rectangle L = 2w -1

Area = L x w

Area = (2w-1)w

this becomes

`\begin{gathered} 2w^2\text{ - w = 21 } \\ 2w^2\text{ -w - 21 = 0} \\ 2w^2\text{ +6w - 7w - 21 = 0} \\ 2w(w+3)\text{ -7(w+3) = 0} \\ (2w-7)(w+3)\text{ = 0} \\ 2w\text{ = 7} \\ \text{w = }(7)/(2) \\ \text{w = 3.5} \\ w\text{ + 3 = 0} \\ \text{w = -3} \\ \text{The width cannot be negative, so we go for w = 3.5} \end{gathered}`

Now length is

`\begin{gathered} L\text{ = 2w -1} \\ L\text{ = 2 x 3.5 -1} \\ L\text{ = 7 -1} \\ L\text{ = 6} \end{gathered}`

Therefore the Width = 3.5 yd and The length = 6 yd