Question

The area of a rectangle is 42 ft squared, and the length of the rectangle is 5 ft more than twice the width. Find the dimensions of the rectangle. length and width.​

Answer 1

length = 12 ft, width = 3.5 ft

Explanation:

w = width

l = length = 2w + 5

A = wl = w(2w + 5) = 42

2w² + 5w - 42 = 0

(w + 6)(2w - 7) = 0

w + 6 = 0, w = -6 (dimension cannot be negative)

2w - 7 = 0, w = 3.5

l = 2(3.5) + 5 = 12

Answer 2

Length = 12 ft

Width =
`(7)/(2) ft`

Explanation:

Given,

Area of rectangle =
`42 \: {ft}^(2)`

Width = X

Length = 2x + 5

Now,

`x(2x + 5) = 42`

`2 {x}^(2) + 5x = 42`

`2 {x}^(2) + 5x - 42 = 0`

`2 {x}^(2) + 12x - 7x - 42 = 0`

`2x(x + 6) - 7(x + 6) = 0`

`(2x - 7)(x + 6) = 0`

Either

`2x - 7 = 0`

`2x = 0 + 7`

`2x = 7`

`x = (7)/(2)`

Or,

`x + 6 = 0`

`x = 0 - 6`

`x = - 6`

Negative value can't be taken.

So, width =
`(7)/(2) ft`

Again,

Finding the value of length,

Length =
`2x + 5`

`2 * (7)/(2) + 5`

`7 + 5`

`12`

Length = 12 ft