Question

The length of a rectangle is 5 yd less than double the width, and the area of the rectangle is 52 yd^2 . How do you find the dimensions of the rectangle?

Answer 1

Width = 6.5 yds, length = 8 yds.

Explanation:

Define the variables first.

We could use two different variables, but we have been told how the length and width are related.

Let the width be `x" width is the smaller side"`
The length = `2x -5`

"Area = l x w" and the area is given to be 52 squ yards.

`A = x(2x-5) = 52`

`2x^2 -5x = 52" quadratic equation"`

`2x^2 -5x -52 = 0`

To factorise, find factors of 2 and 52 which cross-multiply and subtract to give 5.

`color(white)(xxx)(2)" "(52)`
`color(white)(xx.x) 2" 13 "rArr 1xx13 = 13`
`color(white)(xx.x) 1" 4 "rArr2xx4 = 8" "13-8 = 5`

We have the correct factors, now fill in the signs. We need -5.

`color(white)(xxx)(2)" "(-52)`
`color(white)(xx.x) 2" - 13 "rArr 1xx-13 = -13`
`color(white)(xx.x) 1" +4 "rArr2xx+4 = +8" "-13+8 = -5`

`(2x-13)(x+4) = 0`

Each factor could be equal to 0

`x = 6.5 or x = -4` (reject)

The width = 6.5 yards. Now find the length: 6.5 x 2 -5 = 8 yards

Check:
Width = 6.5yds, length = 8yds
Area = 6.5 x 8 = 52

Answer 2

Length`= 8 yd`
Width `= 6.5 yd`.

Explanation:

Let width be `=x`
Therefore, length `=2x -5`

We know that
`"Area" = "Length" xx "Width"`
Inserting given and assumed numbers we get

`52=(2x-5)xx x`
rearranging we obtain

`2x^2 -5x -52 = 0`

To factorize we use split the middle term method. We have two parts of middle term as `-13x and 8x`. The equation becomes

`2x^2-13x+8x-52 = 0`
Paring and taking out common factors we have
`x(2x-13)+4(2x-13) = 0`
`=>(2x-13)(x+4) = 0`

Setting each factor equal to `0`, we have two roots
`(2x-13)=0and (x+4) = 0`
`x = 13/2=6.5`
`x = -4`, rejected as width can not be a `-ve` value

`:.`Width `= 6.5 yd`. And length`= 2xx6.5 -5 = 8 yd`

Check:
Area `= 8xx 6.5 = 52yd^2`