Question

The area of a rectangle is 65yd^2 , and the length of the rectangle is 3yd less than double the width, how do you find the dimensions?

Answer 1

build the equations and solve...

Explanation:

let the area be `A = l * w` where length is `l` and width is `w`

so 1.st equqtion will be

`l*w=65`

and length is 3 yd less than double the width says:

`l = 2w-3` (2nd eq.)

replacing `l` with `2w-3` in first eq. will yield

`(2w-3) * w = 65`
`2w^2-3w = 65`
`2w^2-3w-65 = 0`

now we have a 2nd order equation just find the roots and take the positive one as width can not be negative...

`w=(3+-sqrt(9+4*2*65))/(2*2) = (3+-sqrt(529))/(4) = (3+-23)/4`
`w=-5 , 13/2` so taking `w=13/2 = 6.5 yd`

replacing `w` with `6,5` in second eq. we get

`l=2w-3 = 2*6.5-3 = 13-3 = 10 yd`

`A = l*w = 10 * 6.5 = 65yd` will confirm us...