Question

The length of a rectangle is 5 m less than twice the width, and the area of the rectangle is `33 m^2`. What are the dimensions of the rectangle?

Answer 1

Length`=color(blue)(6)`

Width `= color(blue)(11/2)`

Explanation:

Let `x=` the width of the rectangle.

Then the length is:

`(2x-5)m`

Area is length x width:

`(2x-5)*x=33=> 2x^2-5x-33=0`

Factor:

`(2x-11)(x+3)=0=>x=-3 and x=11/2` ( negative root not applicable )

Dimensions of rectangle:

Length`= 2(11/2)-5=color(blue)(6)`

Width `= color(blue)(11/2)`

Answer 2

`w = 5.5` meters and `l = 6` meters.

Explanation:

We can start by listing what we know and what we don't know.

• `l = 2w - 5`
• `a = 33`
• `w = ?`
• `a = lw`

We know that if A = B and A = C, then B must equal C. We can use this basic logical principle to write ourselves an equation that we can use to find the solution.

if `a = 33 and a = lw` then `lw = 33`
`lw = 33`

We can then use the same principle to simplify this equation so that it uses only one variable. If `l = 2w - 5`, then `lw = (2w-5)w`, so we can say that
`(2w-5)w = 33`.

After that, you need to simplify, factor, and evaluate:

`2w^2 - 5w = 33`
`2w^2 - 5w - 33 = 0`
`2w^2 - 5w - 33 = (2w-11)(w+3)`
`(2w-11)(w+3) = 0`

`((2w-11)(w+3) = 0)/(w+3)` or `((2w-11)(w+3) = 0)/(2w-11)`

`2w - 11 = 0` or `w+3 = 0`
`2w = 11` or `w = -3`
`w = 5.5` or `w = -3`

It is impossible for a rectangle to have a width of -3 meters, so we can eliminate that option. Now we know that our width is 5.5 meters.

Now we can plug the width into our length equation to find the length.

`l = 2(5.5) - 5`
`l = 6`

To check our answers, we can plug our length and width back into our `lw = 33` equation to make sure that `lw` actually equals 33.

`6*5.5 = 33`
`33 = 33`