Question

The length of a rectangle is 4 meters longer than the width. If the area is 22 square meters. find the rectangles dimensions. The width is what? The length is what?

Answer 1

The width is:

`-2+√(26)\text{ meters}\text{ }(\text{or approximately 3.0990 meters})`

And the length is:

`2+√(26)\text{ meters}\text{ } (\text{or approximately 7.0990 meters})`

Explanation:

Recall that the area of a rectangle is given by:

`\displaystyle A = w\ell`

Where w is the width and l is the length.

We are given that the length of a rectangle is four meters longer than the width. Thus:

`\ell = w + 4`

And we also know that the area of the rectangle is 22 square meteres.

Substitute:

`(22)=w(w+4)`

Distribute and isolate the equation:

`w^2+4w-22=0`

The equation isn't factorable, so we can instead use the quadratic formula:

`\displaystyle w = (-b\pm√(b^2-4ac))/(2a)`

In this case, a = 1, b = 4, and c = -22. Substitute:

`\displaystyle w = (-(4)\pm√((4)^2-4(1)(-22)))/(2(1))`

Evaluate:

`\displaystyle\begin{aligned} w &= (-4\pm√(104))/(2)\\ \\ &=(-4\pm√(4\cdot 26))/(2) \\ \\ &=(-4\pm2√(26))/(2) \\ \\ & = -2\pm √(26) \end{aligned}`

Thus, our two solutions are:

`w_1=-2+√(26)\approx 3.0990\text{ or } w_2=-2-√(26)\approx-7.0990`

Since the width cannot be negative, we can ignore the second solution.

Since the length is four meters longer than the width:

`\ell = (-2+√(26))+4=2+√(26)\text{ meters}`

Thus, the dimensions of the rectangle are:

`\displaystyle (2+√(26)) \text{ meters by } (-2+√(26))\text{ meters}`

Or, approximately 3.0990 by 7.0990.