Question

A sandbox has an area of 26 square feet, and a length of 5.5 feet. What is the width of the sandbox?

Answer 1

The width is `approx 4.73 " feet"`.

Explanation:

The area of a rectangular object is the product of its length and its width:

`A=l xx w`

In this question, we are given the area and the length, and we are asked to find the width. The formula above will let us find it, but not in its current form.

Whenever we have a formula for which we know all but one of the variables in it, we can solve for that one unknown variable in terms of the others. In other words, we can rearrange the formula so that the unknown is on its own on one side (i.e. "isolated"), and whatever is left on the other will be some arrangement of just the known values, letting us find a value that the isolated variable is equal to.

This problem is rather straightforward. We know `A` and `l`, and we need to find `w`. So we need to isolate `w`. At the moment, `w` is being multiplied by `l`, so we'd like to "undo" that multiplication. This is done by dividing by `l`.

But wait—we can't just divide the right-hand side by `l`, that wouldn't keep the equation balanced. Whenever we do something to one side, we need to do the same thing to the other. (e.g. if we have `2x= 6`, then we can divide both sides by 2 to get the equivalent statement `(2x)/2 = 6/2`, which reduces to `x=3`.)

We divide both sides of our equation by `l` to get

`A/l = (l xx w)/l`

Because we chose to divide by `l`, the `l`'s on the right side cancel, leaving us with

`A/l=(cancel l xx w)/cancel l=w`

or, putting the unknown on the left,

`w=A/l`

Now, we can just plug in our known values for `A` and `l` to get our answer for `w`:

`w=A/l = (26 " ft"^2)/(5.5" ft")~~4.73" ft"`

You'll also notice how the units reduce: `"ft"^2` is the same as `"ft"xx"ft"`, so when we divide `("ft"xx"ft")/"ft"`, one pair of `"ft"`'s cancels, leaving us with `(cancel"ft"xx"ft")/cancel"ft"="ft"` as the units for `w`. This makes sense, since Area (2-D) is the product of length (1-D) and width (1-D).