Answer:
The width is 3cm, and the length is 7cm.
Step-by-step explanation:
Let's call the length of the rectangle
, the width of the rectangle
, and the area of the rectangle
, and write down everything we know about them.
The length of the rectangle is 11 centimetres less than six times its width, so we know that
.
We also know that
. But what exactly is the area of a rectangle? Well, it's the product of its length and width, <em>i.e. </em>
<em>.</em>
Then we have two equations involving
and
:
, and
.
So how do we find
and
? Let's substitute our first equation into our second, like so:
![l \times w = (6w - 11) \times w\\{}\hspace{0.8cm}= 6w^2 - 11w\\{}\hspace{0.8cm}= 21](https://tex.z-dn.net/?f=l%20%5Ctimes%20w%20%3D%20%286w%20-%2011%29%20%5Ctimes%20w%5C%5C%7B%7D%5Chspace%7B0.8cm%7D%3D%206w%5E2%20-%2011w%5C%5C%7B%7D%5Chspace%7B0.8cm%7D%3D%2021)
We can rearrange this new equation to get everything on the same side, so that we have a standard quadratic equation, like this:
.
We can use the quadratic formula for this. You should have seen this before:
,
where
are the coefficients of our quadratic equation. Plugging the values in, we see
.
With a little further cancellation, we can find our two possible values of
are
and
.
Since
is a width of a rectangle, it can't possibly be negative, and so it must be 3.
We can now plug in this value into our equation from earlier, namely
. We can see that
,
and so we get our answer,
cm,
cm.
We can prove that this is right by multiply 3 and 7 together, and seeing that we get 21, the area we expect.