Answer:
The width is 3cm, and the length is 7cm.
Step-by-step explanation:
Let's call the length of the rectangle [tex]l[/tex], the width of the rectangle [tex]w[/tex], and the area of the rectangle [tex]A[/tex], 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 [tex]l = 6w - 11[/tex].
We also know that [tex]A = 21[/tex]. But what exactly is the area of a rectangle? Well, it's the product of its length and width, i.e. [tex]A = l\times w[/tex].
Then we have two equations involving [tex]l[/tex] and [tex]w[/tex]:
[tex]l = 6w - 11[/tex], and [tex]l\times w = 21[/tex].
So how do we find [tex]l[/tex] and [tex]w[/tex]? Let's substitute our first equation into our second, like so:
[tex]l \times w = (6w - 11) \times w\\{}\hspace{0.8cm}= 6w^2 - 11w\\{}\hspace{0.8cm}= 21[/tex]
We can rearrange this new equation to get everything on the same side, so that we have a standard quadratic equation, like this:
[tex]6w^2 - 11w - 21 = 0[/tex].
We can use the quadratic formula for this. You should have seen this before:
[tex]w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex],
where [tex]a,b,c[/tex] are the coefficients of our quadratic equation. Plugging the values in, we see
[tex]w = \frac{11\pm\sqrt{121+504}}{12}\\\\{}\hspace{0.25cm}=\frac{11\pm\sqrt{625}}{12}\\\\{}\hspace{0.25cm}=\frac{11\pm25}{12}[/tex].
With a little further cancellation, we can find our two possible values of [tex]w[/tex] are
[tex]w = \frac{36}{12} = 3[/tex] and [tex]w = \frac{-14}{12}[/tex].
Since [tex]w[/tex] 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 [tex]l = 6w - 11[/tex]. We can see that
[tex]l = 6\times 3 - 11 = 18 - 11 = 7[/tex],
and so we get our answer, [tex]w = 3[/tex]cm, [tex]l = 7[/tex]cm.
We can prove that this is right by multiply 3 and 7 together, and seeing that we get 21, the area we expect.
