Question

The length of a rectangle is 11 centimeters less than six times its width. Its area is 21 square centimeters. Find the dimensions of the rectangle.

Answer 1

Answer:

The width is 3cm, and the length is 7cm.

Step-by-step explanation:

Let's call the length of the rectangle $l$, the width of the rectangle $w$, and the area of the rectangle $A$, 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 $l = 6w - 11$.

We also know that $A = 21$. But what exactly is the area of a rectangle? Well, it's the product of its length and width, i.e. $A = l\times w$.

Then we have two equations involving $l$ and $w$:

$l = 6w - 11$, and $l\times w = 21$.

So how do we find $l$ and $w$? Let's substitute our first equation into our second, like so:

$l \times w = (6w - 11) \times w\\{}\hspace{0.8cm}= 6w^2 - 11w\\{}\hspace{0.8cm}= 21$

We can rearrange this new equation to get everything on the same side, so that we have a standard quadratic equation, like this:

$6w^2 - 11w - 21 = 0$.

We can use the quadratic formula for this. You should have seen this before:

$w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,

where $a,b,c$ are the coefficients of our quadratic equation. Plugging the values in, we see

$w = \frac{11\pm\sqrt{121+504}}{12}\\\\{}\hspace{0.25cm}=\frac{11\pm\sqrt{625}}{12}\\\\{}\hspace{0.25cm}=\frac{11\pm25}{12}$.

With a little further cancellation, we can find our two possible values of $w$ are

$w = \frac{36}{12} = 3$  and $w = \frac{-14}{12}$.

Since $w$ 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 $l = 6w - 11$. We can see that

$l = 6\times 3 - 11 = 18 - 11 = 7$,

and so we get our answer, $w = 3$cm, $l = 7$cm.

We can prove that this is right by multiply 3 and 7 together, and seeing that we get 21, the area we expect.