# The length of a rectangle is 6m longer than the width. If the area of a rectangle is $84^2$m, find the dimensions of the rectangle.

I'm in grade 11 math.

The length of a rectangle is $$6m$$ longer than the width. If the area of a rectangle is $$84~\text{m}^2$$, find the dimensions of the rectangle. I don't know where to start other than:

$$l = w + 6$$

$$84 = lw$$

• Can you solve $84 = (w+6)w$? – Umberto P. Sep 25 '18 at 22:38
• @Zebert This uses the same technique as this question and this question. – Toby Mak Sep 25 '18 at 22:38

From the given information, you have the following system of equations to solve: \begin{align*} \begin{cases} L = 6 + W\\ LW = 84\\ \end{cases} \end{align*}

Where $$W$$ denotes the width and $$L$$ denotes the length. If we substitute the first relation into the second, it results into the following equation: \begin{align*} (6+W)W = 84 \Longleftrightarrow W^{2} + 6W - 84 = 0 \end{align*}

Can you proceed from here?

EDIT

Observe that $$W^{2} + 6W = (W^{2} + 6W + 9) - 9 = (W+3)^{2} - 9$$, from whence we obtain \begin{align*} W^{2} + 6W - 84 = 0 \Longleftrightarrow (W+3)^{2} - 93 = 0 \Longleftrightarrow (W+3)^{2} = 93 \Longleftrightarrow W = \pm\sqrt{93} - 3 \end{align*}

Since $$W$$ must be positive, we conclude that $$W = \sqrt{93}-3$$ and $$L = \sqrt{93} + 3$$.

Another possible approach is to apply the Bhaskara's formula, which is given by: \begin{align*} ax^{2} + bx + c = 0 \Longleftrightarrow x = \frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \end{align*}

• I tried but I don't know how. Can you please continue? – Zebert Sep 25 '18 at 22:44
• Sure. I will edit my answer. – user1337 Sep 25 '18 at 22:45

Substitute $$w=l-6$$ into $$lw=84$$ we get $$l^2-6l=84\Leftrightarrow l=3\pm \sqrt{93}$$ Since $$l>0$$, we get $$l=3+\sqrt{93}$$. So $$w=\sqrt{93}-3$$