A rectangular lot has a perimeter of 44cm. The area of the lot is 107cm$^2$. Find the dimensions of the lot? I'm in grade 11 math. I need help to solve this.
A rectangular lot has a perimeter of 44cm. The area of the lot is 107cm$^2$. What are the dimensions of the lot?
So far I did this but it doesn't make sense:
$P=2l+2w$ or $44 = 2l+2w$
$A = lw$  or $44 = 2l + 2w$
$44=2l+2w$ all divided by 2 is $22=l+w$ than I isolate the l so it is: $l = 22-w$
sub $l=22-w$ into $107=lw$
$107=(22-w)w$
$107=-w^2 + 22w$
$107=-(w^2-22w)$
$107=-(w^2-22w+121-121)$
$107=-(w^2-22w+121)+121$
but I'm lost from there. Any help would be appreciated.
Thanks!
 A: On these types of word problems, it is best to turn the sentences into equations.
"A rectangular lot has a perimeter of 44cm." would turn into $2l+2w=44$.
"The area of the lot is 107cm$^2$." would turn into $lw=107$. 
Now, we have a system of equations: $$2l+2w=44$$ $$lw=107$$
We can start solving now. Let's first solve for $w$, so $$l=\frac{107}{w}$$ $$\implies 2\cdot\frac{107}{w}+2w=44$$ $$\implies \frac{214}{w}+2w=44$$ Eww. What do we have here? Fractions! Nobody likes those. So let's get rid of them! We multiply the entire equation by $w$ getting $$\implies 214+2w^2=44w$$ $$\implies 2w^2-44w+214=0$$
We have formed a quadratic equation, so thus we can use the quadratic formula. Recall that the quadratic formula is $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. In our equation, $a=2$, $b=-44$, and $c=214$, so we plug these numbers into the formula yielding $$\frac{-(-44)\pm\sqrt{(-44)^2-(4\cdot 2\cdot 214)}}{2\cdot 2}$$ $$\implies\frac{44\pm\sqrt{224}}{4}$$ $$\implies\frac{44\pm 4\sqrt{14}}{4}$$
We now divide by four getting
$$w=11\pm\sqrt{14}$$
Now let's find $l$. $107\div (11\pm\sqrt{14})=11\mp\sqrt{14}$.
Therefore, we can now conclude that when the length of the lot is $11-\sqrt{14}$, then the width is $11+\sqrt{14}$. When the length of the lot is $11+\sqrt{14}$, then the width is $11-\sqrt{14}$.
A: If you choose to solve this by completing the square, observe that $$\begin{align}107 & =-\underbrace{(w^2-22w+121)}_{(w-11)^2}+121 \\ &= -(w-11)^2 + 121 \\ \implies -14 & = - (w-11)^2 \tag{move 121 over}\\ \implies 14 &= (w-11)^2 \tag{divide both sides by $-1$} \\\implies \pm \sqrt{14} &= w - 11,\end{align}$$
which from here you can find both possible widths. Afterwards, you will have two possible lengths as well. Without some more information, you will have two possible solutions.
A: Just Keep going:
$107=-(w^2-22w+121)+121$
$107 - 121 = -(w^2 - 22x + 121)$
$-14= -(w^2 - 22x + 121)$
$14 = w^2 - 22x + 121$
$14 = (w - 11)^2$
$\pm \sqrt {14} = w - 11$
$11 \pm \sqrt{14} = w$.
That doesn't seem right but it could be.  Would the teacher give a problem with an irrational solution?  Well, maybe....
$l = 22 - (11 \pm \sqrt{14}) = 11 \mp \sqrt{14}$.
So $2l + 2w = 2(11\mp \sqrt{14}) + 2(11 \pm \sqrt{14}) = 44$.  So far so good.
$lw = (11 + \sqrt{14})(11 - \sqrt{14} = 11^2 + 11\sqrt{14} -11\sqrt{14} - \sqrt{14}^2 = 121 - 14 = 107$.
Oh, it is correct after all.
====
Here's a little trick.
You have $l+w = 22$.  So the Average of $l$ and $w$ is $11$.  Let $l = $ Average $+d= 11 +d$ and $w = $ Average $-d=11-d$ (We might as well assum it is longer than it is wide.  It doesn't make a difference.)
So $107 = (11 + d)(11 - d) = 11^2 - d^2$ so
$107 = 121 - d^2$ and $d^2 = 121 - 107 = 14$ so $d = \sqrt{14}$ (well, it could be $-\sqrt{14}$ but we assumed $d \ge 0$ when we assumed it was longer than it was wide.  It doesn't make a difference.)
So $l = 11 + \sqrt {14}$ and $w = 11 - \sqrt{14}$.
Just a trick.
