A rectangular lot has a perimeter of 44cm. The area of the lot is 107cm$^2$. Find the dimensions of the lot?

Question

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!

Answer 1

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 $11-\sqrt{14}$.

Answer 2

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.

Answer 3

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.

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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.