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Length, Width and Area

Age 14 to 16 Short Challenge Level:

Using factors of 225
The area is equal to product of the length and the width. So the length and the width are a factor pair of 225.

225 = 5 $\times$ 45, but 5 and 45 do not have a difference of 16
225 = 5 $\times$ 3 $\times$ 15 = 15 $\times$ 15, but 15 and 15 do not have a difference of 16
225 = 5 $\times$ 5 $\times$ 3 $\times$ 3 = 25 $\times$ 9. 25 and 9 do have a difference of 16, so the width is 9.


Using pairs of numbers with a difference of 16
Suppose the width is $w$, then the length is $w+16$.

Then the area is $w\times(w+16) = 225$.

We can try different numbers for $w$.
If $w=1$, then the area would be $1\times17=17$ - too small.
If $w=20$, then the area would be $20\times34=680$ - too big.
If $w=5$, then the area would be $5\times21=105$ - too small.
If $w=10$, then the area would be $10\times26=260$ - too big.
If $w=9$, then the area would be $9\times25=225$ - perfect.

So $w=9$.


Using algebra
Suppose the width is $w$, then the length is $w+16$.

Then the area is $w\times(w+16) = 225$, which means that $w^2+16w=225\Rightarrow w^2+16w-225=0$

We could try to factorise this into two brackets, $(w+\text{__})(w-\text{__})\equiv w^2+16w-225$. However, notice that $225$ would be the product of the two unknown numbers, and $16$ would be their difference - so we are back to the beginning.

We have two other options:

Completing the square
$w^2+16w$ can be written in the form $(w+?)^2-???$.

Notice that $$\begin{split}(w+8)^2=(w+8)(w+8)&=w^2+8w+8w+64\\
&=(w^2+16w)+64\end{split}$$ So $(w+8)^2-64 = (w^2+16w) +64-64 =w^2+16w$ 

So $$\begin{align}w^2+16w=225&\Rightarrow(w+8)^2-64 =225\\
&\Rightarrow (w+8)^2=289\\
&\Rightarrow (w+8)^2=17^2\\
&\Rightarrow (w+8)=\pm17\\
&\Rightarrow w=\pm17 - 8\end{align}$$ $w$ must be positive, so $w=17-8=9$.


Using the quadratic formula
The quadratic formula is that, if $ax^2+bx+c=0$, then $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$.

In this case, $w^2+16w-225=0$, $w$ is the variable $x$, and $a=1$, $b=16$, $c=-225$

So $b^2-4ac = 16^2-4\times1\times-225 =256 + 900 = 1156$

We will need $\sqrt{b^2-4ac}=\sqrt{1156}$.
$30^2=900$ - too small
$40^2=1600$ - too big
$35^2=1050+150+25=1225$ - too big
$34^4 = 1020 + 120 +16=1156$ - perfect

So $w=\dfrac{-16\pm34}{2}$, which gives $w=\dfrac{18}2=9$ or $w=\dfrac{-50}{2}=-25$. Clearly $w>0$ so $w=9$.
You can find more short problems, arranged by curriculum topic, in our short problems collection.