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# Length, Width and Area

**Answer**: 9

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

The area is equal to the product of the length and the width:

Area is $w\times(w+16) = 225$.

So the length and the width are a factor pair of 225.

225 = 5 $\times$ 5 $\times$ 3 $\times$ 3

225 = 5 $\times$ 45, difference = 40

225 = 15 $\times$ 15, difference = 0

225 = 9 $\times$ 25, difference = 16

$\therefore$ width = 9

**Using trial & improvement**

We can try different numbers for $w$ (knowing that it is 16 less than the length):

If $w=1$, then the area would be $1\times 17=17$ - too small.

If $w=20$, then the area would be $20\times 36=720$ - too big.

If $w=5$, then the area would be $5\times 21=105$ - too small.

If $w=10$, then the area would be $10\times 26=260$ - just too big.

If $w=9$, then the area would be $9\times 25=225$ - perfect.

So $w=9$.

**Using algebra**

Area $w^2+16w=225$

*Completing the square*

$(w+8)^2=w^2+8w+8w+64=(w^2+16w)+64$

So $(w+8)^2-64 = w^2+16w\\

\begin{align}\therefore w^2+16w=225&\Rightarrow(w+8)^2-64 =225\\

&\Rightarrow (w+8)^2=289\\

&\Rightarrow (w+8)=\pm17\\

&\Rightarrow w=\pm17 - 8\end{align}$

$w$ must be positive, so $w=17-8=9$.

*Using the quadratic formula*

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$w$ is the variable $x$, and $a=1$, $b=16$, $c=-225$

$\begin{align}w&=\dfrac{-16\pm\sqrt{16^2+4\times225}}{2}\\

&=-8\pm\sqrt{16\times4+225}\\

&=-8\pm\sqrt{289}\\

&=-8\pm17\end{align}$

Need width to be positive so $w=-8+17=9$

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Age 14 to 16

ShortChallenge Level

- Problem
- Solutions

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

The area is equal to the product of the length and the width:

Area is $w\times(w+16) = 225$.

So the length and the width are a factor pair of 225.

225 = 5 $\times$ 5 $\times$ 3 $\times$ 3

225 = 5 $\times$ 45, difference = 40

225 = 15 $\times$ 15, difference = 0

225 = 9 $\times$ 25, difference = 16

$\therefore$ width = 9

We can try different numbers for $w$ (knowing that it is 16 less than the length):

If $w=1$, then the area would be $1\times 17=17$ - too small.

If $w=20$, then the area would be $20\times 36=720$ - too big.

If $w=5$, then the area would be $5\times 21=105$ - too small.

If $w=10$, then the area would be $10\times 26=260$ - just too big.

If $w=9$, then the area would be $9\times 25=225$ - perfect.

So $w=9$.

Area $w^2+16w=225$

$(w+8)^2=w^2+8w+8w+64=(w^2+16w)+64$

So $(w+8)^2-64 = w^2+16w\\

\begin{align}\therefore w^2+16w=225&\Rightarrow(w+8)^2-64 =225\\

&\Rightarrow (w+8)^2=289\\

&\Rightarrow (w+8)=\pm17\\

&\Rightarrow w=\pm17 - 8\end{align}$

$w$ must be positive, so $w=17-8=9$.

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$w$ is the variable $x$, and $a=1$, $b=16$, $c=-225$

$\begin{align}w&=\dfrac{-16\pm\sqrt{16^2+4\times225}}{2}\\

&=-8\pm\sqrt{16\times4+225}\\

&=-8\pm\sqrt{289}\\

&=-8\pm17\end{align}$

Need width to be positive so $w=-8+17=9$

You can find more short problems, arranged by curriculum topic, in our short problems collection.