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 $w=9$. 

Image

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\

$$\begin{array}{rl}\therefore w^2+16w=225& \Rightarrow (w+8{)}^2-64=225\\ & \Rightarrow (w+8{)}^2=289\\ & \Rightarrow (w+8)=\pm 17\\ & \Rightarrow w=\pm 17-8\end{array}$$$

$w$ 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 $a=1$, $b=16$, $c=-225$

$$$\begin{array}{rl}w& ={\displaystyle \frac{-16\pm \sqrt{{16}^2+4\times 225}}{2}}\\ & =-8\pm \sqrt{16\times 4+225}\\ & =-8\pm \sqrt{289}\\ & =-8\pm 17\end{array}$$$

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