Length, width and area
The area of a rectangle is 225 square units. Find its width.
Problem
The area of a rectangle is 225 square units.
Its length is 16 more than its width.
Find its width.
This problem is taken from the World Mathematics Championships
Student Solutions
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
![length, width and area length, width and area](/sites/default/files/styles/large/public/thumbnails/content-id-13724-length%252520width%252520and%252520area.png?itok=OKzNaXq_)
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\
$w$ 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 $a=1$, $b=16$, $c=-225$
$
Need width to be positive so $w=-8+17=9$