# The playing surface in the game of curling is a rectangular sheet of ice with an area of about 225m^2. The width is about 40 m less than the length. How do you find the approximate dimensions of the playing surface?

Express width in terms of length, then substitute and solve to arrive at the dimensions of $L = 45 m$ and $W = 5 m$

#### Explanation:

$A = L W$

We're given the area and we know that the width is 40m less than the length. Let's write the relationship between L and W down:

$W = L - 40$

And now we can solve $A = L W$:

$225 = L \left(L - 40\right)$

$225 = {L}^{2} - 40 L$

I'm going to subtract ${L}^{2} - 40 L$ from both sides, then multiply by $- 1$ so that ${L}^{2}$ is positive:

${L}^{2} - 40 L - 225 = 0$

Now let's factor and solve for L:

$\left(L - 45\right) \left(L + 5\right) = 0$

$\left(L - 45\right) = 0$
$L = 45$

and

$\left(L + 5\right) = 0$
$L = - 5$

So L = 45. Now let's solve for W:

$W = 45 - 40 = 5$

So the dimensions are $L = 45 m$ and $W = 5 m$