# The length of a rectangle is 3 centimeters less than its width. What are the dimensions of the rectangle if its area is 54 square​ centimeters?

Feb 27, 2018

Width$= 9 c m$
Length$= 6 c m$

#### Explanation:

Let $x$ be width, then length is $x - 3$
Let area be $E$. Then we have:
$E = x \cdot \left(x - 3\right)$
$54 = {x}^{2} - 3 x$
${x}^{2} - 3 x - 54 = 0$

We then do the Discriminant of the equation:

$D = 9 + 216$
$D = 225$

${X}_{1} = \frac{3 + 15}{2} = 9$
${X}_{2} = \frac{3 - 15}{2} = - 6$ Which is declined, since we can't have negative width and length.

So $x = 9$
So width $= x = 9 c m$ and length$= x - 3 = 9 - 3 = 6 c m$

Feb 27, 2018

Length is $6 c m$ and the width is $9 c m$

#### Explanation:

In this question, the length is less than the width. It does not matter at all - they are just names for the sides. Usually the length is longer, but let's go with the question.

Let the width be $x$
The length will be $x - 3 \text{ }$ ( it is $3$cm less)

The area is found from $l \times w$

$A = x \left(x - 3\right) = 54$

${x}^{2} - 3 x - 54 = 0 \text{ } \leftarrow$ make a quadratic equation equal to $0$

Factorise: Find factors of $54$ which differ by $3$

$\left(x \text{ "9)(x" } 6\right) = 0$

There must be more negatives: $\text{ }$ because of $- 3 x$

$\left(x - 9\right) \left(x + 6\right) = 0$

Solve for $x$

$x - 9 = 0 \text{ } \rightarrow x = 9$

$x + 3 = 0 \text{ "rarr x =-3" }$ reject as the length of a side.

the width is $9 c m$ and the length is $9 - 3 = 6 c m$