# The width of a rectangle is 5 less than twice its length. If the area of the rectangle is 126 cm^2, what is the length of the diagonal?

Sep 10, 2015

$\sqrt{277} \text{cm" ~~ 16.64"cm}$

#### Explanation:

If $w$ is the width of the rectangle, then we are given that:

$w \left(w + 5\right) = 126$

So we would like to find a pair of factors with product $126$ which differ by $5$ from one another.

$126 = 2 \cdot 3 \cdot 3 \cdot 7 = 14 \cdot 9$

So the width of the rectangle is $9 \text{cm}$ and the length is $14 \text{cm}$

Alternative method

Instead of factoring in this way, we could take the equation:

$w \left(w + 5\right) = 126$

rearrange it as ${w}^{2} + 5 w - 126 = 0$

and solve using the quadratic formula to get:

$w = \frac{- 5 \pm \sqrt{{5}^{2} - \left(4 \times 1 \times 126\right)}}{2 \times 1} = \frac{- 5 \pm \sqrt{25 + 504}}{2}$

$= \frac{- 5 \pm \sqrt{529}}{2} = \frac{- 5 \pm 23}{2}$

that is $w = - 14$ or $w = 9$

We are only interested in the positive width so $w = 9$, giving us the same result as the factoring.

Finding the diagnonal

Using Pythagoras theorem, the length of the diagonal in cm will be:

$\sqrt{{9}^{2} + {14}^{2}} = \sqrt{81 + 196} = \sqrt{277}$

$277$ is prime, so this does not simplify any further.

Using a calculator find $\sqrt{277} \approx 16.64$