Question

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?

Answer 1

`sqrt(277)"cm" ~~ 16.64"cm"`

Explanation:

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

`w(w+5) = 126`

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

`126 = 2 * 3 * 3 * 7 = 14 * 9`

So the width of the rectangle is `9"cm"` and the length is `14"cm"`

Alternative method

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

`w(w+5) = 126`

rearrange it as `w^2+5w-126 = 0`

and solve using the quadratic formula to get:

`w = (-5+-sqrt(5^2-(4xx1xx126)))/(2xx1)=(-5 +-sqrt(25+504))/2`

`=(-5+-sqrt(529))/2=(-5+-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) ~~ 16.64`