Question

The length of a rectangle is twice as long as the width. If the diagonal length of the rectangle

is 60 cm, what is the perimeter of the rectangle (to the nearest centimetre)?
Provide a complete algebraic solution, including a supporting diagram.
I

Answer 1

`60^2 = (2x^2) +x^2`

`3600 = 5x^2`

`x = \sqrt{(3600)/(5)}= 12√(5)`

And we can find the perimeter as:

`P= 2(2x) + 2x`

And replacing the value given for x we got:

`P =2 (2* 12√(5)) + 2* (12√(5)) = 48 √(5) +24√(5) = 72 √(5) cm`

And for this case the perimeter would be approximately
`72 √(5) cm`

Explanation:

For this case we can assume that the lenght is 2x the width x and the diagonal is 60 cm.

From the picture given we have a right tirngle and we can set the following equation:

`60^2 = (2x^2) +x^2`

`3600 = 5x^2`

And solving for x we got:

`x = \sqrt{(3600)/(5)}= 12√(5)`

And we can find the perimeter as:

`P= 2(2x) + 2x`

And replacing the value given for x we got:

`P =2 (2* 12√(5)) + 2* (12√(5)) = 48 √(5) +24√(5) = 72 √(5) cm`

And for this case the perimeter would be approximately
`72 √(5) cm`