Question

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

Answer 1

Answer:

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

$3600 = 5x^2$

$x = \sqrt{\frac{3600}{5}}= 12\sqrt{5}$

And we can find the perimeter as:

$P= 2(2x) + 2x$

And replacing the value given for x we got:

$P =2 (2* 12\sqrt{5}) + 2* (12\sqrt{5}) = 48 \sqrt{5} +24\sqrt{5} = 72 \sqrt{5} cm$

And for this case the perimeter would be approximately $72 \sqrt{5} cm$

Step-by-step 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:

$x = \sqrt{\frac{3600}{5}}= 12\sqrt{5}$

And we can find the perimeter as:

$P= 2(2x) + 2x$

And replacing the value given for x we got:

$P =2 (2* 12\sqrt{5}) + 2* (12\sqrt{5}) = 48 \sqrt{5} +24\sqrt{5} = 72 \sqrt{5} cm$

And for this case the perimeter would be approximately $72 \sqrt{5} cm$