Answer:
[tex] 60^2 = (2x^2) +x^2[/tex]
[tex] 3600 = 5x^2[/tex]
[tex] x = \sqrt{\frac{3600}{5}}= 12\sqrt{5}[/tex]
And we can find the perimeter as:
[tex] P= 2(2x) + 2x[/tex]
And replacing the value given for x we got:
[tex] P =2 (2* 12\sqrt{5}) + 2* (12\sqrt{5}) = 48 \sqrt{5} +24\sqrt{5} = 72 \sqrt{5} cm[/tex]
And for this case the perimeter would be approximately [tex]72 \sqrt{5} cm[/tex]
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:
[tex] 60^2 = (2x^2) +x^2[/tex]
[tex] 3600 = 5x^2[/tex]
And solving for x we got:
[tex] x = \sqrt{\frac{3600}{5}}= 12\sqrt{5}[/tex]
And we can find the perimeter as:
[tex] P= 2(2x) + 2x[/tex]
And replacing the value given for x we got:
[tex] P =2 (2* 12\sqrt{5}) + 2* (12\sqrt{5}) = 48 \sqrt{5} +24\sqrt{5} = 72 \sqrt{5} cm[/tex]
And for this case the perimeter would be approximately [tex]72 \sqrt{5} cm[/tex]
