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Please See The Attached File.

If you draw this octagon inside the circle, the longest diagonal will 2R.
As you see the right-angled triangle in the picture, two other sides of the triangle is the shortest diagonal of the octagon.
---> \(x^{2} + x^{2} = (2R)^{2}\)

\(2x^{2} = 4R^{2}\)

\(x = √2R\)

\(\frac{Diagonal_{Longest}}{ Diagonal_{Shortest} }= \frac{2R}{√2R}= \frac{√2}{1}\)

Answer (D).
Attachments

Octagon.jpg
Octagon.jpg [ 52.02 KiB | Viewed 11217 times ]

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I find the easiest way to consider the octagon circumscribed in a circle. As it is a regular polygon, naturally, the longest distance inside the octagon is the longest length of the circle = diameter. From here it can be seen that longest diagonal and shortest diagonal form one leg and hypotenuse of an isosceles right triangle.

Ratio Sqrt 2 : 1
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