These are also incorrect actually! The middle one will be x(1+rt2), the same as the “circumscribing” square…for the others you will have to do (rather painful) Pythagoras applications and hence you would rarely be expected to find them in an exam scenario (which is why I left it as a homework exercise, more from the point of view of developing a comfort with geometry)
Is it right to say two shortest diagonals and one longest diagonal will make an isosceles right-angled triangle and the ratio of longest diagonal to the shortest diagonal will be √2 :1 ?
CAT-holics 
Longest – 5x^2
Middle – 2x
Sir, how to get the last one ?
These are also incorrect actually! The middle one will be x(1+rt2), the same as the “circumscribing” square…for the others you will have to do (rather painful) Pythagoras applications and hence you would rarely be expected to find them in an exam scenario (which is why I left it as a homework exercise, more from the point of view of developing a comfort with geometry)
regards
J
Longest: sqrt [2x^2 (2+rt2) ]
Smallest: sqrt [x^2 (2+rt2) ]
Middle one is just the summation (side of the square!)
H
In the figure :
Middle one= x(1+sqrt(2))
The longest one would be = x* sqrt[4+2sqrt(2)]
shorter one would be = x* sqrt[3+sqrt(2)]
Is that correct?
Hi J,
Is it right to say two shortest diagonals and one longest diagonal will make an isosceles right-angled triangle and the ratio of longest diagonal to the shortest diagonal will be √2 :1 ?
Perfectly right, as 4 such “shortest diagonals” will form a square whose diagonal will be the “longest diagonal” here.
regards
J