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Explanation:

If ACB is a right angled isosceles triangle and AB is hypotenuse with length 7*($\sqrt{2}$) length, then sides AC and BC = 7 cm
Now, the circle inscribed is an incircle.
Radius of in-circle = {7 + 7 - [7*($\sqrt{2}$)]}/2 = 7 - $\frac{7\sqrt{2}}{2}$
Now, if center of circle is marked as O, then quadrilateral OZCY is a square.
Hence, CY = CZ = 7 - $\frac{7\sqrt{2}}{2}$
Hence, AZ = 7 - 7 - $\frac{7\sqrt{2}}{2}$ = $\frac{7\sqrt{2}}{2}$
Also, BX = BY and BY = BC - CY = $\frac{7\sqrt{2}}{2}$
Hence, BX = $\frac{7\sqrt{2}}{2}$

Hence, the given ratio becomes 1 : 1 : [($\sqrt{2}$) - 1]