Correct Answer: $2–\sqrt{2}: \sqrt{2}$

Solution :
Given: ABC is an isosceles right-angle triangle. $\angle ABC = 90 ^{\circ}$ and AB = 12 cm.
In $\triangle ABC$, $H^2=B^2+P^2$ (Using Pythagoras theorem).
⇒ $H^2=12^2+12^2$
⇒ $H^2=144+144=288$
⇒ $H=\sqrt{288}=12\sqrt2$ cm
Inradius of inscribed circle of right angled triangle = $\frac{(P + B – H)}{2}$.
Circumradius of the circumscribed circle of right angle triangle = $\frac{H}{2}$
where $P$, $B$, and $H$ are the perpendicular, base, and hypotenuse.
Inradius of inscribed circle of right angled triangle = $\frac{(12 + 12 – 12\sqrt2)}{2}=6(2–\sqrt2)$.
Circumradius of the circumscribed circle of the right angle triangle = $\frac{12\sqrt2}{2}=6\sqrt2$.
The ratio of the radius of the circle inscribed in it to the radius of the circle circumscribing triangle ABC = $6(2–\sqrt2):6\sqrt2=2–\sqrt{2}: \sqrt{2}$.
Hence, the correct answer is $2–\sqrt{2}: \sqrt{2}$.