Find the CDF of $\ X $ where $\ X = \sqrt{A^2 + B^2} $

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Say $\ A,B \sim U(0,1) $ and $\ X = \sqrt{A^2 + B^2} $

and I want to find the the value $\ t$ of $\ X $ for which $\ P(X \le t) = 0.5 $

I understand I need to find the CDF of $\ X $ first and that $\ A, B $ distributed uniformly in the unit square. But I can't see how to find the CDF of $\ X $ ?

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It is very easy. Do a drawing and calculate the area $\sqrt{a^2+b^2}<r$ in the unit square $[0;1]\times [0;1]$ then set this probability as 0.5 and solve in $r$

... of course you have to assume independence between $A,B$

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Thus the solution is

$$\frac{\pi r^2}{4}=\frac{1}{2}$$

$$r=\sqrt{\frac{2}{\pi}}$$