Does this needs construction?

The formula we can apply here is $r = \dfrac{A}{s}$ , where $r$ is the radius of the incircle of a given triangle and $A,s$ are respectively the area and the semi-perimeter of the triangle. A proof of this formula is given here .

In this case, $\Delta TSR$ is a right triangle with side lengths $|TR| = 3$ cm and $|SR| = 6$ cm, giving us an area of $A = \frac{1}{2}*3*6 = 9$ sq. cm.. Since the hypotenuse $ST$ has length $\sqrt{3^{2} + 6^{2}} = \sqrt{45} = 3\sqrt{5}$ cm., the semi-perimeter is then $s = \dfrac{1}{2}(9 + 3\sqrt{5})$ cm..

Thus we find that $r = \dfrac{A}{s} = \dfrac{9*2}{9 + 3\sqrt{5}} = \dfrac{3(3 - \sqrt{5})}{2} = \boxed{1.146}$ cm. to 3 decimal places.

Name the points where the circle touches QR as U, where the circle touches SR as V and where the circle touches ST as W. Clearly as TR=3 so UT=3-r (r is the radius of the circle which we need to find). So, TW=3-r( TW and TU are tangents from an external point T). Also, VR=r so VS=WS=6-r. This means that TS=3-r+6-r=9-2r. But TR=3√5 by Pythagoras Theorem in ∆TRS. So, 9-2r=3√5, which means r=1.146cm

In any rt. triangle ABC where /_B=90°, in-radius = (AB+BC -AC)/2. In this case, r=(6+2 -√45)/2~1.1459 cm