If PR $= 15$cm and AC $= 5$cm, use similar triangles to find the radius of the circle.
I understand how to do similar triangles,but an unsure of how to find the appropriate lengths in order to find the radius. I tried to use my triangle rules to no avail. How would I go about solving this? Thanks a lot
First, the angle ACB is a right angle by Thales theorem. Second, the angle QPA is the same as the angle labeled x (parallel lines) which is the same as the angle CBA (sum of angles of the triangle ABC is $180°$). Hence the triangles PQA and ABC are similar and we have $$\frac{AB}{AC}=\frac{PQ}{AQ}$$ Now let $r$ denote the radius of the circle, then this is equivalent to $$\frac{2r}{5}=\frac{15-r}{r}$$ This is a quadratic equation, only the solution $$r=-\frac{5}{4}+\sqrt{\left(\frac{5}{4}\right)^2+\frac{75}{2}}$$ is positive.
Edit: as remarked by Toby Mak, this simplifies to $r=5$.
Edit 2: I assumed that Q is the center of the circle. I don't think there is a unique solution otherwise.