Can you find S T?

Geometry

[@robmaric9097]

Because lines OS and TQ are parallel, you can add them to get 7. OQ would be the hypotenuse. (7)^2 + b^2 = (14)^2. b=√149 =7√3.

[@Grizzly01-vr4pn]

What was the point of the circles?

[@Maths_physicshubs]

Is because of the circles that we applied transverse common tangent theorem. I know you are looking at it from similar triangles.

[@Grizzly01-vr4pn]

@@Maths_physicshubs Well, no. I just extended QT, then a perpendicular from SO to intersect that. You then have a right triangle hypotenuse OQ = 14, one leg = 5 + 2 = 7 and the other leg parallel and equal to ST. So Pythagoras gives you ST = 7√3 Looking into the transverse common tangent theorem, I can see how that would eventually lead you to the same place, but they still seem pretty superfluous to my mind.

[@Maths_physicshubs]

​@@Grizzly01-vr4pnok

[@dogeplayz0919]

to make those who can't see clearly in math blind