@@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.
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.
What was the point of the circles?
Is because of the circles that we applied transverse common tangent theorem. I know you are looking at it from similar triangles.
@@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.
@@Grizzly01-vr4pnok
to make those who can't see clearly in math blind