Why can't I get the kite inscribed in a circle with a radius of 6.5, center O? Then I can connect O with B and get another right triangle and find the other leg. And also for the area, the kite area should be (6+6)(9+4)/2.
@alertsidewinder Good question! That's definitely a valid solution. You correctly found that the circle circumscribing the kite has diameter AC, so the circle has radius 13/2 = 6.5. Connecting O and B gives right triangle OPB with OP = 9-6.5 = 2.5, OB = 6.5. By Pythagoras, you can then find PB as 6 (maybe noticing that the side lengths are in a 5-12-13 ratio to minimize calculations) and you end up with the same answer as the official video's method.