<![CDATA[Module 2 Day 5 Your Turn Part 1]]>https://forum.poshenloh.com/category/605RSS for NodeMon, 25 Sep 2023 05:39:40 GMTTue, 02 Feb 2021 16:44:29 GMT60<![CDATA[Cyclic Quadrilateral]]>@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.
Great job! ðŸ™‚
]]>https://forum.poshenloh.com/topic/736/cyclic-quadrilateralhttps://forum.poshenloh.com/topic/736/cyclic-quadrilateralTue, 02 Feb 2021 16:44:29 GMT<![CDATA[Sometimes special effects can be a bit creepy...]]>It kind of looks like the Nazi symbol
]]>https://forum.poshenloh.com/topic/496/sometimes-special-effects-can-be-a-bit-creepyhttps://forum.poshenloh.com/topic/496/sometimes-special-effects-can-be-a-bit-creepyFri, 25 Sep 2020 04:54:52 GMT