Inscribing a circle, a semicircle, and a quarter circle in a semicircle. A geometry puzzle.

[@SyberMath]

I really like this puzzle because it brings together a circle, a semicircle and a quarter circle! The values are not too hard to find and the numbers are not that complicated. Sorry about the length of the video. I had to work through many equations and I never imagined it would take this long.

[@ashishpradhan9606]

This question is a bit explanatory. So it's worth to take this much time you give it.

[@ashishpradhan9606]

By the way I enjoy a lot 🙂🙂🙂.

[@SyberMath]

Thank you! I enjoyed solving it. 😊😊😊

[@srijanbhowmick9570]

I did exactly the same as you until 6:35 , after that I used Stewart's theorem , because we can find out the lengths of the divided parts of the bottom side of the triangle . They are 15/4 & 5/4 respectively . Also it is much simpler .

[@SyberMath]

Nice! I like it!

[@ronaldovieirafragosoronald2080]

more simple by cosine law triangles 2+r, 3.75 e 4.25 e 2+r, 5 e 3+r , a (angle between 2+r 3,75 and between 2+r 5)

[@barakathaider6333]

👍

[@tarekraja7537]

@syberMath. Cool ! That was very clever to find first 'R'. Thank you very much sir.

[@SyberMath]

Thank you!!!

[@yusufdenli9363]

Başka bir yoldan da yapılabilir.
Çeyrek çemberin olduğu yerde çapı gören çevre açı çizilerek öklit uygulanır. En büyük yarım çemberin çapı bulunabilir. Daha sonra tüm çemberlerin merkezleri birleştirilerek oluşan üçgende stewart teoremi uygulanarak, r bulunabilir.

[@leonhardeuler5211]

This is nice! Thank you, SyberMath.

[@SyberMath]

You're very welcome! I'm glad you're enjoying it. 😊

[@leonhardeuler5211]

@@SyberMath I love it!!!