The Best geometry skills can be learned on your channel. THANK YOU
@PreMath Жыл бұрын
Glad you think so! ❤️ Thanks for your feedback! Cheers! 😀 You are awesome. Keep it up 👍
@loneranger6361 Жыл бұрын
Dear Sir. Thank you so much for your excellent explanation. Thank you you for catering for all students with respect to your excellent manner of teaching
@jairam4459 Жыл бұрын
Please elaborate two tangents theorem with clarity. Thanks
@AmirgabYT218510 ай бұрын
S=27(10-3π)≈15,66
@JSSTyger Жыл бұрын
The 2:58 mark gave me the info I needed. :) Area = 270-81π
@quigonkenny9 ай бұрын
By two tangent theorem, for any two tangents of a circle that converge to the same point, those tangents are the same length. Therefore, PB = AB = 9, and PD = ED = 6, and BD = 9+6 = 15. Triangle ∆DCB: CB² + DC² = BD² (r-9)² + (r-6)² = 15² r² - 18r + 81 + r² - 12r + 36 = 225 2r² - 30r - 108 = 0 r² - 15r - 54 = 0 r² + 3r - 18r - 54 = 0 r(r+3) - 18(r+3) = 0 (r+3)(r-18) = 0 r + 3 = 0 | r - 18 = 0 r = -3 ❌ | r = 18 ✓ r > 0 A = bh/2 = (18-9)(18-6)/2 A = 9(12)/2 = 9(6) = 54 Quarter circle O: A = πr²/4 = π(18²)/4 = 9²π = 81π Green region: A = r² - 81π - 54 A = 18² - 54 - 81π A = 324 - 54 - 81π A = 270 - 81π = 27(10-3π) ≈ 15.53
@arnavkange1487 Жыл бұрын
nice sum enjoyed it
@PreMath Жыл бұрын
Glad you enjoyed Thanks for your feedback! Cheers! 😀
@KAvi_YA666 Жыл бұрын
Thanks for video.Good luck sir!!!!!!!!!!
@PreMath Жыл бұрын
So nice of you ❤️ Thank you! Cheers! 😀
@ybodoN Жыл бұрын
Since OB bisects ∠AOP and OD bisect ∠EOP ⇒ ∠BOD = π / 4. So we can set the equation _arctan (9 / r) + arctan (6 / r) = π / 4._ It will lead to another quadratic equation: r⁴ − 333r² + 2916 = 0. Solutions are r = −18, r = −3, r = 3 and r = 18 (last one accepted).