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
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).
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S=27(10-3π)≈15,66
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
Please elaborate two tangents theorem with clarity.
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The 2:58 mark gave me the info I needed. :)
Area = 270-81π
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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
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).
(r-9)^2+(r-6)^2=(9+6)^2...r=18...Ag=18^2-(π18^2/4+9*12/2)=270-81π
If the blue triangle hadn't been shown, I don't think I would have figured it out
Nice! green shaded region = 27(10 - 3π)
Green area=(18)^2-1/4π(18)^2=1/2(9)(12)
=324--81π-54=270-81π
109.63 sq units
cut out square, weigh, cut out geen part, weigh, subtract green weight from square, convert, done...
Elementary my dear Doyen! Not really, but I just thought I would would say so.
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Gracias profesor, bonito ejemplo.
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(R-9)²+(R-6)²=(9+6)²
(R²-18R+81)+(R²-12R+36)=15²
2R²-30R+117=225
R²-15R-54=0
R = 18 cm
A = R.6 + R.9 - ¼π R²
A = 15R - ¼πR²
A = 15,53 cm² ( Solved √ )
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asnwer=13cm isit
Please, no need to go over every single step. You can skip many obvious steps.