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Question Number 215195 by mr W last updated on 31/Dec/24

Commented by mr W last updated on 31/Dec/24

$f i n d t h e s h a d e d a r e a = ?$
	Answered by mr W last updated on 02/Jan/25

	Commented by mr W last updated on 31/Dec/24

	Commented by mr W last updated on 02/Jan/25

	$R = d + r + 2 r_{1} = 4 + d$ $λ = \frac{R^{2} + r^{2} - d^{2}}{2 R r} = \frac{5 + 2 d}{4 + d}$ $\cosh^{- 1} λ = 2 \ln [\tan (\frac{π}{4} + \frac{π}{2 \times 6})] = \ln 3$ $λ + \sqrt{λ^{2} - 1} = 3$ $λ^{2} - 1 = {(3 - λ)}^{2} = 9 - 6 λ + λ^{2}$ $\Rightarrow λ = \frac{5}{3} = \frac{5 + 2 d}{4 + d}$ $\Rightarrow d = 5$ $\Rightarrow R = 9$ $r_{4} + r + r_{1} = R \Rightarrow r_{4} = 6$ $\frac{d^{2} + {(r + r_{2})}^{2} - {(R - r_{2})}^{2}}{2 d (r + r_{2})} = - \frac{{(r + r_{2})}^{2} + {(r + r_{1})}^{2} - {(r_{1} + r_{2})}^{2}}{2 (r + r_{2}) (r + r_{1})}$ $\Rightarrow r_{2} = \frac{24}{19}$ $\frac{d^{2} + {(R - r_{3})}^{2} - {(r + r_{3})}^{2}}{2 d (R - r_{3})} = - \frac{{(R - r_{3})}^{2} + {(R - r_{4})}^{2} - {(r_{4} + r_{3})}^{2}}{2 (R - r_{3}) (R - r_{4})}$ $\Rightarrow r_{3} = \frac{8}{3}$ $\cos γ = \frac{{(R - r_{4})}^{2} + {(R - r_{3})}^{2} - {(r_{4} + r_{3})}^{2}}{2 (R - r_{4}) (R - r_{3})}$ $= - \frac{13}{19}$ $\Rightarrow \sin γ = \frac{8 \sqrt{3}}{19} \Rightarrow γ = π - \sin^{- 1} \frac{8 \sqrt{3}}{19}$ $\sin α = \frac{(R - r_{3}) \sin γ}{r_{4} + r_{3}} = \frac{(9 - \frac{8}{3})}{6 + \frac{8}{3}} \times \frac{8 \sqrt{3}}{19} = \frac{4 \sqrt{3}}{13}$ $\sin β = \frac{(R - r_{4}) \sin γ}{r_{4} + r_{3}} = \frac{(9 - 6)}{6 + \frac{8}{3}} \times \frac{8 \sqrt{3}}{19} = \frac{36 \sqrt{3}}{247}$ $A_{s h a d e} = \frac{1}{2} [γ R^{2} - (π - α) r_{4}^{2} - (π - β) r_{3}^{2} - (R - r_{4}) (R - r_{3}) \sin γ]$ $= \frac{1}{2} [9^{2} (π - \sin^{- 1} \frac{8 \sqrt{3}}{19}) - 6^{2} (π - \sin^{- 1} \frac{4 \sqrt{3}}{13}) - {(\frac{8}{3})}^{2} (π - \sin^{- 1} \frac{36 \sqrt{3}}{247}) - (9 - 6) (9 - \frac{8}{3}) \frac{8 \sqrt{3}}{19}]$ $= \frac{341 π}{18} - 4 \sqrt{3} - \frac{81}{2} \sin^{- 1} \frac{8 \sqrt{3}}{19} + 18 \sin^{- 1} \frac{4 \sqrt{3}}{13} + \frac{32}{9} \sin^{- 1} \frac{36 \sqrt{3}}{247}$ $\approx 30.512513235$
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