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Question Number 136597 by Dwaipayan Shikari last updated on 23/Mar/21

	Answered by mr W last updated on 23/Mar/21

	Commented by otchereabdullai@gmail.com last updated on 25/Mar/21

	$The Gifted prof W!$
	Commented by mr W last updated on 23/Mar/21

	$A_{0} A_{1} = A_{1} A_{2} = . . . = A_{6} A_{7} = a = 2 \times \sin \frac{π}{14}$ $A_{0} A_{7} = 2$ $A_{0} A_{6} = 2 \times \cos \frac{π}{14}$ $A_{0} A_{5} = 2 \times \cos \frac{2 π}{14}$ $A_{0} A_{4} = 2 \times \cos \frac{3 π}{14}$ $A_{0} A_{3} = 2 \times \cos \frac{4 π}{14}$ $A_{0} A_{2} = 2 \times \cos \frac{5 π}{14}$ $\frac{1}{2} r_{6} (A_{0} A_{7} + A_{0} A_{6} + A_{6} A_{7}) = \frac{1}{2} A_{0} A_{7} \times A_{0} A_{6} \times \sin \frac{π}{14}$ $r_{6} (1 + \cos \frac{π}{14} + \sin \frac{π}{14}) = 2 \cos \frac{π}{14} \sin \frac{π}{14}$ $r_{5} (\cos \frac{π}{14} + \cos \frac{2 π}{14} + \sin \frac{π}{14}) = 2 \cos \frac{2 π}{14} \cos \frac{π}{14} \sin \frac{π}{14}$ $r_{4} (\cos \frac{2 π}{14} + \cos \frac{3 π}{14} + \sin \frac{π}{14}) = 2 \cos \frac{3 π}{14} \cos \frac{2 π}{14} \sin \frac{π}{14}$ $r_{3} (\cos \frac{3 π}{14} + \cos \frac{4 π}{14} + \sin \frac{π}{14}) = 2 \cos \frac{4 π}{14} \cos \frac{3 π}{14} \sin \frac{π}{14}$ $r_{2} (\cos \frac{4 π}{14} + \cos \frac{5 π}{14} + \sin \frac{π}{14}) = 2 \cos \frac{5 π}{14} \cos \frac{4 π}{14} \sin \frac{π}{14}$ $r_{1} (\cos \frac{5 π}{14} + \cos \frac{6 π}{14} + \sin \frac{π}{14}) = 2 \cos \frac{6 π}{14} \cos \frac{5 π}{14} \sin \frac{π}{14}$ $s u m a r e a o f s m a l l c i r c l e s :$ $S = π \underset{n = 1}{\sum^{6}} r_{n}^{2} = 0.4076 5528 1334 5393$ $⌊ 10^{4} S ⌋ = 4076 5528 1334 53$ $s u m o f d i g i t s = 57$
	Commented by Dwaipayan Shikari last updated on 24/Mar/21

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