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Question Number 186976 by mr W last updated on 12/Feb/23

Commented by mr W last updated on 12/Feb/23

$t h e d i s t a n c e s f r o m a p o i n t t o t h r e e$ $v e r t e x e s o f a r e g u l a r h e x a g o n a r e$ $g i v e n . f i n d t h e d i s t a n c e s f r o m t h i s$ $p o i n t t o t h e o t h e r t h r e e v e r t e x e s .$
	Answered by mr W last updated on 12/Feb/23

	$a p p l y i n g t h e c o r r e l a t i o n b e t w e e n t h e$ $d i s t a n c e s w h i c h i f o u n d i n Q 186924 .$ $2 (y^{2} + 5^{2}) = 3 \times 8^{2} + 10^{2}$ $\Rightarrow y^{2} = 121 \Rightarrow y = \sqrt{121} = 11$ $2 (y^{2} + 10^{2}) = 3 z^{2} + 5^{2}$ $\Rightarrow z^{2} = \frac{2 (11^{2} + 10^{2}) - 5^{2}}{2} = 139 \Rightarrow z = \sqrt{139}$ $2 (5^{2} + 10^{2}) = 3 x^{2} + y^{2}$ $\Rightarrow x^{2} = \frac{2 (5^{2} + 10^{2}) - 11^{2}}{3} = 43 \Rightarrow x = \sqrt{43}$
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