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Question Number 173408 by mr W last updated on 11/Jul/22

Commented by mr W last updated on 11/Jul/22

$t h e a r e a s o f t h r e e p a r t s o f a r e g u l a r$ $h e x a g o n a r e g i v e n . f i n d t h e a r e a s o f$ $t h e o t h e r p a r t s .$
	Answered by aleks041103 last updated on 11/Jul/22

	$i f w e e x t e n d t h e s i d e s w e c a n f o r m 2$ $e q u l a t e r a l t r i a n g l e s . I n t h o s e, t h e s u m$ $o f t h e d i s t a n c e s f r o m a n y p o i n t i n s i d e$ $t o t h e s i d e s i s c o n s t a n t i n d e p e n d e n t o f$ $t h e c h o s e n p o i n t .$ $\Rightarrow S_{h e x} = 2 (3 + 4 + 5) = 2.12 = 24$ $T h e s u m o f t h e d i s t a n c e s f r o m a p o i n t$ $t o t w o o p p o s i t e s i d e s i s c o n s t a n t .$ $\Rightarrow 3 + x = 4 + y = 5 + z = S / 3 = 8$ $\Rightarrow x = 5, y = 4, z = 3$
	Commented by aleks041103 last updated on 11/Jul/22

	Commented by mr W last updated on 11/Jul/22

	$g r e a t! t h a n k s s i r!$
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