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Question Number 67514 by mr W last updated on 28/Aug/19

Commented by mr W last updated on 28/Aug/19

$s i d e l e n g t h o f r e g u l a r h e x a g o n i s 1 .$ $f i n d t h e y e l l o w a r e a = ?$
	Answered by mr W last updated on 29/Aug/19

	$s i d e l e n g t h o f b i g r e d e q u i l a t e r a l : \frac{3}{2}$ $a r e a o f b i g r e d e q u i l a t e r a l : \frac{\sqrt{3}}{4} {(\frac{3}{2})}^{2} = \frac{9 \sqrt{3}}{16}$ $a r e a o f e a c h s m a l l r e d t r i a n g l e :$ $\frac{1}{2} \times {(\frac{1}{2})}^{2} \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{32}$ $y e l l o w a r e a = b i g r e d - 3 \times s m a l l r e d$ $= \frac{9 \sqrt{3}}{16} - \frac{3 \sqrt{3}}{32} = \frac{15 \sqrt{3}}{32}$
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