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Question Number 207292 by mr W last updated on 10/May/24

Commented by mr W last updated on 10/May/24

$f i n d t h e s i d e l e n g t h o f t h e r e g u l a r$ $h e x a g o n .$
	Answered by A5T last updated on 10/May/24

	Commented by A5T last updated on 10/May/24

	$D C = A C = s \sqrt{3} \Rightarrow A B = s \sqrt{3} - s = s (\sqrt{3} - 1)$ $\frac{s i n 45 °}{2} = \frac{s i n 75 °}{s (\sqrt{3} - 1)} \Rightarrow s = 2 + \sqrt{3}$
	Commented by mr W last updated on 11/May/24
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	Answered by mr W last updated on 11/May/24

	Commented by mr W last updated on 11/May/24

	$x = \sqrt{3} a - a$ $\frac{2 a - 2}{2} = \frac{2 a}{\sqrt{3} a - a}$ $\Rightarrow a = 2 + \sqrt{3}$
	Answered by mr W last updated on 11/May/24

	Commented by mr W last updated on 11/May/24

	$\frac{y}{2} = \frac{A B}{A C} = \frac{\sqrt{3} a}{2 a} \Rightarrow y = \sqrt{3}$ $a = 2 + \sqrt{3}$
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