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Question Number 193908 by Rupesh123 last updated on 22/Jun/23

	Answered by Subhi last updated on 22/Jun/23

	$p u t e a c h s i d e = l$ $B H = A G = \frac{l}{2}$ $G \hat{A K} = M \hat{B H} = 120 - 90 = 30$ $H M = G K = \frac{l}{4} (o p p o s i t e t o a n g l e 30 i n a r i g h t t r i a n g l e)$ $G H = l + \frac{1}{2} l = \frac{3}{2} l$ $i n Δ G F E :$ $E G = \sqrt{l^{2} + \frac{1}{4} l^{2} - l^{2} c o s (120)} = \frac{\sqrt{7}}{2} l (c o s l a w)$ $\frac{\sqrt{7} l}{2 s i n (120)} = \frac{l}{s i n (x)} ⇛ x = {s i n}^{- 1} (\frac{\sqrt{21}}{7}) = 40.89$ $A B / / G H ⇛ E \hat{G H} = \hat{A} = 120$ $E \hat{G H} = 120 - x = 79.1$ $9 = [R G H] = \frac{1}{2} . \frac{\sqrt{7}}{2} . \frac{3}{2} l^{2} . s i n (79.1)$ $l = 3.039$ $A r e a o f r e g u l a r h e x a g o n = \frac{3 \sqrt{3}}{2} l^{2} = 24$
	Commented by Subhi last updated on 22/Jun/23

	Commented by Rupesh123 last updated on 22/Jun/23
	Nice solution, sir!
	Answered by talminator2856792 last updated on 23/Jun/23

	$hexagon side length a :$ $using 30, 60, 90 triangle arrive at :$ $9 = \frac{3}{4} \times \frac{2 a}{\sqrt{3}} \times \frac{3 a}{2} \div 2$ $a^{2} = 8 \sqrt{3}$ $regular hexagon can be split into six equilateral$ $triangles all with side lengths equal to a .$ $\Rightarrow 6 (\frac{a \times \frac{a}{\sqrt{3}}}{2})$ $= \sqrt{3} a^{2}$ $but a^{2} is already known .$ $\Rightarrow \sqrt{3} a^{2} = \sqrt{3} (8 \sqrt{3})$ $= 24$
	Commented by Rupesh123 last updated on 23/Jun/23
	Perfect ��
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