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Question Number 206554 by cortano21 last updated on 18/Apr/24

	Answered by starsouf last updated on 18/Apr/24


	Answered by mr W last updated on 18/Apr/24

	Commented by mr W last updated on 18/Apr/24

	$S = a r e a o f h e x a g o n$ $x + (4 x - 3) = \frac{S}{3} = (2 x - 1) + (3 x - 2)$ $2 x - 1 + B = \frac{S}{3} \Rightarrow B = 3 x - 2$ $3 x - 2 + A = \frac{S}{3} \Rightarrow A = 2 x - 1$ $(4 x - 3) - x = \frac{a h}{2} \Rightarrow 6 x - 6 = a h$ $2 (3 x - 2) + (4 x - 3) - 2 (2 x - 1) - x = \frac{2 a h}{2}$ $\Rightarrow 5 x - 5 = a h$ $6 x - 6 = 5 x - 5$ $\Rightarrow x = 1 ✓$
	Commented by A5T last updated on 16/May/24

	$2 (3 x - 2) + (4 x - 3) - 2 (2 x - 1) - x = \frac{2 a h}{2}$ $I g u e s s t h i s s h o u l d b e :$ $2 (3 x - 2) + (4 x - 3) - 2 (2 x - 1) - x = 2 (\frac{2 a h}{2}) = 2 a h$ $\Rightarrow 5 x - 5 = 2 a h \Rightarrow 12 x - 12 = 5 x - 5$ $\Rightarrow x = 1$
	Commented by A5T last updated on 16/May/24

	$I u n d e r s t a n d t h a t t h e a r e a i s \frac{2 a \times h}{2} = a h .$ $B u t 2 (3 x - 2) + 4 x - 3 - 2 (2 x - 1) i s t w i c e o f t h a t .$ $I f o n e a p p l i e s t h i s m e t h o d t o y o u r l a t e s t q u e s t i o n,$ $o n e w o u l d g e t ? = 3 .$
	Commented by mr W last updated on 16/May/24

	$y o u a r e r i g h t . i t s h o u l d b e$ $2 (3 x - 2) + (4 x - 3) - 2 (2 x - 1) - x = 2 \times \frac{2 a h}{2}$
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