An hexagon of unit side is drawn on plane. Draw a square having the same area as the hexagon using only unmarked ruler and compass. What if an n−gon with unit edges is given? Is it always possible to draw a square of the same area as n−gon using ruler and compass.


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Question Number 3644 by prakash jain last updated on 17/Dec/15

$An hexagon of unit side is drawn on plane .$ $Draw a square having the same area as the$ $hexagon using only unmarked ruler and$ $compass .$ $What if an n - gon with unit edges is given ?$ $Is it always possible to draw a square$ $of the same area as n - gon using ruler$ $and compass .$
Commented by Rasheed Soomro last updated on 17/Dec/15

$A hexagon of unit side consists of 6$ $triangles of unit side .$ $Area of triangle of unit side (▴) :$ $▴ = \sqrt{\frac{3}{2} {(\frac{3}{2} - 1)}^{3}} = \sqrt{\frac{3}{2} \times \frac{1}{2^{3}}}$ $= \frac{\sqrt{3}}{4}$ $Area of hexagon of unit side (A) :$ $A = 6 ▴ = 6 \times \frac{\sqrt{3}}{4} = \frac{3 \sqrt{3}}{2} = \sqrt{\frac{27}{4}}$ $Square of area \frac{3 \sqrt{3}}{2} has its side (s) equal to$ $\sqrt{\sqrt{\frac{27}{4}}}$ $So ultimately we need to draw a line segment of$ $measure$ $\sqrt{\sqrt{\frac{27}{4}}} units$ $One way may be as under :$ $∙ Extend the side of hexagon 27 times . Say it AB$ $∙ Cut off AC = \frac{1}{4} AB .$ $∙ Cut off CD equal to hexagon side from AB .$ $Continue$
	Answered by Rasheed Soomro last updated on 19/Dec/15

	$* L e t A B i s a s i d e o f h e x a g o n m e a s u r i n g$ $1 u n i t .$ $∙ D r a w a l i n e l p e r p e n d i c u l a r t o A B t h r o u g h$ $A (o r B)$ $∙ C u t o f f A C = 6 A B, C D = A B o n o n e$ $s i d e o f A B a n d A E = A B o n t h e o t h e r$ $s i d e . (A C, C D a n d A E {d o n}^{'} t o v e r l a p$ $o n e a n o t h e r .)$ $∙ T a k e a p o i n t F o n C D s o t h a t C F = \frac{3}{4} C D$ $* A F = A C + C F = 6 + \frac{3}{4} = \frac{27}{4}$ $* E F = A F + E A = \frac{27}{4} + 1 .$ $∙ D r a w s e m i c i r c l e o n E F o n t h e s i d e o f B$ $∙ P r o d u c e A B t o m e e t t h e s e m i c i r c l e s a y a t G .$ $* A G = \sqrt{\frac{27}{4}} .$ $∙ P r o d u c e A B o n t h e s i d e o f A a n d c u t o f f$ $A H = A B$ $* H G = A G + A H = \sqrt{\frac{27}{4}} + 1$ $∙ D r a w s e m i c i r c l e o n H G . L e t t h e l i n e l m e e t s$ $t h e s e m i c i r c l e a t I .$ $* A I = \sqrt{\sqrt{\frac{27}{4}}}$ $∙ T a k i n g A I a s s i d e o f t h e s q u a r e c o m p l e t t h e$ $s q u a r e .$ $* A r e a o f t h e s q u a r e w i l l b e {(\sqrt{\sqrt{\frac{27}{4}}})}^{2} = \sqrt{\frac{27}{4}}$ $w h i c h i s a l s o t h e a r e a o f t h e g i v e n h e x a g o n .$ $S e e c o m m e n t a l s o .$
	Commented by Rasheed Soomro last updated on 17/Dec/15

	$n - g o n c a n b e squared o n l y w h e n \sin \frac{2 π}{n} c a n$ $b e e x p r e s s e d a s r a t i o n a l s o r s u r d s c o n t a i n i n g$ $o n l y s q u a r e r o o t s .$
	Commented by prakash jain last updated on 17/Dec/15

	$It is correct that not all n - gon can be squared .$
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