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

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 .