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# Geometry Proof

Let the side length a regular hexagon be $$a$$. Prove that the area of this regular hexagon is $$\dfrac{3\sqrt3 a^2}2$$.

Note by Rohit Camfar
8 months ago

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@Dan Ley I am notifying you here because only one person is related with this discussion. I have to ask you something really important. Are you an IMO aspirant? · 5 months, 1 week ago

That's quite a question @Rohit Camfar , I have competed in the BMO but that's as far as I've gone, IMO is for the very best! What makes you ask? Besides, I have chosen to study Engineering not maths at University:) Are you an IMO aspirant by any chance? I have another question for you- is 14 your real age? · 5 months, 1 week ago

So, you are not interested in IMO. I asked you because I have created a group on Slack where we prepare for RMO (the way to IMO in India). I am searching for IMO aspirants around the world on Brilliant. Its Okay if you are not an IMO aspirant. · 5 months, 1 week ago

Unfortunately not but thanks for the offer! · 5 months, 1 week ago

Its Ok! But if you find any then pleaseee tell me ! · 5 months, 1 week ago

Comment deleted 5 months ago

Are you the same person? · 5 months, 1 week ago

Sorry for the confusion Yes! That is the account which I use on my mobile. · 5 months, 1 week ago

divide hexagon to 6 - equilateral triangles each with side length (a) by joining the center of hexagon to each vetices.

area of each equilateral triangle = sqrt3/4*a^2

area of hexagon = 6sqrt3a^2 /4 = 3sqrt3a^2 /2 · 8 months ago