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Hexagon

by goyalsau » Sun Nov 21, 2010 9:29 pm
In Hexagons has 9 Diagonals

Things doubt me are these.

But why it will have 3 diagonals of length 3a ( a = side of diagonal )
6 diagonals sqrt 3 a .

I am not able to understand why it that so, and Area of Hexagon will be formed by 6 equilateral triangles.

Area of Hexagon (3 sqrt 3 a ^2) / 2

Guys Can anybody explain why is that so...
Saurabh Goyal
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by shovan85 » Sun Nov 21, 2010 10:06 pm
goyalsau wrote:In Hexagons has 9 Diagonals
Total number of sides = 6

Total number of diagonals = 6C2 - 6 = 15 - 6 = 9
goyalsau wrote: Things doubt me are these.

But why it will have 3 diagonals of length 3a ( a = side of diagonal )
See the RED diagonals below. These are not 3a but 2a.

Proof:

In the diagram,

Length of XY = a

In triangle CBX, CB = a and angle CBX = 60.

So by using 30-60-90 triangle rule CX = a/2

Thus EY = a/2

Hence the diagonal BE = BX + XY + YE = a/2 + a + a/2 = 2a
goyalsau wrote:
6 diagonals sqrt 3 a .
I am not able to understand why it that so, and
See the Green diagonals in the below image. These diagonals are = SQRT(3)*a

Proof:

In triangle DAF,

DA = 2a and AF = a

Thus by Pythagorean Rule,

DF = (DA^2 - AF^2) ^ (1/2) = [ (2a) ^ 2 - (a^2) ] ^ (1/2) = SQRT(3)*a
goyalsau wrote: Area of Hexagon will be formed by 6 equilateral triangles.

Area of Hexagon (3 sqrt 3 a ^2) / 2

Guys Can anybody explain why is that so...
See the triangle AOF in the diagram.

AF = a.

AD is a diagonal of length 2a (RED) => AO = a

CF is a diagonal of length 2a (RED) => CO = a

So all sides are equal => Equilateral Triangle.

Such 6 triangles comprise to the proper hexagon. Thus, Area of Hexagon = 6 * [sqrt(3) * a^2]/4 = (3 sqrt 3 a ^2) / 2
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by goyalsau » Sun Nov 21, 2010 10:30 pm
shovan85 wrote:
goyalsau wrote:In Hexagons has 9 Diagonals
Total number of sides = 6

Total number of diagonals = 6C2 - 6 = 15 - 6 = 9
goyalsau wrote: Things doubt me are these.

But why it will have 3 diagonals of length 3a ( a = side of diagonal )
See the RED diagonals below. These are not 3a but 2a.

Proof:

In the diagram,

Length of XY = a

In triangle CBX, CB = a and angle CBX = 60.

So by using 30-60-90 triangle rule CX = a/2 { it will not be CX it will be BX }

Thus EY = a/2

Hence the diagonal BE = BX + XY + YE = a/2 + a + a/2 = 2a
goyalsau wrote:
6 diagonals sqrt 3 a .
I am not able to understand why it that so, and
See the Green diagonals in the below image. These diagonals are = SQRT(3)*a

Proof:

In triangle DAF,

DA = 2a and AF = a

Thus by Pythagorean Rule,

DF = (DA^2 - AF^2) ^ (1/2) = [ (2a) ^ 2 - (a^2) ] ^ (1/2) = SQRT(3)*a
goyalsau wrote: Area of Hexagon will be formed by 6 equilateral triangles.

Area of Hexagon (3 sqrt 3 a ^2) / 2

Guys Can anybody explain why is that so...
See the triangle AOF in the diagram.

AF = a.

AD is a diagonal of length 2a (RED) => AO = a

CF is a diagonal of length 2a (RED) => CO = a

So all sides are equal => Equilateral Triangle.

Such 6 triangles comprise to the proper hexagon. Thus, Area of Hexagon = 6 * [sqrt(3) * a^2]/4 = (3 sqrt 3 a ^2) / 2
Thanks , thanks................... thanks........................ infinite thanks.............
Saurabh Goyal
[email protected]
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by diebeatsthegmat » Thu Nov 25, 2010 8:48 am
goyalsau wrote:In Hexagons has 9 Diagonals

Things doubt me are these.

But why it will have 3 diagonals of length 3a ( a = side of diagonal )
6 diagonals sqrt 3 a .

I am not able to understand why it that so, and Area of Hexagon will be formed by 6 equilateral triangles.

Area of Hexagon (3 sqrt 3 a ^2) / 2

Guys Can anybody explain why is that so...
]

hey baby, what source you are studying? can you tell me from what source the geometric math you posted was? because its so time consuming :(
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