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ayush94
You mentioned - AB = x + S + x

How did you assume that AP = QB = x??

AbhiroopGhosh


Refer triangle APU and triangle QBR

Angle APU = Angle BQR = 60
Angle PAU = Angle QBR = 90

Therefore
Angle PUA = Angle QRB = 30

As PU = QR (because its a regular hexagon), thus PA = QB.

PA = S/2 and QB = S/2 [PU = QR = S ; APU and BQR are 30-60-90 triangle]
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AbhiroopGhosh
When a regular hexagon is placed inside of a square, a 30 - 60 - 90 triangle is formed with the edge of the square.

In the diagram - angle QBR = 90 ; angle BQR = 60 therefore angle BRQ = 30

[ Note: angle BQR is the exterior angle to the regular hexagon, thus each angle is 360 / 6 = 60 ]

Let the each side of the regular hexagon be 'S'.

AB = AP + PQ + QB
AB = x + S + x
12 = 2x + S ----------------------- (1)

In triangle QBR

QR = S

Therefore QB = x = S / 2 ----------------------- (2)

Therefore

12 = 2 * \(\frac{S}{2}\) + S

12 = 2 * S

S = 6

IMO - B

So according to you S=6 so QR=6 and BR also equals 6? But then that should be an isosceles triangle, right?
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AbhiroopGhosh
When a regular hexagon is placed inside of a square, a 30 - 60 - 90 triangle is formed with the edge of the square.

In the diagram - angle QBR = 90 ; angle BQR = 60 therefore angle BRQ = 30

[ Note: angle BQR is the exterior angle to the regular hexagon, thus each angle is 360 / 6 = 60 ]

Let the each side of the regular hexagon be 'S'.

AB = AP + PQ + QB
AB = x + S + x
12 = 2x + S ----------------------- (1)

In triangle QBR

QR = S

Therefore QB = x = S / 2 ----------------------- (2)

Therefore

12 = 2 * \(\frac{S}{2}\) + S

12 = 2 * S

S = 6

IMO - B

So according to you S=6 so QR=6 and BR also equals 6? But then that should be an isosceles triangle, right?

No, BR would not be equal to 6, BR is opposite to 60 degrees in the triangle QBR.

QR = 6, then BR = 3\(\sqrt{3}\)
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