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If a regular hexagon is drawn in a square of side 12 cm, such that there is at least one vertex of the hexagon on each side of the square, what is the side of the regular hexagon that is drawn?
(A) \(4\sqrt{3}\) cm
(B) \(6\) cm
(C) \(6(\sqrt{6} - \sqrt{2})\) cm
(D) \(6(\sqrt{6}-\sqrt{3})\) cm
(E) \(6(\sqrt{7}-\sqrt{5})\)cm
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(A) \(4\sqrt{3}\) cm
(B) \(6\) cm
(C) \(6(\sqrt{6} - \sqrt{2})\) cm
(D) \(6(\sqrt{6}-\sqrt{3})\) cm
(E) \(6(\sqrt{7}-\sqrt{5})\)cm
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gmatophobia
Quant Chat Moderator
Joined: 22 Dec 2016
Last visit: 09 Sep 2026
Posts: 3,174
Location: India
Concentration: Strategy, Leadership
Posts: 3,174
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
_________________
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
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Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
gmatophobia
Quant Chat Moderator
Joined: 22 Dec 2016
Last visit: 09 Sep 2026
Posts: 3,174
Location: India
Concentration: Strategy, Leadership
Posts: 3,174
ayush94
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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Want to discuss quant questions and strategies : Join the quant chat group today
Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
Posts: 73
AbhiroopGhosh
So according to you S=6 so QR=6 and BR also equals 6? But then that should be an isosceles triangle, right?
gmatophobia
Quant Chat Moderator
Joined: 22 Dec 2016
Last visit: 09 Sep 2026
Posts: 3,174
Location: India
Concentration: Strategy, Leadership
Posts: 3,174
VelvetThunder
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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Want to discuss quant questions and strategies : Join the quant chat group today
Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
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