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08 Dec 2021, 05:15
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
19% (02:06) correct
81%
(02:40)
wrong
based on 16
sessions
History
Date
Time
Result
Not Attempted Yet
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
Are You Up For the Challenge: 700 Level Questions
(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
Are You Up For the Challenge: 700 Level Questions
08 Dec 2021, 05:59
Kudos
Bookmarks
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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