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In a hexagon, the total number of diagonals that can be draw
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07 Jan 2021, 09:47
07 Jan 2021, 09:47
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Question Stats:
71% (00:30) correct
28% (01:16) wrong based on 7 sessions
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In a hexagon, the total number of diagonals that can be drawn exceeds the total number of sides by:
A. 1
B. 2
C. 3
D. 4
E. 5
Kudos for the right answer and explanation
A. 1
B. 2
C. 3
D. 4
E. 5
Kudos for the right answer and explanation
Re: In a hexagon, the total number of diagonals that can be draw
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08 Jan 2021, 07:51
08 Jan 2021, 07:51
1
There is a formula to find the no of diagonals in geometry.
No of diagonals = nC2 - n , where n = no of sides
Here, hexagon has 6 sides.
No of diagonals are = 6C2 - 6 = \(\frac{6*5}{2}\) - 6
= 15 - 6 = 9
Now diagonals that can be drawn exceeds the total number of sides by = 9 - 6 = 3
Answer C
No of diagonals = nC2 - n , where n = no of sides
Here, hexagon has 6 sides.
No of diagonals are = 6C2 - 6 = \(\frac{6*5}{2}\) - 6
= 15 - 6 = 9
Now diagonals that can be drawn exceeds the total number of sides by = 9 - 6 = 3
Answer C
gmatclubot
Re: In a hexagon, the total number of diagonals that can be draw [#permalink]
08 Jan 2021, 07:51
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