Well guys I searched if this has been posted before, and didnt find it. Btw whats the best way to search if a question has been posted before or not. I simply put the whole question in the search, is there a smarter way I am missing?
Anway, here's the question
A Polygon has 20 diagonals, how many sides does it have.
A 12
B 11
C 10
D 9
E 8
Now by taking general cases I was trying to solve and I came up with this equation where No. of diagonals of a polygon with n sides = (n-3)n/2
Its applicable here, but i want to know if its general, and if there is any other shorter way to solve the problem
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Polygon diagonals formula
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tohellandback
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well this is how I solved:
0 diagonals-3 sides
2 diagonals-4 sides
5 diagonals-5sides
9 diagonals-6sides
you see 0+2=2,2+3=5,5+4=9. so next will be 9+5=14, 14+6=20
20 diagonals will have 8 sides
0 diagonals-3 sides
2 diagonals-4 sides
5 diagonals-5sides
9 diagonals-6sides
you see 0+2=2,2+3=5,5+4=9. so next will be 9+5=14, 14+6=20
20 diagonals will have 8 sides
The powers of two are bloody impolite!!
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Naruto
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I didnt understand please elaborate what series are u continuing?tohellandback wrote:well this is how I solved:
0 diagonals-3 sides
2 diagonals-4 sides
5 diagonals-5sides
9 diagonals-6sides
you see 0+2=2,2+3=5,5+4=9. so next will be 9+5=14, 14+6=20
20 diagonals will have 8 sides
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tohellandback
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0-2 (add 0 to 2)
2-5(added 3)
5-9 (added 4)
so next you will add 5
9-14(added 5)
14-20(added six)
2-5(added 3)
5-9 (added 4)
so next you will add 5
9-14(added 5)
14-20(added six)
The powers of two are bloody impolite!!
A Polygon has 20 diagonals, how many sides does it have.Naruto wrote:Well guys I searched if this has been posted before, and didnt find it. Btw whats the best way to search if a question has been posted before or not. I simply put the whole question in the search, is there a smarter way I am missing?
Anway, here's the question
Now by taking general cases I was trying to solve and I came up with this equation where No. of diagonals of a polygon with n sides = (n-3)n/2
Its applicable here, but i want to know if its general, and if there is any other shorter way to solve the problem
A 12
B 11
C 10
D 9
E 8
What is a shorter way? What can be shorter and surer than factoring
n^2 -3n -40=0
(n-8)(n+5)=0
n=8.
With formula this takes less than 15 seconds. There is no shorter way order than already knowing the answer!!
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Naruto
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You misunderstood me, I meant is this formula genuine because i derived it based on eg. for triangel, quadrilateral, pentagon and hexagon. I wanted to know if there is another approach, because deriving that formula took me 4 mins.dtweah wrote:A Polygon has 20 diagonals, how many sides does it have.Naruto wrote:Well guys I searched if this has been posted before, and didnt find it. Btw whats the best way to search if a question has been posted before or not. I simply put the whole question in the search, is there a smarter way I am missing?
Anway, here's the question
Now by taking general cases I was trying to solve and I came up with this equation where No. of diagonals of a polygon with n sides = (n-3)n/2
Its applicable here, but i want to know if its general, and if there is any other shorter way to solve the problem
A 12
B 11
C 10
D 9
E 8
What is a shorter way? What can be shorter and surer than factoring
n^2 -3n -40=0
(n-8)(n+5)=0
n=8.
With formula this takes less than 15 seconds. There is no shorter way order than already knowing the answer!!
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Vemuri
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I solved the question the same way as tohellandback did. But, your question is valid. Even I am curious to know if there is a relationship between sides & diagonals of a polygon. It will greatly help in answering the questions quickly.
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scoobydooby
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there indeed is a general formula to find out the number of sides given the number of diagonals. (dug out my high school math book)
the number of diagonals of a polygon of n sides is: nC2-n
the number of diagonals of a polygon of n sides is: nC2-n
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rah_pandey
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A formal proof of n^2-3n-40=0
for a n sided polygon lets choose one point say A
now there are n-1 points except A itself from which lines can be drawn to A but not all will be diagonals. 2 such lines that can be drawn are the sides passing from A and joining adjoining points
thus effectively n-3 points can be joined to form a diagonal from A thus n-3 diagonals are possible from any given point
for n points total no of such lines =n*(n-3)
but of these n*(n-3) there is overlap(by 2 times)
thus total no of diagonal =n*(n-3)/2
using for 20 diagonals
we get n(n-3)/2=20
=> n=8 or n=-5
for a n sided polygon lets choose one point say A
now there are n-1 points except A itself from which lines can be drawn to A but not all will be diagonals. 2 such lines that can be drawn are the sides passing from A and joining adjoining points
thus effectively n-3 points can be joined to form a diagonal from A thus n-3 diagonals are possible from any given point
for n points total no of such lines =n*(n-3)
but of these n*(n-3) there is overlap(by 2 times)
thus total no of diagonal =n*(n-3)/2
using for 20 diagonals
we get n(n-3)/2=20
=> n=8 or n=-5
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Vemuri
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That's great Scoobydooby !!! Thanks a ton for providing the formula. Here goes my "thanks" to youscoobydooby wrote:there indeed is a general formula to find out the number of sides given the number of diagonals. (dug out my high school math book)
the number of diagonals of a polygon of n sides is: nC2-n
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Ian Stewart
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There are a few good ways to prove the formula, and rah_pandey provided one above. An alternative proof - first starting with a simpler question:rah_pandey wrote:A formal proof of n^2-3n-40=0
* If you draw n dots on a page, how many different lines can be drawn connecting pairs of these dots? This is the same question as: If n people attend a party and all shake hands, how many handshakes take place? Since we need to choose 2 dots to make a line, and order doesn't matter, the answer is nC2.
* Now, looking at the n corners of an n-sided shape, we can draw nC2 different lines connecting pairs of these corners. All of these lines are diagonals *except* the n edges, so we have nC2 - n diagonals. That might look different from the formula in the post above, but it turns out to be equal if you expand.
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