Which of the following could be the number of diagonals of an n-polygo

Question

Which of the following could be the number of diagonals of an n-polygon?

A. 3 
B. 6 
C. 8 
D. 12 
E. 20

thyagi · Answer

I think the question is asking the maximum number of diagonals a polygon can have.... An octagon has 20 diagonals, that's why the OA is E.... the next polygon is enneagon with 27 diagonals.... But the maximum number given in answer choices is 20... And that's why it's E ..

ankitahyd2013 · Answer

An 8 sided polygon will havde 20 diagonal
diagonal of a polygon=n(n-3)/2
put integers values of n from 4 to 8
only option E is possible
IMO-E

fskilnik · Answer

?\,\,\,:\,\,\,\# \,d\,\,\underline {{m{could}}\,\,{m{be}}} \,\,\,\,\left( {N \ge 3\,\,{\mathop{m int}} \,\,\left( * ight)\,,\,\,N{m{ - polygon}}} ight)

d\,\, 	o \,\,\,\left\{ \matrix{
 \,{m{each}}\,\,{m{vertex}}\,\,{m{with}}\,\,\left( {{m{not}}\,\,{m{itself}}} ight)\,\,{m{nor}}\,\,\left( {{m{next \,\, to}}\,\,{m{it}}\,\,{m{vertex}}} ight) \hfill \cr 
 \,\left \,\,{m{diagonal}}\,\,{m{is}}\,\,{m{the}}\,\,{m{same}}\,\,{m{of}}\,\,\left \,\,{m{diagonal}} \hfill \cr} ight.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,d = {{N\left( {N - 3} ight)} \over 2}

Alternate argument: (Explain!)

d = C\left( {N,2} ight) - N = {{N!} \over {2!\left( {N - 2} ight)!}} - N = {{N\left( {N - 1} ight)} \over 2} - {{2N} \over 2} = {{N\left( {N - 3} ight)} \over 2}

\left. \matrix{
 \left( A ight)\,\,\,{{N\left( {N - 3} ight)} \over 2} = 3\,\,\,\,\, \Rightarrow \,\,\,\,\,N\left( {N - 3} ight) = 6\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * ight)} \,\,\,\,\,{m{impossible}} \hfill \cr 
 \left( B ight)\,\,\,{{N\left( {N - 3} ight)} \over 2} = 6\,\,\,\,\, \Rightarrow \,\,\,\,\,N\left( {N - 3} ight) = 12\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * ight)} \,\,\,\,\,{m{impossible}} \hfill \cr 
 \left( C ight)\,\,\,{{N\left( {N - 3} ight)} \over 2} = 8\,\,\,\,\, \Rightarrow \,\,\,\,\,N\left( {N - 3} ight) = 16\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * ight)} \,\,\,\,\,{m{impossible}} \hfill \cr 
 \left( D ight)\,\,\,{{N\left( {N - 3} ight)} \over 2} = 12\,\,\,\,\, \Rightarrow \,\,\,\,\,N\left( {N - 3} ight) = 24\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * ight)} \,\,\,\,\,{m{impossible}}\,\,\, \hfill \cr} ight\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( E ight)\,\,{m{by}}\,\,{m{exclusion}}\,\,\,\,\left( {**} ight)

\left( {**} ight)\,\,\,\,\left( E ight)\,\,\,{{N\left( {N - 3} ight)} \over 2} = 20\,\,\,\,\, \Rightarrow \,\,\,\,\,N\left( {N - 3} ight) = 40\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * ight)} \,\,\,\,\,N = 8

We follow the notations and rationale taught in the  GMATH  method.

Regards, 
Fabio.

JS1290 · Answer

Number of diagonals can be found using n(n-3)/2 formula. Except E, all options fail to satisfy this equation.

MathRevolution · Answer

=>

The number of diagonals of an n-polygon is  n C 2  – n =  \frac{n(n-1)}{2} – n=\frac{n(n-3)}{2}. 
If  n = 4 , then the number of diagonals is  4 C 2  – 4 = 6 – 4 = 2.
If  n = 5 , then the number of diagonals is  5 C 2  – 5 = 10 – 5 = 5.
If  n = 6 , then the number of diagonals is  6 C 2  – 6 = 15 – 6 = 9.
If  n = 7 , then the number of diagonals is  7 C 2  – 7 = 21 – 7 = 14.
If  n = 8 , then the number of diagonals is  8 C 2  – 8 = 28 – 8 = 20.

Therefore, E is the answer.
Answer: E

Archit3110 · Answer

diagonal of polygon ; n * (n-3)/2
so solve using given options
say
20= n*(n-3)/2
we get = n= 8 ,-5
n = 8 is max possible value of n 
IMO E

Fdambro294 · Answer

If you know the formula for how to find the count of the diagonals in a polygon, you can arrive at the answer rather quickly.

Each vertex can make a diagonal with every other vertex except itself and the 2 adjacent vertices.

(N) (N - 3)

However, since we are counting from each N vertex, we will be double-counting each diagonal: once from one end, a second time from the other end. Thus, we should divide the answer by 2.

(N) (N - 3)
________
(2)

And 8 sided octagon has 20 diagonals

(8) (8 - 3) / (2) = (8) (5) / 2 = 20

E is the answer

Or you can look for the Pattern in the terms

A 4 sided quadrilateral ———-> 2 diagonals

5 sided pentagon ———-> 5 diagonals

6 sides ———> 9 diagonals

7 sides ———> 14 diagonals

By this point you can see the pattern emerging OR notice that all of the answer choices have been skipped over. All the remains is E - 20

The pattern: For each increase of +1 side, the difference between consecutive polygons goes up by + 1

5 - 2 = 3

9 - 5 = 4

14 - 9 = 5

So the next polygon after the 7 sided polygon will have 6 more diagonals

14 + 6 = 20

E is the answer

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Which of the following could be the number of diagonals of an n-polygo

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Which of the following could be the number of diagonals of an n-polygo

Prep Toolkit

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