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polygon with 18 sides

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polygon with 18 sides

by arora007 » Fri Jul 30, 2010 8:51 am
How many diagonals does a polygon with 18 sides have if three of its vertices do not send any diagonal?

a) 90
b) 126
c) 210
d) 264
e) 306

[spoiler] Hint... n(n-3)/2 gives the formula for number of diagonals with polygon of n sides... [/spoiler]
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by nithi_mystics » Fri Jul 30, 2010 9:34 am
Is it 90?

No of diagonals of a polygon = n(n-3)/2
where n is the no of vertices.

here n =15 (since 3 vertices do not send any diagonals)

15*12/2 = 90
Thanks
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by arora007 » Fri Jul 30, 2010 9:55 am
you are right!
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by sumanr84 » Fri Jul 30, 2010 10:23 am
arora007 wrote:How many diagonals does a polygon with 18 sides have if three of its vertices do not send any diagonal?

a) 90
b) 126
c) 210
d) 264
e) 306

[spoiler] Hint... n(n-3)/2 gives the formula for number of diagonals with polygon of n sides... [/spoiler]
Arora bhai..you scared me with this new formulas but good to know before test..
I am on a break !!
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by arora007 » Fri Jul 30, 2010 10:29 am
even i got scared... then wrote down the logic... its actually pretty simple...

"A vertex can form diagonals with all vertices except
|1| itself,
|2|3| vertices adjacent to it.
So we get the (n-3) ,
but a diagonal is made with a pair of points... so divide by to 2 to avoid duplication

thus the formula n(n-3)/2"
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by sumanr84 » Fri Jul 30, 2010 10:58 am
I, infact, used some other technique to work around it.

We have 15 points. Obviously, we cannot have diagonals with 2 neighboring points. So, we have 12 points with which a single point can form a diagonal.

Each of these 15 points can form 12 diagonals. So, 12 * 15 = 180 diagonals

But, we know that a diagonal from x-->y is same as from y-->x.

So, divide 180 / 2 = 90 ,

But formula is real simple to apply..thanks
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