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3 votes
3 votes

A $diagonal$ in a polygon is a straight line segment that connects two non-adjacent vertices, and is contained in the interior of the polygon (except for its points). Two such diagonals are said to cross if they have a point in common in the interior of the polygon. In one such polygon with $n$ vertices, a certain number (say $k$) of  non-crossing diagonals were drawn to cut up the inside of the polygon into regions, each of which was a quadrilateral. how many diagonals were drawn, that is, what is $k$?

  1. cannot be determined from the information given
  2. $\frac{n}{2}-2$
  3. $\frac{n}{4}-1$
  4. $n-4$
  5. $n^2 - 9.5 n +22$
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3 Answers

4 votes
4 votes

We have drawn k diagonals on a n-sided polygon.

Now, total edges = n(sides)+k(diagonals) = n+k

For each diagonal we draw, one extra region will be formed. Initially, there are 2 regions (entire region inside polygon and the external region)

total regions = k+2

Now, using the law, sum of degrees of all regions = 2*no of edges.

Here, r-1 regions will have degree of 4 (all the regions formed inside polygon) and the external region has degree (n+k) as it is bounded by (n) edges (sides of the polygon).

So, 4*(r-1)+(n)*1=2*e
      4*(k+2-1)+(n)*1=2*(n+k)
      4k+4+n=2n+2k
k=(n-4)/2
Option B is correct

1 votes
1 votes
Each region is quadrilateral i.e bounded by 4 sides.

 

So we can write 4r<=2e

 

A n-gon has n sides. Also K diagonals are drawn, so total sides will be n+k.

 

r=n+k-n+2=k+2

 

Therefore, 4(k+2)<=2(n+k)

=> k<=n-4
0 votes
0 votes

Way to approach :- Let suppose our polygon contains 'n' vertices. Then we can observe that minimum number of vertices must be 5 , to have at least on such diagonal that cut the polygon into at least one quadrilateral region.

Now to find total number of such diagonals , we can follow this approach

match Vn to V3 , match Vn-1 to V4 , match Vn-2 to V5 ......match Vn-(x-3) to Vx

When we follow this approach we will come to know that when our 'n' is even there would be 4 vertices that can't be map in that manner and when 'n' is odd there would be 3 such vertices.

so that's the overall approach

attaching image.

Answer:

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