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Number of quadrilatera in polygon

jacks

Well-known member
Apr 5, 2012
226
Number of Quadrilateral that can be made using the vertex of a polygon of 10 sides as

there vertices and having Exactly $2$ sides common with the polygon

My solution:

first we will take $2$ sides common is $=10$ ways now if we take two sides as $A_{1}A_{2}$ and $A_{1}A_{3}$ then we

will not take vertex $A_{4}$ and $A_{10}$ So we will choose $1$ vertes from $5$ vertices which can be done by

$\displaystyle \binom{5}{1}$ ways So Total No. of ways in which $2$ sides common is $=10 \times \displaystyle\binom{5}{1} =50$ but answer given

is $=75$.

But I did not understand it. Would anyone like to explain me where i am wrong

Thanks
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Re: number of quadrilatera in polygon

Number of Quadrilateral that can be made using the vertex of a polygon of 10 sides as

there vertices and having Exactly $2$ sides common with the polygon

My solution:

first we will take $2$ sides common is $=10$ ways now if we take two sides as $A_{1}A_{2}$ and $A_{1}A_{3}$ then we

will not take vertex $A_{4}$ and $A_{10}$ So we will choose $1$ vertes from $5$ vertices which can be done by

$\displaystyle \binom{5}{1}$ ways So Total No. of ways in which $2$ sides common is $=10 \times \displaystyle\binom{5}{1} =50$ but answer given

is $=75$.

But I did not understand it. Would anyone like to explain me where i am wrong

Thanks
You can also have a quadrilateral $A_1A_2A_4A_5$. It too has exactly $2$ of its sides common with the polygon. There, of course, many more such quadrilaterals.
 

soroban

Well-known member
Feb 2, 2012
409
Hello, jacks!

I think I've got it . . .


Number of quadrilateral that can be made using the vertices
of a polygon of 10 sides as their vertices and having
exactly 2 sides common with the polygon.

Answer: 75
Code:
              A   B

          J           C


        I               D

 
          H           E

              G   F
Case 1: the two sides are not adjacent.

The first side can be any of the 10 adjacent vertex pairs:
. . [tex]AB, BC, CD, \text{ . . . } JA[/tex]

Suppose the first side is [tex]AB.[/tex]
Then the second side has 5 choices:
. . [tex]DE, EF, FG, GH, HI[/tex]

It seems there are [tex]10\times 5 \,=\,50[/tex] such quadrilaterals.
But this list includes [tex]\{AB,FG\}[/tex] and [tex]\{FG,AB\}[/tex]

Hence, there are: [tex]\tfrac{50}{2}\,=\,25[/tex] such quadrilaterals.


Case 2: The two sides are adjacent.
There are 10 triples of vertices:
. . [tex]ABC,BCD,CDE,\text{ . . . }JAB[/tex]

Suppose the triple is [tex]ABC.[/tex]
Then the fourth vertex can be: [tex]\{E,F,G,H,I\}[/tex]

Hence, there are: [tex]10\times5\,=\,50[/tex] such quadrilaterals.


Therefore, there are: [tex]25 + 50 \,=\,75[/tex] quadrilaterals.