How many triangles are there in a polygon, if the polygon has 44 diagonals?
Hello, sureshrju!
Please check the wording of the problem . . .
A convex polygon with $\displaystyle n$ sides has: .$\displaystyle \frac{n(n-3)}{2}$ diagonals.How many triangles are there in a polygon, if the polygon has 44 diagonals?
. . But .$\displaystyle \frac{n(n-3)}{2} \:=\:44$ .has no integral solutions. . . . . . wrong!
I did some very bad algebra . . . sorry!
Hi Saroban,
There is any polygon of concave? My friend ask this question to me, So i can not check the question again, i asked him about the question he said it is the correct one.
n(n-3)/2=44
n(n-3)=88
n^2-3n=88
n^2-3n-88=0
Then, (n+8)(n-11)=0
n=-8 or n=11
Since n denotes side, it can not be negative. So, n=11.
So, it is a polygon of 11 sides.
Can you please help me in finding out the number of triangles in the 11 sided polygon?
In fact $\displaystyle \frac{n(n-3)}{2}=44$ does have integral solutions: $\displaystyle 11~\&~-8$.
Thus we have 11 vertices.
The number of triangles on 11 non-collinear points is $\displaystyle \binom{11}{3}$.
If you are asking for the total number of triangle formed by the diagonals and/or sides that is a different matter.