Question

The number of diagonals and triangles formed in an octagon are
A.$20,56$
B.$32,58$
C.$30,58$
D.$32,56$

Answer 1

Hint: Number of diagonals in $n$ sided polygon = Total number of lines connected two points – Total number of sides of polygon. Equation to calculate number of lines connected two points is ${}^{n}{{c}_{2}}$, formula for combination is${}^{n}{{c}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. After substituting the equation for total number of diagonals in $n$ sided polygon is $d=\dfrac{n\left( n-3 \right)}{2}$. One triangle is formed by a combination of three points from eight vertices. The equation for calculating the number of triangles is $t={}^{8}{{c}_{3}}$.

Complete step-by-step answer:
Number of diagonals in $n$ sided polygon = Total number of lines connected two points – Total number of sides of polygon
If we have $n$ points so total number of lines is
$\Rightarrow number\, of\, lines\,={}^{n}{{c}_{2}}$
Formula for combination is,
$\Rightarrow {}^{n}{{c}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Because we choose $2$ from $n$ points
Total number of sides of polygon is represented as $n$
Let the number of diagonals is represented as $d$
Formula for finding the number of diagonals in $n$ sided polygon is
$\Rightarrow d=\dfrac{n\left( n-1 \right)}{2}-n$
Simplifying the above equation,
$\begin{align}
  & \Rightarrow d=\dfrac{n\left( n-1 \right)-2n}{2} \\
 & \Rightarrow d=\dfrac{n\left( n-1-2 \right)}{2} \\
 & \Rightarrow d=\dfrac{n\left( n-3 \right)}{2}.....(1) \\
\end{align}$
Here, number of sides of octagon is$8$
$\Rightarrow n=8$
Substituting the value of $n$ in equation (1)
$\begin{align}
  & \Rightarrow d=\dfrac{8\left( 8-3 \right)}{2} \\
 & \Rightarrow d=\dfrac{8\times 5}{2} \\
 & \Rightarrow d=\dfrac{40}{2} \\
 & \Rightarrow d=20 \\
\end{align}$
Number of diagonals of octagon is 20
Let ${{v}_{1}},{{v}_{2}},.....,{{v}_{8}}$ be the vertices of the octagon. One triangle is formed by three points from these 8 vertices. Therefore, the number of triangles is equal to the number of combinations of three points formed from eight points.
Number of triangles is represented as $t$
$\Rightarrow t={}^{8}{{c}_{3}}$
Solving it,
$\begin{align}
  & \Rightarrow t=\dfrac{8!}{3!\left( 8-3 \right)!} \\
 & \Rightarrow t=\dfrac{8!}{3!5!} \\
 & \Rightarrow t=\dfrac{8\times 7\times 6}{3\times 2} \\
 & \Rightarrow t=8\times 7 \\
 & \Rightarrow t=56 \\
\end{align}$
Number of triangles of an octagon is $56$
Therefore, the number of diagonals and triangles formed in an octagon are (A)$20,56$.

Note: Number of diagonals in $n$ sided polygon = Total number of lines connected two points – Total number of sides of polygon. Equation to calculate number of lines connected two points is ${}^{n}{{c}_{2}}$, formula for combination is${}^{n}{{c}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. After substituting the equation for total number of diagonals in $n$ sided polygon is $d=\dfrac{n\left( n-3 \right)}{2}$. One triangle is formed by a combination of three points from eight vertices. The equation for calculating the number of triangles is$t={}^{8}{{c}_{3}}$.