Question

How many diagonals are there in a 14 – sided polygon?
(a) 47
(b) 57
(c) 67
(d) 77

Answer 1

At first find total number of lines formed by 2 points from 14 points. After finding it, subtract it by 14 lines as they represent the sides which are not considered as diagonals.

Complete step-by-step answer:
In the question we are asked about how many diagonals does a 14 – sided polygons have.
Now before finding the exact count let’s understand briefly what a diagonal is.
In geometry a diagonal is a line segment joining two vertices of a polygon or polyhedron when those vertices are not on the same edge. Informally any sloping line is called diagonal.
Diagonals are formed by two points having condition that they do not form a side of the polygon.
So diagonals can be counted by selecting two points from 14 points which can be done in ${}^{14}{{C}_{2}}$ ways as ${}^{n}{{C}_{r}}$ represents the number of ways of choosing r items from n given items.
Also ${}^{n}{{C}_{r}}$ represents $\dfrac{n!}{\left( n-r \right)!\times r!}$.
So, ${}^{14}{{C}_{2}}$ is equal to $\dfrac{14!}{\left( 14-2 \right)!\times 2!}$ or $\dfrac{14\times 13}{2}$ which is equal to 91.
Out of these 91 possible lines connecting 14 points on a plane, no three points are collinear which means that we have to subtract 14 lines which represent sides of polynomials.
So, the number of diagonals are 91 – 14 or 77.

So, the correct answer is “Option d”.

Note: The formula we used in the question is ${}^{n}{{C}_{2}}-n$ where n represents the number of sides of a polygon. We know the value of ${}^{n}{{C}_{2}}$ is $\dfrac{n\left( n-1 \right)}{2}$, then we can write ${}^{n}{{C}_{2}}-n$ as $\dfrac{n\left( n-1 \right)}{2}-n$ which can also be written as $\dfrac{{{n}^{2}}-n-2n}{2}$ or $\dfrac{{{n}^{2}}-3n}{2}$ or $\dfrac{n\left( n-3 \right)}{2}$. Thus one can also use the formula $\dfrac{n\left( n-3 \right)}{2}$ to get the answer.