Question

. How many diagonals are there in a polygon having 12 sides?
Each interior angle of a polygon is 135º. How many sides does it have?
Is it possible to have a regular polygon each of whose interior angles is 45°?​

Answer 1

Answer:

12 sided polygon has 12 vertices

choose any two vertices to form a diagonal

number of possible choice is

$\binom{12}{2}$

$= \frac{12 \times (12 - 1)}{2}$

$= 66$

Answer 2

Answer:

1. Any n sided polygon where n>=3 , convex or concave has (n(n-3))/2 diagonals.

As each vertex has diagonals to all other vertices except itself and the two adjacent vertices, or n − 3 diagonals, and each diagonal is shared by two vertices

Step-by-step explanation:

n=12

diagonal =( 12(12-3))/2

= (12*9)/2

=108/2

=54

2. Sum of all interior angles= (n-2) x 180

According to the question,

sum of all interior angles=n x 135

so, we get,

n x 135=(n-2) x 180

180n-135n=360

45n=360

n=8

3.

If the polygon has n sides, then the sum of all the interior angles = (2n-4)×90° and by the above hypothesis this is equal to n×45°. Hence 180n -360 =45n ==> 135n =360 or n = (360/135) =8/3. As n should be a positive integer, this shows that it is not possible to have such a polygon