Question

Sum of interior angles of triangle is ${180}^o$, and that for a quadrilateral is ${360}^o$, then sum of interior angles of a polygon having n sides is

• A. ${180}^o$$\times$(n +1 )
• B. ${180}^o$$\times$(n - 2 )
• C. ${180}^o$$\times$(n - 1 )
• D. ${180}^o$$\times$n

Answer 1

The correct option is B ${180}^o$$\times$(n - 2 )

Triangle has 3 sides
sum of interior angles
= ${180}^o$(3 - 2 ) = ${180}^o$
Where 3 is the number of sides.

Quadrilateral has 4 sides
sum of interior angles
= ${180}^o$(4 - 2 ) = ${180}^o$$\times$2 = ${360}^o$
Where 4 is the number of sides.
So hence if polygon has n sides
sum of interior angles of n-sided polygon is
= ${180}^o$$\times$(n - 2 )