If the size of the interior angles of a polygon are 2a, 4a, 6a, 6a, 8a & 10a. What would be the largest and smallest interior angle of the polygon, respectively?
Answers
Given :- If the size of the interior angles of a polygon are 2a, 4a, 6a, 6a, 8a & 10a. What would be the largest and smallest interior angle of the polygon, respectively ?
Answer :-
we know that ,
- sum of interior angles of a regular polygon with n sides is = (n - 2) * 180° .
given that,
- Polygon has = 6 sides .
so,
→ 2a + 4a + 6a + 6a + 8a + 10a = (6 - 2) * 180°
→ 36a = 4 * 180°
→ a = 20 .
then,
- Largest angle = 10a = 10 * 20 = 200°
- Smallest angle = 2a = 2*20 = 40° .
[ Note :- we can find sum of interior angles if all sides are equal , or regular polygon and in that case all interior angles will also be equal .]
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