Subject :IMO Class : Class 8
The sum of the interior angles of a polygon is 1440°. Find the number of sides of the polygon.
A14
B12
C10
D5
Can you please clarify the answer
Ans 1:
Class : Class 10
Answer: C
Solution
If a polygon has 'n' sides, then it is divided into (n – 2) triangles. The sum of the angles of a triangle = 180°. Therefore, the sum of interior angles of a polygon having n sides is (n – 2) x 180.
(n - 2) x180 = 1440
n - 2 = 1440/180 = 8
n = 8 + 2 = 10
Solution
If a polygon has 'n' sides, then it is divided into (n – 2) triangles. The sum of the angles of a triangle = 180°. Therefore, the sum of interior angles of a polygon having n sides is (n – 2) x 180.
(n - 2) x180 = 1440
n - 2 = 1440/180 = 8
n = 8 + 2 = 10
Ans 2:
Class : Class 8
the answer should be 10 because=
(n-2)*180=1440
Ans 3:
Class : Class 8
it would be 10.
[n-2]180=1440
180n-360=1440
180n=1800
n=1800/180
= 10