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Prove by induction that the sum of the interior angles of a convex n-sided polygon is (n-2)*180?
How do you solve the following question:
"Prove by induction that for n≥3, the sum of the interior angles of a convex n-sided polygon is (n-2)*180"
Thanks
2 Answers
- JuanLv 67 years agoFavorite Answer
sum_(n = 3)^(n) ((n-2)*180) = 90*((n^2) - 3n + 2)
then we can say
sum_(n = 3)^(n+1) (180*(n - 2)) = 90*n*(n - 1)
Now comes the proof
S_n = 180 + 360 + 540 + ... + 180*(n - 2) = 90*((n^2) - 3n + 2)
S_(n + 1) = 180 + 360 + 540 + ... + 180*(n - 2) + 180*(n - 1) = 90*n*(n - 1)
S_(n + 1) = 90*((n^2) - 3n + 2) + 180*(n - 1) = 90*n*(n - 1)
S_(n + 1) = 90*(n^2) - 270*n + 180 + 180*n - 180 = 90*n*(n - 1)
S_(n + 1) = 90*(n^2) - 90*n = 90*n*(n - 1)
S_(n + 2=1) = 90*n*(n - 1) = 90*n*(n - 1)
therefore true
- Anonymous7 years ago
1) All polygons with n sides ≥3 can be composed with n-2 triangles.
2) All triangles contain interior angles = 180 degrees
3) Therefore, all polygons with n sides ≥3 have a sum of (n-2)*180 degrees.
