Answer: check explanation
Step-by-step explanation:
Let K(n) be the sum of the interior angles in any n-sided convex polygon which is exactly 180(n −2)
degrees.
CASE: n = 3. A 3-sided polygon is a triangle, whose interior angles were shown always to sum to be 180 degree
INDUCTION HYPOTHESIS: Suppose that K(b) holds for some b ≥ 3. Which means that the interior angles in any b-sided convex
polygon is exactly 180(n −2) degrees.
INDUCTION STEP: We need to show that K(n is greater than or equals to 3) . That is, the interior angles of any b+1-sided convex polygon
is exactly 180(b−2) = 180(b −1) degrees,
Let X be any (b+1)-vertex convex polygon, say with successive vertices x1, x2,..., xb+
Now Y is also a convex polygon , so by the induction hypothesis K(b), the sum of the interior
angles of Y is 180(k −2).
Now let T be the triangle with vertices xk , xb+1, x1. The sum of the interior angles in X is the sum of those
in Y plus the sum of those in T .
So the sum of the interior angles in X is
180(b −2)+180 = 180((b +1)−2) = 180(b −1).
Since X was arbitrary, we conclude that the sum of the interior angles of any (b +1)-sided convex polygon
is 180((b−2)+1) = 180(b−1). That is, P(b +1) holds.
And which means that n is greater or equals to 3
, where radius is given as 4
, where a is side length of square