Answer:
61.2 in
Step-by-step explanation:
Draw a segment from O perpendicular to AB forming a right triangle.
Call the point of intersection J.
m∠HAB = (8 - 2)180/8 = 135
m∠BAO = 135/2 = 67.5
AJ/10 = cos 67.5
AJ = 10 cos 67.5
AB = 2(10 cos 67.5)
Perimeter = 8(AB) = 8(2)(10 cos 67.5) = 61.2 in.
The perimeter of a regular octagon that has a distance from its center to any of its vertices of 10 inches would be 61.2 inches.
A polygon is a closed figure made up of three or more line segments connected end to end.
First, we draw a segment from O perpendicular to AB forming a right triangle.
Let say the point of intersection J.
∠HAB = (8 - 2)180/8
= 135
∠BAO = 135/2
= 67.5
AJ/10 = cos 67.5
AJ = 10 cos 67.5
AB = 2(10 cos 67.5)
Perimeter = 8(AB)
= 8(2)(10 cos 67.5)
= 61.2 in.
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