A regular octagon is inscribed in a circle with a radius of 10 cm. What is the length of one side of the octagon?
2 answers:
Answer:
The length of one side of the octagon is 7.65 cm
Step-by-step explanation:
The parameters given are;
A regular octagon inscribed in a circle of radius, r, of 10 cm.
The length of each side is found from the isosceles triangle formed by the radius and one side of the octagon
The sum of interior angles in a polygon, ∑θ = 180 × (n - 2)
Where;
n = The number of sides of the polygon
θ = The interior angle of the polygon
For the octagon, we have;
n = 8, therefore;
∑θ = 180 × (8 - 2) = 1080
Given that there are eight equal angles in a regular octagon, we have;
∑θ = 8 × θ
= 1080
θ = 1080/8 = 135°
The sum of angles at the center of the circle = 360
Therefore, the angle at the center (tip angle) of the isosceles triangle formed by the radius and one side of the octagon = 360/8 = 45°
The base angles of the isosceles triangle is therefore, (180 - 45)/2 = 67.5° = θ/2
The length of the base of the isosceles triangle formed by the radius and one side of the octagon = The length of one side of the octagon
From trigonometric ratios, the length of the base of the isosceles triangle is therefore;
2 × r × cos(θ/2) = 2×10 × cos(67.5°) = 7.65 cm
The length of the base of the isosceles triangle = 7.65 cm = The length of one side of the octagon.
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Answer:
Ö + θ ( (k/m) + (g/l)) = 0
Step-by-step explanation:
Use the FBD attached:
Apply Newtons 2 nd Law in tangential direction:
Sum ( Ft ) = m*a
Sum of all tangential forces is:
m*g*sin(θ) + k*l*sin(θ)*cos(θ) = - m*l*Ö
Using small angle approximations:
sin (θ) = θ
cos (θ) = 1
Ö = angular acceleration.
m*g*θ + k*l*θ = -m*l*Ö
Ö + θ ( (k/m) + (g/l)) = 0
