The area of standard stop sign regular octagon is 742.3 squared inches, if the standard stop sign has a 16.2 inch radius.
Step-by-step explanation:
The given is,
Radius of standard stop sign is 16.2 inch radius
Step:1
The regular octagon is equal to the 16 right angled triangle,
Angle of right angle triangle = [tex]\frac{360}{16}[/tex] = 22.5°
Ref attachment,
From the OAB right angle triangle,
Trigonometric ratio,
sin ∅ = [tex]\frac{Opp}{Hyp}[/tex]
Where, ∅ = 22.5°
Hyp = Radius = 16.2 inches
Ratio becomes,
sin 22.5°= [tex]\frac{b}{16.2}[/tex]
b = (0.374607)(16.2)
b = 6.19947 inches
Trigonometric ratio,
cos ∅ = [tex]\frac{Adj}{Hyp}[/tex]
Where, ∅ = 22.5°
Hyp = Radius = 16.2 inches
Adj = h
Ratio becomes,
cos 22.5°= [tex]\frac{h}{16.2}[/tex]
h = (0.9238795)(16.2)
h = 14.967 inches
Step:2
From the triangle OAC,
Area, [tex]A = \frac{1}{2} (Height )( Base)[/tex]
Where, Height = 14.967 inches
Base = b + x = 6.19947 + 6.19947
= 12.3989 inches
From the values equation becomes,
A = [tex]\frac{1}{2} (14.967)(12.39894)[/tex]
A = 92.7875 Squared inches
Step:3
The octagon is equal to sum of 8 triangles
Area of octagon = 8 × Area of triangle
= 8 × 92.7875
= 742.3 squared inches
Result:
The area of standard stop sign regular octagon is 742.3 squared inches, if the standard stop sign has a 16.2 inch radius.
