Asked by Hannah
The area of a regular octagon is 25. What is the area of a regular octagon with sides five times as large as the sides of the first octagon?
Answers
Answered by
Anonymous
Area of octagon:
A=2*[1+sqroot(2)]*a^2
A1=Area of first octagon
A1=25
A2=Area of five time lager side octagon
A1=2*[1+sqroot(2)]*a^2
A2=2*[1+sqroot(2)]*(5a)^2
A2=2*[1+sqroot(2)]*25*a^2
A2/A1=2*[1+sqroot(2)]*25*a^2/2*[1+sqroot(2)]*a^2= 25
A2/A1=25
A2=25*A1= 25*25= 625
A=2*[1+sqroot(2)]*a^2
A1=Area of first octagon
A1=25
A2=Area of five time lager side octagon
A1=2*[1+sqroot(2)]*a^2
A2=2*[1+sqroot(2)]*(5a)^2
A2=2*[1+sqroot(2)]*25*a^2
A2/A1=2*[1+sqroot(2)]*25*a^2/2*[1+sqroot(2)]*a^2= 25
A2/A1=25
A2=25*A1= 25*25= 625
Answered by
Anonymous
In google type:
Octagon
When you see list of results click on wikipedia
Octagon
When you see list of results click on wikipedia
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