Challenge problem

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Challenge problem

by mirantdon » Fri Jul 22, 2011 7:39 am
If there is a cyclic 8 sided polygon with four sides of length a and four sides of length b, (in some order) then what is the angle formed between the adjacent sides a and b .?

A 90
B 45
C 120
D 135


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by HeintzC2 » Fri Jul 22, 2011 8:10 am
mirantdon wrote:If there is a cyclic 8 sided polygon with four sides of length a and four sides of length b, (in some order) then what is the angle formed between the adjacent sides a and b .?

A 90
B 45
C 120
D 135


p.S :Chocolates await :-P
Answer is D

The most common "cyclic 8 sided polygon" is an Octagon, which in this case a = b. Regardless of the length of a and b, for a cyclic 8 sided polygon, the angle will remain the same between a and b.

We can divide the inside of the octagon into 8 equivalent triangles. The point of these 8 triangles form a 360 degree circle, and since all of the triangles are equivalent, each point is equal to 45 degrees (360 degrees/8 triangles).

Since each triangle has 2 congruent angles and one odd angle, we know the other two angles must both equal each other, as well that all 3 angles must add to 180 degrees.
This results in 180-45 = angle1+angle2, where angle1 = angle2 = 67.5 degrees.

Since all of the angles on the outside ring of the octagon are equal to one another, we know the angle formed between edge a and edge b must equal angle1+angle2 or 135 degrees.

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by mirantdon » Fri Jul 22, 2011 8:24 am
Nailed it .!

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by MBA.Aspirant » Fri Jul 22, 2011 12:51 pm
interior angle of a regular polygon = (n-2*180)/n

((8-2)*180)/8 = 135