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#### Dividing circle into geometrically different thirds

72 viewed last edited 2 years ago
Deepika Visala
0
A question that I was recently thinking of and couldn't work out an answer to was to determine a general solution as to the size of the shapes needed to make a circle divided into equal size parts using this shape: http://imgur.com/a/0Z5t0 where the white dot in the middle is the centre of the circle. Thanks :)
Vivekanand Vellanki
1

In the attached figure, FE is parallel to AB. Let, the angle OAD be \theta degrees. It is also known that \theta < 30 since if angle OAD is 30, the area of the sector AOC will be 1/3rd the area of the circle. The area of ADC is given to the 1/3 area of the circle. Assuming unit circle, Area of ADC = Area of sector AOC + area of triangle AOD. Area of sector AOC = \pi\frac{90+\theta}{360} To find area of triangle AOD. OD=\cos{\theta} and AD=\sin{\theta} Area of triangle AOD = \frac{\cos{\theta}\sin{\theta}}{2} This gives the equation: \pi\frac{90+\theta}{360} + \frac{\cos{\theta}\sin{\theta}}{2} = \pi/3 \frac{\cos{\theta}\sin{\theta}}{2} = \pi(1/3 - \frac{90+\theta}{360}) \frac{\cos{\theta}\sin{\theta}}{2} = \pi\frac{30 - \theta}{360} \cos{\theta}\sin{\theta} = \pi\frac{30 - \theta}{180}

Mahesh Godavarti
0
Do you have a geometric construction for this?
Sangeetha Pulapaka
0

Draw a circle of any radius and mark the center of the circle as O. Mark some random point on the circumference of the circle and name it as A. Join OA. Keep the protractor on line OA and measure 120 ^\circand mark it as point B. Now keep the protractor on line OB and measure 120 ^\circ again, this time mark the next point as C.The circle is divided this way into three equal parts. Join the points ABC to get an equilateral triangle with angles 60 ^\circ each.This means that all the three sides are of equal length.