0=1 due to a polygon?

For any nn sided regular polygon inscribed in a circle of radius rr, our objective is to calculate the perimeter of the polygon.

The figure represents the given situation. OO is the center of the circle and ABAB,BCBC are edges of a regular polygon. OLOL and OMOM are perpendiculars drawn on line segments ABAB and BCBC respectively. Since perpendiculars from center bisect the chord, thus we have,



OLB=OMB\angle OLB = \angle OMB

Therfore by RHS Criteration of congruency, ΔOLBΔOMB\Delta OLB \cong \Delta OMB.

Thus by CPCT, we have

LBO=OBM\angle LBO = \angle OBM

For any regular polygon of nn sides, each angle is given by 180(n2)n\frac{180(n-2)}{n}.

Therefore, LBM=180(n2)n\angle LBM = \frac{180(n-2)}{n}. Thus LBO=180(n2)2n\angle LBO = \frac{180(n-2)}{2n}.


BLr=cos(90(n2)n°)\frac{BL}{r} = cos(\frac{90(n-2)}{n}°)

 BL=rcos(90(n2)n°)\ BL = rcos(\frac{90(n-2)}{n}°) and thus,

BC=2BL=2rcos(90(n2)n°)BC = 2BL = 2rcos(\frac{90(n-2)}{n}°)

Since it is a regular polygon,

Perimeter = (n)(BA)=2nrcos(90(n2)n°)(n)(BA) = 2nrcos(\frac{90(n-2)}{n}°) .........................(1).........................(1)

But as nn approaches infinity the polygon tends to coincide the circle in which it is inscribed. Thus in that case the perimeter of the polygon becomes equal to the circumference of the circle in which it is inscribed.

Also As nn approaches infinity,

cos(90(n2)n°)cos(\frac{90(n-2)}{n}°) becomes approximately cos(90°)=0cos(90°) = 0 .........................(2).........................(2) .

Also comparing the formula we obtained in equation 1 , to the circumference of circle,

ncos(90(n2)n°)=πncos(\frac{90(n-2)}{n}°) = \pi .........................(3).........................(3)

From (2) and (3),

π=0\pi = 0

Dividing by π\pi on both sides we get,

0=10 =1

Find the mistake!

Note by Akshat Joshi
4 years, 5 months ago

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1 vote

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Hint: There is something wrong with infinity

Akshat Joshi - 4 years, 5 months ago

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You should make this into a problem.

Zoe Codrington - 2 years, 7 months ago

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