Page:Elementary Trigonometry.djvu/107

 $A'B' = 2A'K = 2OK \tan A'OK = 2r \tan \frac{\pi}{n} ,$

$\text{triangle}\ A'OB'= \frac{1}{2} A'B'.OK=r^{2} \tan \frac{\pi}{n} ;$

$\therefore \text{the perimeter of the polygon} = 2nr \tan \frac{\pi}{n}.$

$\text{and the area of the polygon} = nr^2 \tan \frac{\pi}{n}.$

11. Circumference and area of a circle.

It may be assumed that the circumference of the circle is intermediate in length to the perimeters of the inscribed and circumscribed polygons, and that the smaller the sides of the polygons are made, the nearer do their perimeters approach each other and the circumference of the circle in magnitude.

The circumference then of the circle is intermediate in magnitude to

$2nr \sin \frac{\pi}{n}, \text{and}\ 2r \tan\frac{\pi}{n};$

that is, to

$ 2\pi r \frac{\sin \frac{\pi}{n}}{\frac{\pi}{n}}$, and $2\pi r \frac{\tan \frac{\pi}{n} } {\frac{\pi}{n}}$
 * undefined

Now, when n is made very great, $$\frac{\pi}{n}$$ is very small; and, when n is made infinite, $$\frac{\pi}{n}$$ becomes zero.

But (Chap. Art. 14) when $$\frac{\pi}{n}$$ is zero,

$$ \frac{\sin \frac{\pi}{n}}{\frac{\pi}{n}}=1=\frac{\tan \frac{\pi}{n}}{\frac{\pi}{n}}$$