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24. (a) Let $A_{n}$ be the area of a polygon with $n$ equal sidesinscribed in a circle with radius $r .$ By dividing thepolygon into $n$ congruent triangles with central angle$2 \pi / n,$ show that$$\begin{array}{c}{A_{\kappa}=\frac{1}{2} n r^{2} \sin \left(\frac{2 \pi}{n}\right)} \\ {\text { (b) Show that } \lim _{x \rightarrow \infty} A_{n}=\pi r^{2} \cdot[H i n t : \text { Use Equation } 2.4 .6} \\ {\text { on page } 133 .}\end{array}$$

a) $$A_{n}=\frac{1}{2} n r^{2} \sin \left(\frac{2 \pi}{n}\right)$$b) $$\lim _{n \rightarrow \infty} A_{n}=\pi r^{2}$$

Calculus 1 / AB

Calculus 2 / BC

Chapter 5

Integrals

Section 1

Areas, Distances, and Pathogenesis

Integration Techniques

Missouri State University

Idaho State University

Boston College

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Okay, So if we have any equal scientific radius are we have to show that the area is equal to 1/2 and our square times sign off to pi over and so this fairly easy to show. So let's suppose we have a triangle with three sides. Okay, so therefore, the area of the triangle with three sides has the equal 1/2 1/2 times, three times the really square times sign off to high over three. Simply because if you said this is equal to the radius and then the sine of the angle is thehe mount of the circle that is being covered so we could set this The amount of circle that it's being covered and we have pie are square to be the area of the whole circle. So you basically can get an area of the amount of the circle there is being covered multiplying by. Of course, you have to cut it off because if this is a circle, you're not counting the red part. So therefore this theory must be approved

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The number 2 is also the smallest & first prime number (since every other even number is divisible by two).

If you write pi (to the first two decimal places of 3.14) backwards, in big, block letters it actually reads "PIE".

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