Question

(Continuation of Exercise 37\()\) (a) Inscribe a regular \(n\) -sided polygon inside a circle of radius 1 and compute the area of one of the \(n\) congruent triangles formed by drawing radii to the vertices of the polygon. (b) Compute the limit of the area of the inscribed polygon as \(n \rightarrow \infty\) (c) Repeat the computations in parts (a) and (b) for a circle of radius \(r .\)

Short answer

\nThe area of one of the \(n\) congruent triangles formed by drawing radii to the vertices of the polygon is \( \frac{1}{n}\sin^2(\frac{\pi}{n}) \). The limit of the area of the inscribed polygon as \(n \rightarrow \infty\) is 0. These results also apply to a circle of any radius \(r\).\n

Step-by-step solution

Compute the Area of a Triangle

Start by deducing the geometry of the situation. A regular \(n\)-sided polygon inscribed in a circle of radius 1 will divide the circle into \(n\) congruent triangles with a common vertex at the center of the circle. This triangle has base equal to the side length of the polygon and height equal to the radius of the circle. Let's denote the side length by \(s\), the radius by \(r\) and the angle opposite to \(s\) by \(\theta\). Thus, the area of one of the triangles can be denoted as \( \frac{1}{2}rs\sin\theta \). Since \(\theta = \frac{2\pi}{n}\) and \(s = 2\sin(\frac{\theta}{2})\), substitute these values into the equation to get the area of one triangle as \( \frac{1}{2}\cdot1\cdot2\sin(\frac{\theta}{2})\sin(\frac{2\pi}{n}) = \frac{1}{n}\sin^2(\frac{\pi}{n}) \).

Compute the Total Area of the Polygon

Since there are \(n\) such congruent triangles forming the polygon, we can say that the total area \(A\) of the polygon is \(\frac{1}{n}\sin^2(\frac{\pi}{n}) \cdot n = \sin^2(\frac{\pi}{n})\).

Compute the Limit

We are tasked to compute the limit of the area as \(n \rightarrow \infty\). Using the limit formulas, we have \(\lim_{n \rightarrow \infty} \sin^2(\frac{\pi}{n}) = \lim_{n \rightarrow \infty} \left(\frac{\pi}{n}\right)^2 = 0^2 = 0\).

Repeat the Computations for a Circle of Radius \(r\)

Finally, we repeat the computations for a circle of radius \(r\). The area of one triangle becomes \( \frac{1}{2}r^2s\sin\theta = \frac{r^2}{n}\sin^2(\frac{\pi}{n}) \) and the total area of the polygon is again \( \sin^2(\frac{\pi}{n}) \). The limit as \(n \rightarrow \infty\) is also equal to 0, so the result generalizes to circles of any radius.