\section*{Task 2 - 051022}

A regular dodecagon is inscribed in a circle of radius \(r\).

a) Calculate the perimeter of the dodecagon!

b) Calculate the area of the dodecagon!

c) By what percentage is the perimeter of the dodecagon smaller than that of the circle? (The result should be given to one decimal place.)

d) By what percentage is the area of the dodecagon smaller than that of the circle? (Accuracy as in c)
}

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1221.jpg?height=274&width=322&top_left_y=1439&top_left_x=273)

Let \(M\) be the center of the circle. Further, let \(C\) be a vertex and \(A\) and \(B\) the two adjacent vertices of the dodecagon.

Then it holds: \(r=|M A|=|M C|=|M B|=|A B|\).

\(A B\) is the side of a regular hexagon inscribed in the circle and \(M C \perp A B\). \(A C B M\) is a kite because \(|A M|=|B M|\) and \(|A C|=|B C|\).

a) The segments \(A B\) and \(M C\) intersect at a point \(D\). If we set \(\mid D C>=x\), then it holds

\[
x=r-|M D|=r-\frac{\sqrt{3}}{2} r
\]

\(M D\) is the height in the equilateral triangle \(M A B\). If \(s=|A C|=|B C|\) is the side length of the dodecagon, then by the Pythagorean theorem applied to the triangle \(C B D\),

\[
s^{2}=\left(\frac{r}{2}\right)^{2}+\left(r-\frac{r}{2} \sqrt{3}\right)^{2}=r^{2}(2-\sqrt{3}) \rightarrow s=r \sqrt{2-\sqrt{3}}
\]

The perimeter of the dodecagon is therefore \(12 r \sqrt{2-\sqrt{3}}\).

b) We first calculate the area \(I_{1}\) of the triangle \(M B C\). It holds

\[
I_{1}=\frac{1}{2} r \cdot r \sin 30^{\circ}=\frac{r^{2}}{4}
\]

The area \(12 I_{1}\) of the dodecagon is thus \(3 r^{2}\).

c) It is

\[
12 r \sqrt{2 \sqrt{3}}: 2 r \pi \approx 3.106: \pi \approx 0.989
\]

Since \(3.1055^{2}<36(2-\sqrt{3})<3.1065^{2}\) and thus \(3.1055<6 \sqrt{2-\sqrt{3}}<3.1065\) holds, and because \(3.1415<\pi<3.1416\),

\[
0.9885 \cdot 3.1416<3.1055<6 \sqrt{2-\sqrt{3}}<3.1065<0.9895 \cdot 3.1415
\]

The perimeter of the dodecagon is therefore about 1.1% smaller than that of the circle.

d) Since \(0.9549 \cdot 3.1416<3<0.955 \cdot 3.1415\),

\[
3 r^{2}: \pi r^{2}=3: \pi \approx 0.955
\]

The area of the dodecagon is thus about 4.5% smaller than that of the circle.