a) The square $A B C D$ with a side of $1 \mathrm{~cm}$ is "tipped" around point $D$ as shown in the following figure. The dotted lines indicate the area occupied by the square during its tipping motion. What is the total area occupied by the square from the beginning to the end of its tipping?

![](https://cdn.mathpix.com/cropped/2024_05_01_4747cb56a5540358921fg-09.jpg?height=451&width=1116&top_left_y=891&top_left_x=570)

b) Similarly, as in the case of the square in the previous item, a regular hexagon $A B C D E F$ with a side of $1 \mathrm{~cm}$ is "tipped" around point $F$ as shown in the following figure. What is the total area occupied by the hexagon from the beginning to the end of its tipping?

![](https://cdn.mathpix.com/cropped/2024_05_01_4747cb56a5540358921fg-09.jpg?height=631&width=1136&top_left_y=1758&top_left_x=563)
Solution

a) See that the figure formed can be divided into two right triangles, $\triangle A B D$ and $\triangle D B^{\prime} C^{\prime}$, and a circular sector $D B B^{\prime}$ with an opening of $90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_4747cb56a5540358921fg-10.jpg?height=459&width=1125&top_left_y=490&top_left_x=363)

The radius of the sector can be calculated using the Pythagorean Theorem

$$
R=B D=\sqrt{B C^{2}+C D^{2}}=\sqrt{1^{2}+1^{2}}=\sqrt{2}
$$

Its opening of $90^{\circ}$ is equivalent to $\frac{1}{4}$ of the area of a circle. Thus, the area will be given by

$$
\begin{aligned}
{[B A D]+\left[B^{\prime} D C^{\prime}\right]+\left[\widehat{B B^{\prime}}\right] } & =\frac{1 \cdot 1}{2}+\frac{1 \cdot 1}{2}+\frac{1}{4} \cdot \pi(\sqrt{2})^{2} \\
& =1+\frac{\pi}{2} \mathrm{~cm}^{2}
\end{aligned}
$$

b) See that the figure formed can be divided into 6 equilateral triangles with a side length of 1 and a circular sector with a radius of 2 and an opening of $60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_4747cb56a5540358921fg-10.jpg?height=637&width=1150&top_left_y=1600&top_left_x=356)

Since the height of an equilateral triangle with a side length of 1 is $\sqrt{3} / 2$, its area is

$$
\frac{1 \cdot \frac{\sqrt{3}}{2}}{2}=\frac{\sqrt{3}}{4}
$$

A sector of $60^{\circ}$ is equivalent to one-sixth of the area of a circle, so the area sought is

$$
6 \cdot \frac{\sqrt{3}}{4}+\frac{1}{6} \cdot \pi \cdot 2^{2}=\frac{3 \sqrt{3}}{2}+\frac{2 \pi}{3}=\frac{9 \sqrt{3}+4 \pi}{6} \mathrm{~cm}^{2}
$$