Collective Noun: herd

The collective noun 'herd' is used to describe a group of grazing animals like sheep and cattle. Gulls are a type of seabird, which is why it is the odd ones out.

Option A: Chamois are a type of goat-antelope that live in mountainous regions. Since they are grazing animals, they are correctly referred to as a herd.

Option C:  While walruses are marine mammals, they are often referred to as a herd when grouped together on land or ice. 

Option D: Wrens are small birds. The correct collective nouns for wrens include a "flock" or a "chime", not a "herd."

Solution

Since $$ABCDEF$$ is a regular hexagon of side $$2$$ cm, its area is

$$\begin{aligned}A_1 = 6\times \frac{\sqrt{3}}{4}\times 2^2\ =6\sqrt{3}\text{ cm}^2\end{aligned}$$

The lines $$AB, SP, RQ$$ and $$ED$$ are parallel and equally spaced. The perpendicular distance between $$AB$$ and $$ED$$ is $$2\sqrt{3}$$ cm.

Therefore, the distance between two consecutive parallel lines is

$$d=\frac{2\sqrt{3}}{3}$$

Using the symmetry of the regular hexagon, the area of hexagon $$PCQRFS$$ is

$$A_2=\frac{22\sqrt{3}}{9}\text{ cm}^2$$

Hence, the required ratio is

$$\begin{aligned}\frac{A_2}{A_1}&=\frac{\frac{22\sqrt{3}}{9}}{6\sqrt{3}}\ =\frac{22}{54}\ =\frac{11}{27}\end{aligned}$$

Therefore, the required ratio is

$$\boxed{11:27}$$

The side of the regular hexagon is (2) cm. A regular hexagon can be divided into 6 equilateral triangles, each having a side of (2) cm.

The area of one equilateral triangle is

$$\begin{aligned}A&=\frac{\sqrt{3}}{4}\times (\text{side})^2\&=\frac{\sqrt{3}}{4}\times 2^2\&=\sqrt{3}\text{ cm}^2\end{aligned}$$

Therefore, the area of the entire hexagon is

$$\begin{aligned}A_1&=6\times\sqrt{3}\&=6\sqrt{3}\text{ cm}^2\end{aligned}$$

Step 2: Find the distance between the parallel lines

The lines (AB), (SP), (RQ) and (ED) are parallel and equally spaced.

In a regular hexagon with horizontal top and bottom sides, the perpendicular distance between (AB) and (ED) is twice the height of an equilateral triangle of side (2) cm.

The height of an equilateral triangle is

$$\begin{aligned}h&=\sqrt{2^2-1^2}\&=\sqrt{4-1}\&=\sqrt{3}\text{ cm}\end{aligned}$$

Hence, the perpendicular distance between (AB) and (ED) is

$$2h=2\sqrt{3}\text{ cm}$$

There are three equal gaps between the four parallel lines. Therefore, the distance between two consecutive lines is

$$\begin{aligned}d&=\frac{2\sqrt{3}}{3}\&=\frac{2\sqrt{3}}{3}\text{ cm}\end{aligned}$$

Step 3: Find the area of hexagon (PCQRFS)

The required hexagon can be divided into a central rectangle and two congruent triangles.

The rectangle has height equal to the distance between (SP) and (RQ), which is

$$h_r=\frac{2\sqrt{3}}{3}\text{ cm}$$

Its length is (\frac{10}{3}) cm.

Therefore, its area is

$$\begin{aligned}A_r&=\frac{10}{3}\times\frac{2\sqrt{3}}{3}\&=\frac{20\sqrt{3}}{9}\text{ cm}^2\end{aligned}$$

Each of the two triangles has a base of (\frac{2\sqrt{3}}{3}) cm and a perpendicular height of (\frac13) cm.

The combined area of the two triangles is

$$\begin{aligned}A_t&=2\times\frac12\times\frac{2\sqrt{3}}{3}\times\frac13\&=\frac{2\sqrt{3}}{9}\text{ cm}^2\end{aligned}$$

Adding these areas, we get

$$\begin{aligned}A_2&=A_r+A_t\&=\frac{20\sqrt{3}}{9}+\frac{2\sqrt{3}}{9}\&=\frac{22\sqrt{3}}{9}\text{ cm}^2\end{aligned}$$

Step 4: Calculate the required ratio

The required ratio is the area of hexagon (PCQRFS) divided by the area of hexagon (ABCDEF).

$$\begin{aligned}\text{Required ratio} &= \frac{A_2}{A_1} \\[5pt]&= \frac{\dfrac{22\sqrt{3}}{9}}{6\sqrt{3}} \\[5pt]&= \frac{22\sqrt{3}}{54\sqrt{3}} \\[5pt]&= \frac{22}{54} \\[5pt]&= \frac{11}{27}\end{aligned}$$

Hence, the required ratio is

$$\boxed{11:27}$$