The given shape is made up of a rectangle, a square and an equilateral triangle. Calculate the perimeter of the shape.

The diameter of the circle is given as $20 \text{ cm}$.
The radius $r$ is half of the diameter:

$$r = \frac{20 \text{ cm}}{2} = 10 \text{ cm}$$

$$A = \pi (10 \text{ cm})^2$$ $$A = 100\pi \text{ cm}^2$$

Using $\pi \approx 3.14159$:

$$A \approx 100 \times 3.14159 \text{ cm}^2$$ $$A \approx 314.16 \text{ cm}^2$$

The area of the circle is $\boxed{100\pi \text{ cm}^2}$ or approximately $\boxed{314.16 \text{ cm}^2}$.

From the diagram and problem description:

• The rectangle is CDEF. Given $CD = 3 \text{ cm}$. Since it's a rectangle, $EF = CD = 3 \text{ cm}$.
• The square is ABFG. Given $FG = 8 \text{ cm}$. Since it's a square, all its sides are equal: $AB = BF = FG = GA = 8 \text{ cm}$.
• The equilateral triangle is ABC. Since $AB$ is a side of the square, $AB = 8 \text{ cm}$. As it's an equilateral triangle, all its sides are equal: $AB = BC = CA = 8 \text{ cm}$.
• The side $CF$ of the rectangle CDEF is aligned with the side $BF$ of the square ABFG. Thus, $CF = BF = 8 \text{ cm}$.
• Since CDEF is a rectangle, its opposite sides are equal, so $DE = CF = 8 \text{ cm}$.

The outer edges are:

• $DC = 3 \text{ cm}$ (from rectangle CDEF)
• $CB = 8 \text{ cm}$ (from equilateral triangle ABC)
• $CA = 8 \text{ cm}$ (from equilateral triangle ABC)
• $AG = 8 \text{ cm}$ (from square ABFG)
• $GF = 8 \text{ cm}$ (from square ABFG)
• $FE = 3 \text{ cm}$ (from rectangle CDEF)
• $ED = 8 \text{ cm}$ (from rectangle CDEF)

$$\text{Perimeter} = DC + CB + CA + AG + GF + FE + ED$$ $$\text{Perimeter} = 3 \text{ cm} + 8 \text{ cm} + 8 \text{ cm} + 8 \text{ cm} + 8 \text{ cm} + 3 \text{ cm} + 8 \text{ cm}$$ $$\text{Perimeter} = 46 \text{ cm}$$

The perimeter of the shape is $\boxed{46 \text{ cm}}$.