Calculate the area and perimeter of a shape that is a semicircle with a straight base of diameter 4 m.

The image shows a shape that is a semicircle with a straight base.
The length of the straight base, which is the diameter of the semicircle, is given as 4 m.
The radius $r$ of the semicircle is half of its diameter.

$$r = \frac{4 \text{ m}}{2} = 2 \text{ m}$$

The area of a semicircle is given by the formula $\frac{1}{2} \pi r^2$.
Substitute the value of $r$:

$$\text{Area} = \frac{1}{2} \pi (2 \text{ m})^2$$ $$\text{Area} = \frac{1}{2} \pi (4 \text{ m}^2)$$ $$\text{Area} = 2\pi \text{ m}^2$$

Using the approximation $\pi \approx 3.14$:

$$\text{Area} \approx 2 \times 3.14 \text{ m}^2 = 6.28 \text{ m}^2$$

The area is $\boxed{2\pi \text{ m}^2 \approx 6.28 \text{ m}^2}$.

The perimeter of this shape consists of two parts: the curved arc length of the semicircle and the straight diameter.
The arc length of a semicircle is $\pi r$.
The diameter is $2r$.

$$\text{Perimeter} = \pi r + 2r$$

Substitute the value of $r$:

$$\text{Perimeter} = \pi (2 \text{ m}) + 2(2 \text{ m})$$ $$\text{Perimeter} = 2\pi \text{ m} + 4 \text{ m}$$ $$\text{Perimeter} = (2\pi + 4) \text{ m}$$

Using the approximation $\pi \approx 3.14$:

$$\text{Perimeter} \approx (2 \times 3.14 + 4) \text{ m} = (6.28 + 4) \text{ m} = 10.28 \text{ m}$$

The perimeter is $\boxed{(2\pi + 4) \text{ m} \approx 10.28 \text{ m}}$.