To find the area of triangle ABC, rt. angled at C, we can use the formula:
$$Area = \frac{1}{2}bh$$

where $$b$$ is the base of the triangle, which is 12 m and $$h$$ is the height. Since angle A is 30 degrees and AB is the hypotenuse, we can use trigonometry to find the height of the triangle:
$$ Tan \text{ A} = \frac{\text{Side opposite to angle A}}{\text{Side adjacent to angle A}}$$

$$ Tan(30) = \frac{h}{12}$$

$$\frac{1}{\sqrt{3}} = \frac{h}{12}$$

$$ \frac{12}{\sqrt{3}} = h$$

On rationalizing the denominator, we get
$$h = 4\cdot \sqrt{3}$$

Substituting these values into the formula for the area of a triangle, we get:
$$Area = \frac{1}{2}(12)(4\cdot \sqrt{3}) = 24\sqrt{3} \approx 41.57 \text{ m}^2$$

Therefore, the area of triangle ABC is approximately 41.57 square meters.