To find the area of triangle \( \triangle UVW \) with sides \( a = 29 \) cm, \( b = 49 \) cm, and \( c = 30 \) cm, we can use Heron's formula.
First, we calculate the semi-perimeter \( s \) of the triangle:
\[ s = \frac{a + b + c}{2} = \frac{29 + 49 + 30}{2} = \frac{108}{2} = 54 \text{ cm} \]
Now we can apply Heron's formula, which states that the area \( A \) of the triangle is given by:
\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]
We first calculate the values of \( s - a \), \( s - b \), and \( s - c \):
\[ s - a = 54 - 29 = 25 \] \[ s - b = 54 - 49 = 5 \] \[ s - c = 54 - 30 = 24 \]
Now substitute these values into Heron's formula:
\[ A = \sqrt{54 \times 25 \times 5 \times 24} \]
Calculating the product step by step:
- \( 54 \times 25 = 1350 \)
- \( 1350 \times 5 = 6750 \)
- \( 6750 \times 24 = 162000 \)
Now we find the square root:
\[ A = \sqrt{162000} \]
Calculating \( \sqrt{162000} \):
\[ \sqrt{162000} = \sqrt{162 \times 1000} = \sqrt{162} \times \sqrt{1000} = \sqrt{162} \times 31.622776 (approx) \]
Calculating \( \sqrt{162} \):
\[ \sqrt{162} = 12.727922 (approx) \]
So:
\[ A \approx 12.727922 \times 31.622776 \approx 402.7 \text{ cm}^2 \]
Thus, the area of triangle \( \triangle UVW \) is approximately:
\[ \boxed{402.7} \text{ cm}^2 \]
If you would like the exact integer value, calculate \( \sqrt{162000} \) directly:
\(\sqrt{162000} \approx 402.7\) rounded to one decimal place remains \( \boxed{402.7} \).