Let the length of the first train be \( L \). The relative speed is \( 48 + 42 = 90 \, \text{km/h} \). Converting to meters per second: \( 90 \times \frac{5}{18} = 25 \, \text{m/s} \). The total distance is \( L + \frac{L}{2} = \frac{3L}{2} \).
With a time of 12 seconds, \( \frac{3L}{2} = 25 \times 12 \), resulting in \( L = 200 \, \text{meters} \). The platform's length is then \( \frac{200}{45} \times 1000 = 400 \, \text{meters} \).