From \(\Delta\) BCD,
\(\sin \theta = \frac{50}{56}\)
\(\theta = \sin^{-1} (0.8929)\)
\( < BCD = 63.23°\)
\(\therefore < ACB = 63.23° \times 2 = 126.46°\)
\(< ACB (reflex) = 360° - 126.46°\)
= \(233.54°\)
Perimeter of the arc = \(\frac{233.54}{360} \times 2 \times \frac{22}{7} \times 56\)
= \(228.35 m\)
Perimeter of the railway tunnel = Perimeter of the arc + |AB|
= 228.35 m + 100 m
= 328.35 m.