The problem asks to calculate the perimeter of the cross-section of a railway tunnel. The cross-section consists of a line segment $AB$ and a circular arc. We are given that $|AB| = 100$ meters and the radius of the arc is $r = 56$ meters. We need to find the perimeter of the cross-section, which is the sum of the length of the line segment $AB$ and the length of the arc.

1. Problem Description

The problem asks to calculate the perimeter of the cross-section of a railway tunnel. The cross-section consists of a line segment $AB$ and a circular arc. We are given that $|AB| = 100$ meters and the radius of the arc is $r = 56$ meters. We need to find the perimeter of the cross-section, which is the sum of the length of the line segment $AB$ and the length of the arc.

2. Solution Steps

First, we need to find the angle $\theta$ subtended by the arc at the center of the circle. We can use the law of cosines in the triangle formed by the center of the circle and the points A and B. Let O be the center of the circle. Then $OA = OB = r = 56$ meters, and $AB = 100$ meters. Using the law of cosines:

$AB^2 = OA^2 + OB^2 - 2(OA)(OB)\cos{\theta}$

$100^2 = 56^2 + 56^2 - 2(56)(56)\cos{\theta}$

$10000 = 3136 + 3136 - 6272\cos{\theta}$

$10000 = 6272 - 6272\cos{\theta}$

$3728 = -6272\cos{\theta}$

$\cos{\theta} = -\frac{3728}{6272} = -\frac{233}{392} \approx -0.5943877$

Now, we find the angle $\theta$:

$\theta = \arccos(-\frac{233}{392}) \approx 2.205$ radians

The arc length $s$ is given by the formula:

$s = r\theta$

where $r$ is the radius and $\theta$ is the angle in radians.

$s = 56 \times 2.205 \approx 123.48$ meters

The perimeter $P$ of the cross-section is the sum of the length of $AB$ and the arc length $s$:

$P = AB + s = 100 + 123.48 = 223.48$ meters

Rounding to the nearest meter, the perimeter is approximately 223 meters.

3. Final Answer

223 meters