The diagram shows the cross- section of a railway tunnel. If |AB| = 100m and the radius of the arc is 56m, calculate, correct to the nearest metre, the peri...
Question 1 Report
The diagram shows the cross- section of a railway tunnel. If |AB| = 100m and the radius of the arc is 56m, calculate, correct to the nearest metre, the perimetre of the cross- section.
Answer Details
Perimeter of the tunnel cross-section.
The cross-section is made up of the straight base \(AB=100\text{ m}\) together with the arc that domes over the top. The arc has radius \(r=56\text{ m}\), and the diagram shows the arc is the major arc (it rises above and beyond a semicircle).
Step 1: Angle subtended by the chord at the centre.
Let \(O\) be the centre and let the perpendicular from \(O\) bisect the chord \(AB\). Half the chord is \(50\text{ m}\). If \(\theta\) is half the central angle of the minor arc:
\[\sin\theta=\frac{50}{56}=0.8929\Rightarrow\theta=63.26^{\circ}\]
So the minor-arc central angle is:
\[2\theta=126.52^{\circ}\]
Step 2: Reflex angle for the major (tunnel) arc.
\[\text{Reflex angle}=360^{\circ}-126.52^{\circ}=233.48^{\circ}\]
Step 3: Length of the major arc.
\[\text{Arc}=\frac{233.48}{360}\times 2\pi r=\frac{233.48}{360}\times 2\times\frac{22}{7}\times 56\]
\[=\frac{233.48}{360}\times 352=0.6486\times 352=228.3\text{ m}\]
Step 4: Perimeter.
\[\text{Perimeter}=AB+\text{arc}=100+228.3=328.3\text{ m}\]
Correct to the nearest metre, the perimeter of the cross-section is \(328\text{ m}\).
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