Question

Fig. 15.87, shows the cross-section of railway tunnel. The radius OA of the circular path is 2 m. If $\angle AOB={90}^{\circ }$, calculate the perimeter of the cross-section.

Answer 1

Given Radius$=2m$ and $\angle AOB={90}^{\circ }$

Perimeter of circle-arc length$=2\pi r-\frac{{90}^{\circ }}{{360}^{\circ }}\times 2\pi r$

In $△AOB$

$A{B}^2=A{O}^2+O{B}^2$ [Pythagoras theorem ]

Perimeter of cross-section = AB + Perimeter of circle-arc length

$=\sqrt{{OA}^2+{OB}^2}+2\pi r-\frac{{90}^0}{{360}^0}\times 2\pi r$

$=\sqrt{2^2+2^2}+2\pi \times 2-\frac{1}{2}\times \pi \times 2=(2\sqrt{2}+3\pi )$

$=(3\pi +2\sqrt{2})m$