Question

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

(i) The height of the tunnel
(ii) The perimeter of the cross-section
(iii) The area of the cross-section.

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Answer 1

1-
OA=2m

AB= $\sqrt{(}{2}^2+2^2)$
= $2\sqrt{2}$
Let the height be h
Area of $△AOB=\frac{1}{2}\times 2\times 2$
$\frac{1}{2}\times 2\sqrt{2}\times OM=2$
OM=$\sqrt{2}$
h=2+$\sqrt{2}$ m

(ii) The perimeter of the cross-section
=Major arcAB+AB
=$2\times \pi \times 2\frac{3}{4}+2\sqrt{2}$
=$3\pi +2\sqrt{2}m$


(iii) The area of the cross-section.

=$\frac{3}{4}\times Areaofcircle+areaof△AOB$
=$\pi \left(2^2\right)\frac{3}{4}+\frac{1}{2}\times 2\times 2$= $(3\pi +2)m^2$