Question

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

Answer 1

In right $△AOB,{AO}^2+{BO}^2={AB}^2$

$\Rightarrow AB=\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}$

Now, ar, $△AOB=\frac{1}{2}\times OA\times OB=\frac{1}{2}\times OM\times AB$

$\Rightarrow \frac{1}{2}\times 2\times 2=\frac{1}{2}\times OM\times 2\sqrt{2}\Rightarrow OM=\frac{2}{\sqrt{2}}=\sqrt{2}$

Now, height of tunnel = Radius + OM $=(2+\sqrt{2})m$