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The figure shows the cross section of a railway tunnel. The radius of the tunnel is $3.5$m, i.e $OA = 3.5$m. $\angle AOB=90^\circ$. Calculate

1) the height of the tunnel

2) the perimeter of it's cross-section, including base

3)the area of cross section

4)the internal surface area of the tunnel, excluding base, if it's length is $50$m

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We need the figure in question for the references to make more sense. –  Shaktal Aug 2 '12 at 17:56

1 Answer 1

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Because $\angle AOB=90^\circ$, the length of $AB$ is $\sqrt{(3.5)^2+(3.5)^2}=3.5\sqrt{2}$ (Pythagorean Theorem).

The perpendicular distance from $O$ to the line $AB$ is equal to $(1/2)AB$. You can see this by noting that $\angle OAB=45^\circ$. Now you can find the height of the tunnel.

The perimeter of cross-section is the straight part $AB$, which we have found, plus three-quarters of the circumference of a circle of radius $3.5$. It is three-quarters because $\angle AOB=90^\circ$, and $90^\circ$ is one-quarter of $360^\circ$.

The cross-sectional area is three-quarters of the area of a circle of radius $3.5$, plus the area of $\triangle AOB$. The area of this triangle is $(1/2)(3.5)(3.5)$.

Finally, the perimeter of the curvy part of cross-section is three-quarters of the circumference of a circle of radius $3.5$. Multiply by $50$ to find the surface area of the curvy part of the tunnel.

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