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Diagram of tunnel entrance

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


  1. the height of the tunnel;

  2. the perimeter of its cross-section, including base;

  3. the area of cross section;

  4. the internal surface area of the tunnel, excluding base, if its length is 50m.

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closed as off-topic by Normal Human, drhab, Math1000, TravisJ, graydad Aug 17 at 14:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – drhab, Math1000, graydad
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We need the figure in question for the references to make more sense. –  Shaktal Aug 2 '12 at 17:56
@NormalHuman: The link is not dead, the image is there, you just have to click "Download image". No need to close a reasonable question for this. –  Alex M. Aug 17 at 14:14
@AlexM. Actually, I don't see any question here, reasonable or not. It's a command: calculate... –  Normal Human Aug 17 at 14:22
@NormalHuman: You have deleted your previous comment and now mine looks out of context... Please also note that the question is 3 years old and has received an accepted answer. –  Alex M. Aug 17 at 14:24

1 Answer 1

up vote 1 down vote accepted

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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