# Calculate using cross-section of tunnel [closed]

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

Calculate

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, graydadAug 17 at 14:28

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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

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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