Help with a problem? Can't solve it.

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sofalugo

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A king wanted to build a tower to store all the water for him and his whole family and court.

The base of the tower is in the shape of a regular triacontakaihenagon, with each side measuring one metre on the exterior, and the tower rises vertically to a height of 31 metres, and it has a flat roof. The walls and roof are 30 centimetres thick.

How many litres of water are required to fill the entire tower?

Need it fast! Help please :(
 
sofalugo said:
A king wanted to build a tower to store all the water for him and his whole family and court.

The base of the tower is in the shape of a regular triacontakaihenagon, with each side measuring one metre on the exterior, and the tower rises vertically to a height of 31 metres, and it has a flat roof. The walls and roof are 30 centimetres thick.

How many litres of water are required to fill the entire tower?

Need it fast! Help please :(
Click to expand...

triacontakaihenagon - how many sides does it have.

What is the expression for the volume of a regular prism?

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.
 
It's a whole lot easier to say "31-gon". I was bored. Here's almost all of it. I did not rely on any known formula for regular n-gons. That certainly would be easier.

Let's take just one of the 31 section.
----- Exterior side = 1m. We'll use this as the base of an isosceles triangle with
----- Interior sides of unknown length. Let's call this length "L".
----- The interior angle is quite simple. It must be 360º / 31. I'll just leave it in that form.
----- This makes the two interior angles at the base exactly 2610º / 31. Check it out. 2610*2 + 360 = 31*180.

That's enough of that. Now, let's cut it in half. Construct a perpendicular bisector on the base. It will hit the apex angle and create two congruent right triangles.
----- Each of these right triangles (there are 62 if we go all the way around the 31-gon) has base ½ m, hypotenuse L, and height that we have not named. Let's call it 'a'.
----- It should be obvious that L^2 = a^2 + (½)^2
----- It should also be obvious that tan(2610º/31) = a/(½) = 2a or sin(2610º/31) = a/L.
----- Use either of these to solve for 'L'. L = a/sin(2610º/31)
----- Use this to solve for 'a'. [a/sin(2610º/31)]^2 = a^2 + (½)^2
----- Reality check, a and L should be about the same! Why?
----- You will notice that I did not tell you the values of 'a' and "L". Let's see what you get.
----- Now, if we REALLY wanted, we could easily calculate the area of the entire base. Too bad this is the exterior dimension.

Off to the interior dimension. Let's keep our focus on just one of the 62 right triangles.
----- If we shorten up 'a' by 30 cm (0.30 m), we get a smaller but SIMILAR triangle.
----- We know enough about it that some simple proprotions will do.
----- Shortened 'a', let's call it 'p', and the shortened base, call it 'b', have this relationship a/p = (½)/b
----- Since we know p = a - 0.30 m, we can solve for 'b' without much problem.
----- Reality check. 'b' should be a little less than 1/2. Why?
----- With p and b, we are ready to solve the problem. Calculate the area of the right triangle with 'b' as base and 'p' as height. Multiply this by 62.

You're almost done.

Staying organized and directed - this is the real hard part of this problem.
 
Thank you for your help :). I just tried and I don't seem to figure it out at all. I kind of solve other problems but this one just overdue me. I finished in summer school last year :p.

If you could express like just the numbers and like the operations, I'd be more than pleased. I'd try to make it.
 
Not a prayer for that. What fun would you get out of it?! :wink: The description is all there.
 
Please do not offer any further help on this problem. This is a question that is being used in an online competition.
 
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