Theatre- arithmetic progression

[Johulus]

We are doing arithmetic progression (series) at the moment and this is a task that was among the all others, so I guess it has got something to do with arithmetic series. Anyway, the situation is the following:




Also, there is one more parameter, the width of a seat which is 55 cm.

Questions/tasks are:

1. Of how many rows does auditorium consist?
2. How many seats does the first row have, and how many the last?
3. How many seats are there in other rows?
4. What is the total number of seats in auditorium?
5. What is the surface are of the theatre?

1. I guess this could be read from the picture. So, there are 12 rows.
2. What I did here is : number of seats in the first row \(\displaystyle \dfrac{16.5 \cdot 100}{55}=30\, seats \) number of seats in the last row \(\displaystyle \dfrac{23.1 \cdot 100}{55}=42\, seats \)
3. I don't know how to solve this. Since these seats should be organised by some arithmetic series I guess, then :
\(\displaystyle a_n=a_1+(n-1)d \\ if\, a_n=42\, a_1=30\, n=12\, (auditorium\, consists\, of\, 12\, rows) \, 42=30+11d \, d=\dfrac{12}{11} \)
That means that every next row has \(\displaystyle \dfrac{12}{11} \) seats more than the previous. But then again, how could a number of seats be other than integer. You don't have 0.5 seats, so I don't understand what should be done here. What's the problem.
4. I would be able do solve this using the formula for the sum of a arithmetic series if the previous task was correct.
5. What I did here is : \(\displaystyle r_1\, (radius\, of\, the\, circle\, on\, which\, the\, first\, curve\, is)\, r_2\, (radius\, of\, the\, last\, row-\, from\, the\, center\, of\, all\, curves\, to\, the\, last\, row)\, l=\dfrac{r\pi\alpha}{180}\, (length\, of\, a\, circle\, arc) \\ \dfrac{l_1}{l_2}=\dfrac{\dfrac{r_1\pi\alpha}{180}}{\dfrac{r_2\pi\alpha}{180}} \, \, r_2=r_1+20m \\ \dfrac{16.5}{23.1}=\dfrac{r_1}{r_1+20} \\ r_1=50m \, \, r_2=70m \\ l_1=\dfrac{r_1\pi\alpha}{180} \, \, \alpha=\dfrac{180l_1}{r_1\pi}\, \, \alpha=18.9076^{°} \\ P_{theatre}=\dfrac{(r_2^{2}-r_1^{2})\pi\alpha}{360} P_{theatre}=395.999 \dotso m^2 \\ P_{theatre}\approx 396m^2\)

I would appreciate it if someone would put me on the right path.

 

[stapel]

We are doing arithmetic progression (series) at the moment and this is a task that was among the all others, so I guess it has got something to do with arithmetic series. Anyway, the situation is the following:




Also, there is one more parameter, the width of a seat which is 55 cm.

Questions/tasks are:

1. Of how many rows does auditorium consist?
2. How many seats does the first row have, and how many the last?
3. How many seats are there in other rows?
4. What is the total number of seats in auditorium?
5. What is the surface are of the theatre?

1. I guess this could be read from the picture. So, there are 12 rows.

2. What I did here is : number of seats in the first row \(\displaystyle \dfrac{16.5 \cdot 100}{55}=30\, seats \) number of seats in the last row \(\displaystyle \dfrac{23.1 \cdot 100}{55}=42\, seats \)

3. I don't know how to solve this. Since these seats should be organised by some arithmetic series I guess, then :
\(\displaystyle a_n=a_1+(n-1)d \\ if\, a_n=42\, a_1=30\, n=12\, (auditorium\, consists\, of\, 12\, rows) \, 42=30+11d \, d=\dfrac{12}{11} \)
That means that every next row has \(\displaystyle \dfrac{12}{11} \) seats more than the previous. But then again, how could a number of seats be other than integer. You don't have 0.5 seats, so I don't understand what should be done here. What's the problem.

My guess would be that you're expected to find the mathematical values (being, as you pointed out, unrealistic) and then, from these, find the real-world answers. For instance, if the next row should, mathematically, have 12/11 more seats, then you'll put one more seat in that row. (Remember that the seating is likely staggered in some manner, so the inconsistent seat-addition will likely be largely unnoticed by patrons.)

Once you get to where you're adding at least 22/11 = 2 seats, add exactly two. Then when you reach 33/11 = 3 seats, add three. And so forth.

You'll be keeping a separate, "real world" running total of the seats in each row, which you'll use in the next part.

I'm not sure I understand what they're asking for the fifth part. "Surface area" applies to three-dimensional surfaces, but your working suggests that they're only asking about square footage. Also, "the theatre" would be the entire structure, or at least the entire space, but you seem to be working only with the drawn area...?

 

[Johulus]

My guess would be that you're expected to find the mathematical values (being, as you pointed out, unrealistic) and then, from these, find the real-world answers. For instance, if the next row should, mathematically, have 12/11 more seats, then you'll put one more seat in that row. (Remember that the seating is likely staggered in some manner, so the inconsistent seat-addition will likely be largely unnoticed by patrons.)

Once you get to where you're adding at least 22/11 = 2 seats, add exactly two. Then when you reach 33/11 = 3 seats, add three. And so forth.

You'll be keeping a separate, "real world" running total of the seats in each row, which you'll use in the next part.

I'm not sure I understand what they're asking for the fifth part. "Surface area" applies to three-dimensional surfaces, but your working suggests that they're only asking about square footage. Also, "the theatre" would be the entire structure, or at least the entire space, but you seem to be working only with the drawn area...?

The problem is the translation because I am not a native English speaker and I don't know all those technical terms in English. Considering what you have just said, yes, they're asking for square footage. They are asking for the area of this figure, like for example the area of a square is a^2. I didn't know that you say square footage for that.

Since all those circle arcs are part of concentric circles, I just got the radius of the smallest curve and the largest, calculated the area of the bigest circle and subtracted the area of a small circle. By doing so I got the area of the annulus and ,since this figure on the picture is the part of that annulus, I just divided the area of the annulus with 360 degrees and mltiplied by 18.... degrees, which corresponds to all those curves and which I previously got. Hope that's all right. And thank you very much on your feedback.