MHB Find Area of Similar Polygons (in Terms of n)

[nicodemus1]

Good Day,

I can't solve the following problem because I don't know how to find the length of the first polygon. That's why my expression for the total area has 2 variables instead of just n.

Any help/ advice on how I can get an expression for the total area in terms of n will be greatly appreciated.

Thanks & Regards

View attachment 2492

 

[soroban]

Re: Geometry and Series

Hello, nicodemus!

There are n similar polygons. The lengths of polygons form an arithmetic sequence with common difference d = 1.

If the area of the smallest polygon is 25 cm², find,
in terms of n, the total area of the n similar polygons.

For simplicity, consider squares.

\begin{array}{ccccc}\text{Square} & \text{Length} & \\ \text{number} & \text{of side} & \text{Area} \\ \hline 1 & 5 & 25 \\ 2 & 6 & 36 \\ 3 & 7 & 49 \\ 4 & 8 & 64 \\ \vdots & \vdots & \vdots \\ n & n+4 & (n+4)^2 \\ \hline \end{array}

We have: .S \;=\;5^2 + 6^2 + 7^2 + 8^2 + \cdots + (n+4)^2The sum of the first n squares is: .\frac{n(n+1)(2n+1)}{6}

The sum of the first n+4 squares is: .\frac{(n+4)(n+5)(2n+9)}{6}

The sum of the first n+4 squares minus the first four squares is:

S \;=\;\frac{(n+4)(n+5)(2n+9)}{6} - (1^2+2^2+3^2+4^2)

 

[nicodemus1]

Good Day.

I still have not been able to solve this question. I have not had a look at it for the past year due to other priorities.

soroban's reply did not help me at all.

I assumed that all n polygons are regular polygons. Therefore, the formula I used for the area was $A={\frac{a^2N}{4}}$cot(${\frac{\pi}{N}}$), where $a$ is the length of a side of the polygon and $N$ is the number of sides of the polygon.

The summation I obtained using this formula was $\sum_{i=0}^{n-1}$${\frac{(a+i)^2N}{4}}$cot(${\frac{\pi}{N}}$), which can also be expressed as ${\frac{N cot({\frac{\pi}{N}})}{4}}\sum_{i=0}^{n-1}(a+i)^2$.

Since the area of the smallest polygon is 25cm$^2$, I got $a = \frac{10}{\sqrt{N cot ({\frac{\pi}{N}})}}$.

But I do not think that I am anywhere near obtaining the solution, which is ${\frac{n(2n^2 + 27n + 121)}{6}}$.

Any form of guidance would be appreciated.

 

[Opalg]

Good Day.

I still have not been able to solve this question. I have not had a look at it for the past year due to other priorities.

soroban's reply did not help me at all.

I assumed that all n polygons are regular polygons. Therefore, the formula I used for the area was $A={\frac{a^2N}{4}}$cot(${\frac{\pi}{N}}$), where $a$ is the length of a side of the polygon and $N$ is the number of sides of the polygon.

The summation I obtained using this formula was $\sum_{i=0}^{n-1}$${\frac{(a+i)^2N}{4}}$cot(${\frac{\pi}{N}}$), which can also be expressed as ${\frac{N cot({\frac{\pi}{N}})}{4}}\sum_{i=0}^{n-1}(a+i)^2$.

Since the area of the smallest polygon is 25cm$^2$, I got $a = \frac{10}{\sqrt{N cot ({\frac{\pi}{N}})}}$.

But I do not think that I am anywhere near obtaining the solution, which is ${\frac{n(2n^2 + 27n + 121)}{6}}$.

Any form of guidance would be appreciated.

Given that $\frac{(n+4)(n+5)(2n+9)}{6} - (1^2+2^2+3^2+4^2)$ simplifies to ${\frac{n(2n^2 + 27n + 121)}{6}}$, you might find it instructive to take another look at soroban's reply.

 

[nicodemus1]

Good Day.

I made a mistake by not taking a closer look and working on soroban's expression.

This was because I assumed that soroban considered a specific scenario only, which is squares in this case, and that it might not answer the question in general.

But after Opalg's post, I believe that the approach soroban used was to first work with squares and then proceed to develop a more general solution which would answer the question.

Is this correct?

 

[Opalg]

View attachment 4397

I made a mistake by not taking a closer look and working on soroban's expression.

This was because I assumed that soroban considered a specific scenario only, which is squares in this case, and that it might not answer the question in general.

But after Opalg's post, I believe that the approach soroban used was to first work with squares and then proceed to develop a more general solution which would answer the question.

Is this correct?

The difficulty I have with this problem is that it is very badly worded. For a start, what is meant by the "length" of a polygon? Is the length of the perimeter, or the length of one side? If the polygon is not regular, then different sides may have different lengths, so which length should we choose? Soroban found that the problem has a neat solution if the polygon is a square, and that "length" means the length of a side of the square.

The next step ought to be to investigate other shapes, to see whether the result holds for them. You have already looked at the case of a regular polygon with $N$ sides, and found that the length of a side of the first polygon would have to be $\frac{10}{\sqrt{N\cot\bigl(\frac\pi N\bigr)}}.$ If $N\ne4$, you will find that this fails to give the claimed formula for the sum of the areas of the first $n$ polygons.

What I conclude from this is that the question is atrociously badly worded. Where it says "length" it should say "length of side", and where it says "polygons" it should say "squares".

 

[nicodemus1]

When I first read the problem, I assumed that "length" meant the "length of a side".

I faced the same dilemma as to what I should do in the case of a polygon which is not regular. That was another reason why I found faced difficulties accepting soroban's solution.

The question following this in the text is worded similarly except that the lengths form a geometric sequence with a common ratio r. I used soroban's method to find the sum of the areas of n similar polygons and obtained the correct answer.