Question

Urgent, It is a Calculus question and I’ll appreciate your help. Thanks

Answer 1

Answer:

$\large \boxed{\sf \ \ \pi r^2 \ \ }$

Step-by-step explanation:

Hello,

We will follow the instructions and then we need first to find the area of any regular n-gon.

I attached one graph so that it is easier to understand.

The n-gon can be divided in n similar isosceles triangles.

So we can find the area of one of these triangles and then multiply by n to get the total area, right?

Let's focus on the triangle OAB then. OA = OB = r, right?

The area of this triangle is the (altitude * AB ) / 2

The angle AOB is $\dfrac{2\pi}{n}$ by construction of the regular n-gon.

So half this angle is $\dfrac{\pi}{n}$ and we can use cosine rule to come up with the altitude:

   $\boxed{ altitude = r\cdot cos(\dfrac{\pi}{n}) }$

Using the sine rule we can write that

   $\boxed{ AB=2r\cdot sin{\dfrac{\pi}{n}} }$

So, the area of the triangle is

   $\dfrac{1}{2}\cdot r\cdot cos(\dfrac{\pi}{n}) } \cdot 2r\cdot sin(\dfrac{\pi}{n})\\\\=r^2\cdot cos(\dfrac{\pi}{n})\cdot sin(\dfrac{\pi}{n})$

Ok, but wait, we know how to simplify. We can use that

$sin(2\theta)=2cos(\theta)sin(\theta)$

So it gives that the area of one triangle is:

   $\dfrac{1}{2}r^2sin(\frac{2\pi}{n})$

Last step, we need to multiply by n.

$\large \boxed{\sf \ \ A(n)=\dfrac{1}{2}nr^2sin(\dfrac{2\pi}{n}) \ \ }$

Now, let's use a result from Calculus:

$\displaystyle \lim_{x\rightarrow0} \dfrac{sin(x)}{x}=1$

How to use it here? Just notice that

$\displaystyle \lim_{n\rightarrow +\infty} \dfrac{sin(\dfrac{2\pi}{n})}{\dfrac{2\pi}{n}}=1\\\\\lim_{n\rightarrow +\infty} \dfrac{n}{2\pi}sin(\dfrac{2\pi}{n})=1\\\\\lim_{n\rightarrow +\infty} nsin(\dfrac{2\pi}{n})=2\pi$

So, finally

$\Large \boxed{\sf \lim_{n\rightarrow +\infty} A(n)=\dfrac{1}{2}r^22\pi=\pi r^2 }$

Hope this helps.

Do not hesitate if you need further explanation.

Thank you