Question

Pg 24, q51. How do I figure this out? thanks

Answer 1

`A_n = piR^2 sin((2pi)/n)/((2pi)/n)`

Explanation:

Consider for the `n`-sided regular polygon the triangle limited by one side and the radii joining the center with the two consecutive vertices limiting the side.

For symmetry reasons, the area of the polygon is `n` times the area of the triangle.

Moreover the triangle is isosceles and the angle at the top is:

`alpha = (2pi)/n`

so that we can see that the height of the triangle is:

`h= R cos(alpha/2) = R*cos(pi/n)`

while the length of the side is:

`l= 2Rsin(alpha/2) = 2Rsin(pi/n)`

The area of the triangle is then:

`a_n = (l*h)/2 = (2R^2sin(pi/n)cos(pi/n))/2 = R^2/2 sin((2pi)/n)`

and the area of the polygon is:

`A_n= (nR^2)/2 sin((2pi)/n)`

If we write this as:

`A_n = piR^2 sin((2pi)/n)/((2pi)/n)`

We can see that as `n` increases, `A_n` tends to the area of the circumscribed circle:

`lim_(n->oo) A_n = piR^2 lim_(n->oo)sin((2pi)/n)/((2pi)/n) = piR^2`

Answer 2

See below

Explanation:

The regular polygons are able to be split into equal isosceles triangles.

The number of these triangles is equal to the number of sides in the polygon, so for `n=5` there are 5 equal isosceles triangles.

The area for one of these triangles is equal to `1/2ab sinC`, however, since `a=b=1`, the area is `sinC/2`

`C` is easy to find, by doing `360^circ/n` or `(2pi)/n`. This is because for 1 full circle, it is `360^circ = 2pi`. And in terms of full triangles, these triangles fit into the circle `n` times.

`360^circ = 2pi:"full circle"`

`"full circle"/n:360^circ/n = (2pi)/n`

N.B. If an equation uses `pi` then it is in radians, if it uses `""^circ` then it is in degrees. The equations work either way

So, we know have an area for one triangle, `sin((2pi)/n)/2=sin(360^circ/n)/2`

However, we need to total area.

`"Total area"=Sigmasin(360^circ/n)/2=nsin(360^circ/n)/2`
`"Total area"=Sigmasin((2pi)/n)/2=nsin((2pi)/n)/2`

`color(white)(l)`
`color(white)(l)`
`color(white)(l)`

n=5:
`"Total area"=5sin(360^circ/5)/2~~2.38`
`"Total area"=5sin((2pi)/5)/2~~2.38`

n=6:
`"Total area"=6sin(360^circ/6)/2=(3sqrt(3))/2`
`"Total area"=6sin((2pi)/6)/2=(3sqrt(3))/2`

n=7:
`"Total area"=7sin(360^circ/7)/2~~2.74`
`"Total area"=7sin((2pi)/7)/2~~2.74`

n=8:
`"Total area"=8sin(360^circ/8)/2=2sqrt(2)`
`"Total area"=8sin((2pi)/8)/2=2sqrt(2)`

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`color(white)(l)`
`color(white)(l)`

In general, as the value of `n` increases to `oo`, the size of the triangles will become much smaller, and will be more likely to take on the shape of a circle. Therefore, the combined are will slowly reach `pi`.

With radians: `lim_(n->+oo)(nsin((2pi)/n))/2=pi`
With degrees: `lim_(n->+oo)(nsin(360^circ/n))/2=pi`

This effect can be explored further, using just a value for the angle, here. Keep in mind that the notes sections don't allow for formatting, i.e. fractions.