Question

Find the area of a regular `15`-gon whose each side is `14mm.`?

Answer 1

Area of a regular `15`-gon with `14mm` sides is `3458` `mm^2`

Explanation:

Any regular polygon with `n` sides can be divided into `n` congruent isosceles triangles by joining its vertices with the center. Let us say we have a polygon of `n` sides with each side measuring `a` units.

Let `AB` be one such side and `O` be center of the regular polygon. Completing the triangle `OAB` and drawing `OM_|_AB`, area of triangle is `(ABxxh)/2`. The figure appears as shown below.

Observe that `m/_AOB=360^@/n`, as there are `n` congruent triangle around `O` and `m/_MOB=180^@/n`

Using trigonometry `(OM)/(MB)=cot(180^@/n)`

i.e. `h=OM=MBxxcot(180^@/n)=a/2xxcot(180^@/n)`

and area of triangle is `1/2xxaxxa/2xxcot(180^@/n)`

= `a^2/4cot(180^@/n)`

and area of polygon is `(na^2)/4cot(180^@/n)`

Therefore area of the regular `15`-gon with `14mm` sides is

`(15xx14^2)/4xxcot12^@` `mm^2`

= `(15xx196)/4xx4.705` `mm^2`

= `3458` `mm^2`

Answer 2

`A= 3,458mm^2`

Explanation:

The size of an interior angle of a regular 15-sided polygon can be calculated:

`theta = (180(n-2))/15 = 156°`

All regular polygons can be divided into isosceles triangles by joining the vertices to the centre. The base angles of the triangles are half of the interior angles of the polygon.

`156° div 2 = 78°`
The angles of the triangles in a 15-sided polygon are `78°, 78° and 24°`

The area of a triangle is: `A = 1/2bh`

To be able to find the height, divide an isosceles triangle into 2 right-angled triangles.
The angles are `90°, 12° and 78°` and the base is `14/2 = 7mm`

The height of the triangle is the side opposite the `78°` angle.

Using trig, we have

`o/a = tan 78" "rarr o = 7tan78`

The area of one of the 15 isosceles triangles is therefore:

`A= 1/2 xx 14xx 7tan78 = 49 tan78" "larr` this is the height

There are 15 such triangles, so the area of the polygon is:

`A = 15 xx 49tan78`

`A=3,457.9mm^2`