Question

Please help me find the area of this polygon

Answer 1

`216.4 {mm}^(2)`

Explanation

The regular polygon has 9 sides.

Each central angle is

`(360)/(n) = (360)/(9) = 40 \degree`

The area of each isosceles triangle is

`(1)/(2) {r}^(2) \sin( \theta)`

We substitute the radius and the central angle to get:

`(1)/(2) * {8.65}^(2) * \sin(40) = 24.05 {mm}^(2)`

We multiply by 9 to get the area of the regular polygon

`9 * 24.05 = 216.4 {mm}^(2)`

Answer 2

The area of the polygon = 216.4 mm²

Explanation:

* Lets talk about the regular polygon

- In the regular polygon all sides are equal in length

- In the regular polygon all interior angles are equal in measures

- When the center of the polygon joining with its vertices, all the

triangle formed are congruent

- The measure of each vertex angle in each triangle is 360°/n ,

where n is the number of its sides

* Lets solve the problem

- The polygon has 9 sides

- We can divide it into 9 isosceles triangles all of them congruent,

if we join its center by all vertices

- The two equal sides in each triangle is 8.65 mm

∵ The measure of the vertex angle of the triangle = 360°/n

∵ n = 9

∴ The measure of the vertex angle = 360/9 = 40°

- We can use the area of the triangle by using the sine rule

∵ Area of the triangle = 1/2 (side) × (side) × sin (the including angle)

∵ Side = 8.65 mm

∵ The including angle is 40°

∴ The area of each triangle = 1/2 (8.65) × (8.65) × sin (40)°

∴ The area of each triangle = 24.04748 mm²

- To find the area of the polygon multiply the area of one triangle

by the number of the triangles

∵ The polygon consists of 9 congruent triangles

- Congruent triangles have equal areas

∵ Area of the 9 triangles are equal

∴ The area of the polygon = 9 × area of one triangle

∵ Area of one triangle = 24.04748 mm²

∴ The area of the polygon = 9 × 24.04748 = 216.42739 mm²

* The area of the polygon = 216.4 mm²