find the area of a regular 18-gon with radius of 14 mmA≈ __ mm squared do not found until the final answer, then round to the nearest tenth as needed.

[tex]\begin{gathered} \text{ }\theta\text{ }=\text{ }\frac{180\text{ \lparen18 - 2\rparen}}{18} \\ \\ \theta\text{ }=\text{ }\frac{180\text{ }\times\text{ 16}}{18} \\ \\ \theta\text{ }=\text{ }\frac{2880}{18} \\ \theta\text{ }=\text{ 160}\degree \end{gathered}[/tex]

Step 2; Find the base angle of the triangle

Recall, that all regular polygon can be divided into isosceles triangle by joining the vertices to the center. Hence, the base angle can be calculated below as

[tex]\begin{gathered} \text{ Base angle = }\frac{160}{2} \\ \text{ Base angle = 80}\degree \end{gathered}[/tex]

Step 3; Find the height of triangle using trigonometric

[tex]\begin{gathered} \text{ tan }\theta\text{ }=\text{ }\frac{\text{ opposite}}{\text{ adjacent}} \\ \text{ } \end{gathered}[/tex]

Since the radius of the polygon is 14mm, therefore, the base length is

[tex]\begin{gathered} \text{ Base length = }\frac{14}{2} \\ \text{ Base length = 7mm} \end{gathered}[/tex][tex]\begin{gathered} \text{ Tan 80 = }\frac{h}{7} \\ \text{ cross multiply} \\ \text{ h = tan 80 }\times\text{ 7} \\ \text{ h = 7tan 80} \\ \text{ tan 80 = 5.671} \\ \text{ h = 7 }\times\text{ 5.671} \\ \text{ h = 39.697 mm} \end{gathered}[/tex]

Step 4; Find the area of the triangle

[tex]\begin{gathered} \text{ Area of a triangle = }\frac{1}{2}bh \\ \text{ Area of a trianlge = }\frac{1}{2}\times7\times39.697 \\ \text{ Area of a triangle = }\frac{277.879}{2} \\ \text{ Area of a triangle = 138.94 mm}^2 \end{gathered}[/tex]

Step 5; Find the area of the polygon

Since there are 18 triangles in the polygon, then calculate the area of the 18-gon

Area of 18-gon = 18 x 138.94

Area of 18-gon = 2500.9 mm^2

Therefore, the area of the 18-gon is 2500.