SOLUTION:
Case: Area of a regular hexagon
Method: The method to be used will split the regular hexagon into 6 equal equilateral triangles.
Each equilateral triangle will be
The area of each equilateral triangle is given by the formula:
[tex]\begin{gathered} \text{Area,} \\ A=\frac{1}{2}bc\times\sin A \\ A=\text{ }\frac{1}{2}(10)(10)\times\sin 60 \\ A=\text{ }\frac{1}{2}(10)(10)\times\frac{\sqrt[]{3}}{2} \\ A=\text{ }25\sqrt[]{3} \end{gathered}[/tex]
Therefore the total area of 6 similar equilateral triangles will be:
[tex]\begin{gathered} \text{Total area,} \\ T_a=\text{ 6}\times25\sqrt[]{3} \\ T_a=\text{ 1}50\sqrt[]{3} \\ T_a=259.8\text{ (nearest tenth)} \end{gathered}[/tex]
Final answer:
The area of the hexagon to the nearest tenth is 259.8 sq cm
