Question

Find the area of a regular hexagonwith a side length of 10 cm. Round tothe nearest tenth.a10 cm[?] cm2

Answer 1

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:

`\begin{gathered} \text{Area,} \\ A=(1)/(2)bc*\sin A \\ A=\text{ }(1)/(2)(10)(10)*\sin 60 \\ A=\text{ }(1)/(2)(10)(10)*\frac{\sqrt[]{3}}{2} \\ A=\text{ }25\sqrt[]{3} \end{gathered}`

Therefore the total area of 6 similar equilateral triangles will be:

`\begin{gathered} \text{Total area,} \\ T_a=\text{ 6}*25\sqrt[]{3} \\ T_a=\text{ 1}50\sqrt[]{3} \\ T_a=259.8\text{ (nearest tenth)} \end{gathered}`

Final answer:

The area of the hexagon to the nearest tenth is 259.8 sq cm