Question

A regular heptagon with $7$ sides is inscribed in a circle with radius $9$ mm.

What is the area of the figure?

Answer 1

Step-1: Finding the angle subtended by two radii.

• Let the radius be $r$, and the angle subtended be $\phi$.
• Let the number of sides be $n$.
• As it is a heptagon, then $n=7$.
• The formula for angle subtended is $\frac{360}{n}$.

Putting the value accordingly:

$$\begin{gathered} \phi =\frac{360}{n} \\ \Rightarrow \phi =\frac{360}{7} \\ \Rightarrow \phi =51.43° \end{gathered}$$

Step-2: Finding the area of the triangle between the radii.

The adjacent two radii and the chord between the two radii form a triangle.
Let the area of the formed triangle be $A$:

$A=\frac{r^2\sin \phi }{2}$

Step-3: Finding the area of the heptagon.

Seven such triangles will form the heptagon.

So, the area of the heptagon can be written as $7A$:

$$\begin{gathered} A_t=7A \\ \Rightarrow A_t=\frac{7r^2\sin \phi }{2} \\ \Rightarrow A_t=\frac{7{\left(9\right)}^2\sin 51.43°}{2} \\ \Rightarrow A_t=\frac{7\times 81\times 0.7818}{2} \\ \Rightarrow A_t=\frac{443.3}{2} \\ \Rightarrow A_t=221.65 \end{gathered}$$

Therefore, the area of the heptagon is $221.65$ mm sq.