The Area of the perfect star shape is,

\begin{equation}

A =\frac{1}{2} nbD \>\>\>\>(1)

\end{equation}

Where n is a number of vertices, D is a diagonal from original o to vertex c and b is the base of triangles.

Here the perfect start shapes mean that the diagonals in all triangles to original are equal.

Proof:

The area is,

\begin{equation}

A_1 =\frac{1}{2} {bh} +\frac{1}{2} {bh’}

\end{equation}

\begin{equation*}

A_1=\frac{1}{2} b(h+h’)

\end{equation*}

Where, h+h’=D, Thus,

\begin{equation} \label{eq:mak}

A_1 = \frac{1}{2} bD \>\>\>\>(2)

\end{equation}

Then the Area of Star shape is a total of all triangles which is,

\begin{equation}

A=\frac{1}{2} {nbD}

\end{equation}

The area for four vertices is,

\begin{equation} \label{eq:mako}

A=\frac{1}{2} (4 bD)={2bD} \>\>\>\>(3)

\end{equation}

The area for five vertices is,

\begin{equation} \label{eq:makoo}

A=\frac{5}{2} {bD} \>\>\>\>(4)

\end{equation}