\n A=\\frac{5}{2} {bD} \\>\\>\\>\\>(4)
\n\\end{equation}<\/p>\n\n\n\n
![\"\"](\"http:\/\/mako-sawin.com\/wp-content\/uploads\/2022\/01\/starfour-e1642549708618.jpg\")
<\/p>\n\n\n
The Area of the perfect star shape is,<\/p>\n
\\begin{equation}
\n\u00a0 A =\\frac{1}{2} nbD \\>\\>\\>\\>(1)
\n\\end{equation}
\nWhere n is a number of vertices, D is a diagonal from original o to vertex c and b is the base of triangles.<\/p>\n
Here the perfect start shapes mean that the diagonals in all triangles to original are equal.<\/p>\n
Proof: <\/p>\n
The area is,<\/p>\n
\\begin{equation}
\n A_1 =\\frac{1}{2} {bh} +\\frac{1}{2} {bh’}
\n\\end{equation}<\/p>\n
\\begin{equation*}
\n A_1=\\frac{1}{2} b(h+h’)
\n\\end{equation*}<\/p>\n
Where, h+h’=D, Thus,<\/p>\n
\\begin{equation} \\label{eq:mak}
\nA_1 = \\frac{1}{2} bD \\>\\>\\>\\>(2)
\n\\end{equation}<\/p>\n
Then the Area of Star shape is a total of all triangles which is,<\/p>\n
\\begin{equation}
\n A=\\frac{1}{2} {nbD}
\n\\end{equation} <\/p>\n
The area for four vertices is, <\/p>\n
\\begin{equation} \\label{eq:mako}
\n A=\\frac{1}{2} (4 bD)={2bD} \\>\\>\\>\\>(3)
\n \\end{equation} <\/p>\n
The area for five vertices is,<\/p>\n
\\begin{equation} \\label{eq:makoo}
\n A=\\frac{5}{2} {bD} \\>\\>\\>\\>(4)
\n\\end{equation}<\/p>\n\n\n\n