MHB Find the area of the "Dodecagon"

[Albert1]

$12\,\, points\,$ $A_1,A_2,A_3,------,A_{12}$$\,\,are\,\, on\,\, a \,\,circle\,\, O\,\,$$(with\,\, radius\,\, r)$

$(for\,\,simplicity:A_1,------,A_{12}\,\, arranged\,\, in\,\ clockwise\ manner)$

$given :$

$\overline{A_1A_2}=\overline{A_3A_4}=\overline{A_5A_6}=\overline{A_8A_9}=\overline{A_{10}A_{11}}=\overline{A_{12}A_{1}}=a$

$\overline{A_2A_3}=\overline{A_4A_5}=\overline{A_6A_7}=\overline{A_7A_8}=\overline{A_{9}A_{10}}=\overline{A_{11}A_{12}}=b$

$Find \,\,the\,\, area \,\,of\,\, the\,\,$ $"Dodecagon"\, A_1A_2-----A_{12}$ $(expressed\,\, in\,\, a\,\,and\,\,b)$

 

[Opalg]

[sp][TIKZ]\coordinate [label=right: $A_3$] (A3) at (0:5cm) ;
\coordinate [label=right: $A_2$] (A2) at (40:5cm) ;
\coordinate [label=above right: $A_1$] (A1) at (60:5cm) ;
\coordinate [label=above: $A_{12}$] (A12) at (80:5cm) ;
\coordinate [label=above left: $A_{11}$] (A11) at (120:5cm) ;
\coordinate [label=left: $A_{10}$] (A10) at (140:5cm) ;
\coordinate [label=left: $A_9$] (A9) at (180:5cm) ;
\coordinate [label=left: $A_8$] (A8) at (200:5cm) ;
\coordinate [label=below left: $A_7$] (A7) at (240:5cm) ;
\coordinate [label=below: $A_6$] (A6) at (280:5cm) ;
\coordinate [label=below: $A_5$] (A5) at (300:5cm) ;
\coordinate [label=below right: $A_4$] (A4) at (340:5cm) ;
\coordinate [label=above right: $A_1$] (A1) at (60:5cm) ;
\draw (A1) -- (A2) -- (A3) -- (A4) -- (A5) -- (A6) -- (A7) -- (A8) -- (A9) -- (A10) -- (A11) -- (A12) -- cycle ;
\draw (0,0) circle (5cm) ;
\fill (0,0) circle (2pt);
\draw (A9) -- (A3) -- (A1) -- (A7) ;
\draw (A5) -- (A11) ;
\draw (A3) -- (A5) -- (A7) -- (A9) -- (A11) -- (A1) ;
\draw (3.7,3.8) node {$a$} ;
\draw (4.8,2) node {$b$} ;
\draw (2.5,-0.22) node {$r$} ;
\draw (-0.3,-0.2) node {$O$} ;
[/TIKZ]

The dodecagon consists of six equilateral triangles like $OA_1A_3$ together with six triangles like $A_1A_2A_3$ (or its mirror image).

The area of each of the equilateral triangles is $\frac{\sqrt3}4r^2$.

The angle $A_1A_2A_3$ is $150^\circ$, because the angle $A_1A_9A_3$ is $30^\circ$ (being half the angle $A_1OA_3$), and opposite angles of a cyclic quadrilateral are supplementary. So the area of the triangle $A_1A_2A_3$ is $\frac12ab\sin150^\circ = \frac14ab.$

The cosine rule in the triangle $A_1A_2A_3$ says that $r^2 = a^2 + b^2 - 2ab\cos150^\circ = a^2 + b^2 + \sqrt3ab.$

Therefore the area of the dodecagon is $6\bigl(\frac{\sqrt3}4(a^2 + b^2 + \sqrt3ab) + \frac14ab\bigr) = \frac{3\sqrt3}2(a^2+b^2) + 6ab.$

[/sp]