Area of Regular Six-Pointed Star

From the figure shown, angles ADC, AOB, and BOC are equal; all are denoted by θ. See the relationship between inscribed and central angles for detailed explanation about the equality of these angles.
 

$2\theta = \frac{1}{6}(360^\circ)$

$\theta = 30^\circ$
 

$\tan \theta = \dfrac{h}{r/2}$

$\tan 30^\circ = \dfrac{h}{20/2}$

$h = 5.7735 \, \text{ cm }$
 

Area of triangle ABO:
$A_{ABO} = \frac{1}{2}rh$

$A_{ABO} = \frac{1}{2}(20)(5.7735)$

$A_{ABO} = 57.735 \, \text{ cm}^2$
 

Area of Hexagram (the six-pointed star)
$A = 12\,A_{ABO}$

$A = 12(57.735)$

$A = 692.82 \, \text{ cm}^2$           answer