Geometry

Let $A$ be the top vertex of the star, $O$ be the center of the star, (and circle), and $B$ be the vertex of the star immediately below and to the right of $A$ . Thus $AO = 10$ .

Since the interior angle of each of the 'points' of the star is one-half the central angle of a pentagon, we have that $\angle OAB = (\frac{1}{2})(\frac{1}{2}) 72^{\circ} = 18^{\circ}$ .

Now the star is composed of $10$ triangles congruent to $\Delta AOB$ , and thus $\angle AOB = (\frac{1}{10}) 360^{\circ} = 36^{\circ}$ .

This means that $\angle ABO = 126^{\circ}$ . Now use the Sine Law on $\Delta AOB$ to find that $AB = AO * \dfrac{\sin(36^{\circ})}{\sin(126^{\circ})} = 10*\tan(36^{\circ})$ .

Thus $\Delta AOB$ has a base length of $10$ and a height of $10*\tan(36^{\circ})\sin(18^{\circ})$ , and so the area of the star is

$A = 10*(\frac{1}{2})*10*10*\tan(36^{\circ})\sin(18^{\circ}) = 500*\tan(36^{\circ})\sin(18^{\circ})$ .

The solution is then $8.841*A = \boxed{992.464}$ to $3$ decimal places.

(This differs slightly from the posted answer of $992.524$ , (probably due to rounding), so I'm guessing that any answer that rounds to $992.5$ will be considered correct.)

By sine law, we have

$\dfrac{x}{\sin 36}=\dfrac{10}{\sin 126}$

$x \approx 7.2654$

$A=10\left(\dfrac{1}{2}\right)(x)(10)(\sin 18)=5(7.2654)(10)(\sin 18) \approx112.257$

The desired answer is $8.841(112.257)=$ $\boxed{992.464}$