Area of triangle inside a $30-60-90$ triangle 
In a right angle triangle ABC with hypotenuse BC and C=60 degrees, M and N are the middles of AB and AC respectively. Draw ND perpendicular to BC (D is point on the side BC). If MD=7cm calculate the area of triangle DNM.


I solved this question in the following way:
$NM//=\frac{CB}{2}$ (Thales theorem)
ABC=30 degrees hence $CA=\frac{CB}{2}$ and hene $NM=CA$
$sinNCD=\frac{ND}{CN}$
$\frac{\sqrt{3}}{2}*CN=ND$
$CN=\frac{2}{\sqrt{3}}*ND$
$AC=\frac{4}{\sqrt{3}}*ND$
$MN=\frac{4}{\sqrt{3}}*ND$
since NM//CB then DNM=90 degrees.
Hence, $DN^2+DN^2*\frac{16}{9}=49$ (Pythagoras), so $DN^2*\frac{25}{9}=49$, so $DN*\frac{5}{3}=7$, so $DN=\frac{21}{5}$
From this we have that $MN=\frac{4}{\sqrt{3}}*\frac{21}{5}=\frac{84}{5\sqrt{3}}$.
Hence $Area=\frac{MN*ND}{2}=\frac{42}{5\sqrt{3}}*\frac{21}{5}=\frac{21*42}{25\sqrt{3}}=\frac{882}{25\sqrt{3}}$
Could you please tell me if I am correct and if I am wrong how much would it be reasonable to assume I would get in an olympiad for my work?

You have done a mistake here
$$ DN^2+DN^2*\frac{16}{\color{red}{3}}=49$$
You wrote $9$.
With correct value, I got $$ [DNM] = \dfrac{98\sqrt{3}}{19}$$
