14. Answer: 300
Let $P$ be the intersection of $C M$ and $B N$, so that $P$ is the centroid of $\triangle A B C$. Then $B P=2 P N$ and $C P=2 P M$. Let $P N=P M=y$, so that $B P=C P=2 y$. Since $\angle B P C=90^{\circ}$, we have $B C=20=\sqrt{(2 y)^{2}+(2 y)^{2}}$ and thus $y=\sqrt{50}$. Now
$$
A B^{2}=4 B M^{2}=4\left((2 y)^{2}+y^{2}\right)=20 y^{2}=1000 .
$$

Thus the altitude of $\triangle A B C$ is $\sqrt{A B^{2}-10^{2}}=30$. Hence the area of $\triangle A B C$ is $\frac{1}{2} \times 20 \times 30=300$.
14. In the diagram below, $\triangle A B C$ is an isosceles triangle with $A B=A C$, and $M$ and $N$ are the midpoints of $A B$ and $A C$ respectively. It is given that $C M$ is perpendicular to $B N, B C=20 \mathrm{~cm}$, and the area of $\triangle A B C$ is $x \mathrm{~cm}^{2}$. Find the value of $x$.