Let BE=h and CE=x, so EA=14-x. We have two right-angled triangles back to back: ABE and BCE.
By Pythagoras h^2+x^2=64 and h^2+(14-x)^2=144, or h^2+x^2-28x+196=144.
But h^2+x^2=64 from the first equation, so 64-28x+196=144 and 28x=64+196-144=116.
Therefore x=116/28=29/7. So we can find h=√(64-(29/7)^2)=6.8437cm approx.
The area of the triangle is (1/2)14h=47.9062 sq cm approx.