Length of medians through acute angles of a right angled triangle are 3 and 4. If area of triangle is $\frac{4}{3}\sqrt{\mathrm{a}} \mathrm{then} \mathrm{a}=\mathrm{\_\_\_\_}$

${\mathrm{BE}}^2=9=\frac{1}{4}\left(2{\mathrm{a}}^2+2{\mathrm{c}}^2-{\mathrm{b}}^2\right) \mathrm{and} {\mathrm{a}}^2={\mathrm{b}}^2+{\mathrm{c}}^2$

$$\begin{gathered} {\mathrm{CF}}^2=16=\frac{1}{4}\left(2{\mathrm{a}}^2+2{\mathrm{b}}^2-{\mathrm{c}}^2\right) \\ {\mathrm{BE}}^2+{\mathrm{CF}}^2=25=\frac{1}{4}\left(4{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2\right)=\frac{5{\mathrm{a}}^2}{4}\Rightarrow {\mathrm{a}}^2=20 \\ \mathrm{Sub} {\mathrm{BE}}^2, {\mathrm{CF}}^2 \mathrm{find} {\mathrm{b}}^2=\frac{16}{3}, {\mathrm{c}}^2=\frac{44}{3} \\ Ar\mathrm{ea}=\frac{1}{2}\mathrm{bc}=\frac{1}{2}\times \frac{4}{\sqrt{3}}\times \frac{2\sqrt{11}}{\sqrt{3}}=\frac{4}{3}\sqrt{11} \end{gathered}$$