If 15 and 9 are lengths of two medians of a triangle, what is the...

Area of $\triangle \mathrm{BDG}$
\begin{aligned} & \text { S }=\frac{11+x}{2} \\ & \begin{aligned} \text { Area }= & \sqrt{\frac{11+x}{2}\left(\frac{11+x}{2}-x\right)\left(\frac{11+x}{2}-5\right)\left(\frac{11+x}{2}-6\right)} \\ & =\sqrt{\frac{11+x}{2}\left(\frac{11-x}{2}\right)\left(\frac{x+1}{2}\right)\left(\frac{x-1}{2}\right)} \\ & =\sqrt{\frac{\left(121-x^2\right)\left(x^2-1\right)}{16}}\end{aligned} \\ & \text { Let } \mathrm{y}=\frac{\left(121-\mathrm{x}^2\right)\left(\mathrm{x}^2-1\right)}{16} \\ & \text { For } \mathrm{max} \text { area } \frac{\mathrm{dy}}{\mathrm{dx}}=0 \\ & -2 \mathrm{x}\left(\mathrm{x}^2-1\right)+\left(121-\mathrm{x}^2\right) 2 \mathrm{x}=0 \\ & -\mathrm{x}^2+1+121-\mathrm{x}^2=0\end{aligned}$\begin{aligned} & 2 x^2=122 \\ & x^2=61 \\ & \text { Area of } \Delta \mathrm{BDG}=\sqrt{\frac{(121-61)(61-1)}{16}} \\ & =\sqrt{\frac{60 \times 60}{16}}=\frac{60}{4}=15 \end{aligned}$$Area of$\triangle \mathrm{ABC}=6 \times 15=90$