Consider the figure.
From the triangle ABD, we have,
[tex]BD=\sqrt[]{16^2-6^2}=14.83[/tex]
Therefore,
[tex]BC=14.83-10=4.83[/tex]
From the traingle ABC, we have,
[tex]AC=\sqrt[]{4.83^2+6^2}=7.7[/tex]
Thus, the area of the traingle is,
[tex]\frac{1}{2}\times7.7\times10=38.5[/tex]
