Perpendicular Medians

Is given $CE$ and $BD$ are medians, so $O$ is the centroid.

Then, $CO=\frac { 2 }{ 3 } CE = 8$ and $BO=\frac { 2 }{ 3 } CD = \frac { 16 }{ 3 }$ .

Also, triangle $BOC$ is right and its area is $[BOC]=\frac { 1 }{ 2 } \cdot 8\cdot \frac { 16 }{ 3 } =\frac { 64 }{ 3 }$ .

As O is the centroid, $[BOC] = \frac { 1 }{ 3 }[ABC]$ , because they have same base $BC$ and it height is $\frac { 1 }{ 3 }$ of triangle $ABC$ height.

So, we get $[ABC]=3\cdot \frac { 64 }{ 3 } = 64$ .

finaly, $x = 64$ and $(x+8)=\boxed{72}$

The medians are perpendicular, so consider EC as the base of triangles BEC and DEC. The sum of the areas of BEC and DEC is (BD)(EC)/2. Also, triangle AED is similar to triangle ABC with ratio 1:2, so the area of triangle AED is one fourth the area of ABC.

Our final equation is: (8*12/2)(4/3) = 72, which is the answer.