Suppose we have a trapezium cut into two triangles along the diagonal. The top and bottom sides are given in terms of
\[x\]
and the height is 4.

The area of BCD is three times the area of ABC. The height of both triangles is 4.
\[\frac{1}{2} (x+4) \times 4=3 \times \frac{1}{2} x \times 4\]
\[6x=2x+8\]
\[4x=8 \rightarrow x=2\]