jmc

The formula for the area of a trapezoid is $\frac{1}{2}h\times (b_1+b_2)$, with $h$ being the height, $b_1$ being the shorter base, and $b_2$ being the longer base. We can find the height of this particular trapezoid with algebra: $$\begin{array}{rl}300& =\frac{1}{2}h\times (20+30)\\ 600& =h\times 50\\ h& =12\end{array}$$Now that we know the height of the trapezoid, we can find the area of triangle $ADC$, whose base is $30$ (the longer base of the trapezoid), and whose height is $12$. Therefore, the area of triangle $ADC=\frac{1}{2}\cdot 30\times 12=180$. We can use this information to find that the area of triangle $ABC$, or the upper portion of the trapezoid, is $300-180=120$. Now we need to separate the area of $BXC$ from $AXB$, knowing that $ABC=120$. Because trapezoid $ABCD$ is not necessarily an isosceles trapezoid, nothing can be assumed about the diagonals, except that they will cut each other, and the height, in the same ratio as the bases, or $2:3$. The height of the trapezoid, $12$ units, is therefore divided into the heights of triangles $DXC$ and $AXB$. We can find these heights with the equation, letting $x$ be the height of triangle $DXC$: $$\begin{array}{rl}\frac{2}{3}\cdot x+x& =12\\ x\left(\frac{2}{3}+1\right)& =12\\ \frac{5}{3}x& =12\\ x& =7.2\end{array}$$So, the height of triangle $AXB$ is $\frac{2}{3}\times 7.2=4.8$. We know that $AB$, the base of $AXB$, is $20$ units, so the area of $AXB=\frac{1}{2}(20)\times 4.8=48$. Therefore, the area of triangle $BXC=120-48=\boxed{72}$ square units.