jmc

First drop perpendiculars from $D$ and $C$ to $\overline{AB}$. Let $E$ and $F$ be the feet of the perpendiculars to $\overline{AB}$ from $D$ and $C$, respectively, and let $$h=DE=CF,x=AE,\text{and}y=FB.$$ Then $$25=h^2+x^2,144=h^2+y^2,\text{and}13=x+y.$$So $$144=h^2+y^2=h^2+(13-x{)}^2=h^2+x^2+169-26x=25+169-26x,$$which gives $x=50/26=25/13$, and $$h=\sqrt{5^2-{\left(\frac{25}{13}\right)}^2}=5\sqrt{1-\frac{25}{169}}=5\sqrt{\frac{144}{169}}=\frac{60}{13}.$$Hence $$\text{Area }(ABCD)=\frac{1}{2}(39+52)\cdot \frac{60}{13}=\boxed{210}.$$$$\text{OR}$$Extend $\overline{AD}$ and $\overline{BC}$ to intersect at $P$. Since $△PDC$ and $△PAB$ are similar, we have $$\frac{PD}{PD+5}=\frac{39}{52}=\frac{PC}{PC+12}.$$So $PD=15$ and $PC=36$. Note that $15$, $36$, and $39$ are three times $5$, $12$, and $13$, respectively, so $\angle APB$ is a right angle. The area of the trapezoid is the difference of the areas of $△PAB$ and $△PDC$, so $$\text{Area}(ABCD)=\frac{1}{2}(20)(48)-\frac{1}{2}(15)(36)=\boxed{210}.$$ $$\text{OR}$$Draw the line through $D$ parallel to $\overline{BC}$, intersecting $\overline{AB}$ at $E$. Then $BCDE$ is a parallelogram, so $DE=12$, $EB=39$, and $AE=52-39=13.$ Thus $D{E}^2+A{D}^2=A{E}^2$, and $△ADE$ is a right triangle. Let $h$ be the altitude from $D$ to $\overline{AE}$, and note that $$\text{Area}(ADE)=\frac{1}{2}(5)(12)=\frac{1}{2}(13)(h),$$so $h=60/13$. Thus $$\text{Area}(ABCD)=\frac{60}{13}\cdot \frac{1}{2}(39+52)=\boxed{210}.$$