Solution:

Here $AB = 3 + 9 = 12,$ $AC = 10 + 27 = 37,$ and angle $B$ is right, so $BC = \sqrt{37^2 - 12^2} = 35$ and $[ABC] = \frac{1}{2} \cdot 12 \cdot 35 = 210.$ The quadrilateral is triangle $ADC$ with triangle $AGF$ removed, where $G$ is the intersection of $\overline{AD}$ and $\overline{EF}.$

By the angle bisector theorem in triangle $ABC,$ $BD : DC = AB : AC = 12 : 37,$ so $[ADC] = \frac{37}{49} \cdot 210 = \frac{1110}{7}.$ In triangle $AEF,$ ray $AG$ bisects the same angle, so $EG : GF = AE : AF = 3 : 10$ and $[AGF] = \frac{10}{13}\,[AEF].$ Also $[AEF] = \frac{AE}{AB} \cdot \frac{AF}{AC}\,[ABC] = \frac{3}{12} \cdot \frac{10}{37} \cdot 210 = \frac{525}{37}.$

Therefore $$[DCFG] = \frac{1110}{7} - \frac{10}{13} \cdot \frac{525}{37} = \frac{1110}{7} - \frac{5250}{481} \approx 158.57 - 10.92 = 147.66,$$ and the closest integer is $148.$