Area of Triangle made from Incircle and Circumcircle

Using a combination of Heron's Formula and another formula for the area of a triangle, $rs$ (with $r$ the inradius and $s$ the semiperimeter), we can find the radius of $\Gamma_{1}$ to be $4$ . We now have a side-length of $IDE$ , $ID=4$ . If we can find the altitude from $ID$ , we'll be done.

Notice that because $A,$ $I,$ and $E$ are all concurrent, and because $I$ is the incenter of $ABC$ , then by definition, $E$ bisects $\stackrel \frown{BC}$ . Dropping a perpendicular from $BC$ to $E$ , we get the following: Since $E$ bisects $\stackrel \frown{BC}$ then F bisects $BC$ . Because $BC=14,$ $CF=7$ . Now define point of tangencies of $\Gamma_{1}$ on $AB$ and $AC$ , $G$ and $H$ respectively. Then we have $AG=AH=x,$ $BD=BG=y,$ and $CD=CH=z$ . We have $x+y=13$ , $y+z=14$ , $x+z=15$ . Solving gives $x=7$ , $y=6$ , and $z=8$ . We only care about $CD=z=8$ , because we get $FD=CD-CF=8-7=1$ . We now have the altitude and can find the area: $\frac{1\times4}{2}=\boxed{2}.$