AMC 12 2004b Problem 14

In $\triangle ABC$, $AB=13$, $AC=5$, and $BC=12$. Points $M$ and $N$ lie on $AC$ and $BC$, respectively, with $CM=CN=4$. Points $J$ and $K$ are on $AB$ so that $MJ$ and $NK$ are perpendicular to $AB$. What is the area of pentagon $CMJKN$?

unitsize(0.5cm);
defaultpen(0.8);
pair C=(0,0), A=(0,5), B=(12,0), M=(0,4), N=(4,0);
pair J=intersectionpoint(A--B, M--(M+rotate(90)*(B-A)) );
pair K=intersectionpoint(A--B, N--(N+rotate(90)*(B-A)) );
draw( A--B--C--cycle );
draw( M--J );
draw( N--K );
label("$A$",A,NW);
label("$B$",B,SE);
label("$C$",C,SW);
label("$M$",M,SW);
label("$N$",N,S);
label("$J$",J,NE);
label("$K$",K,NE);

$\mathrm{(A)}\ 15 \mathrm{(B)}\ \frac{81}{5} \mathrm{(C)}\ \frac{205}{12} \mathrm{(D)}\ \frac{240}{13} \mathrm{(E)}\ 20$