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 $CMJKN$? $\mathrm{(A)}\ 15 \qquad$ $ \mathrm{(B)}\ \frac{81}{5} \qquad$ $ \mathrm{(C)}\ \frac{205}{12} \qquad$ $ \mathrm{(D)}\ \frac{240}{13} \qquad$ $ \mathrm{(E)}\ 20$
Solution
$\angle C=90^\circ$ since $5^2+12^2=13^2$, and $AM=5-4=1$, $BN=12-4=8$.
$\triangle BKN\sim \triangle BCA$ (right angles at $K$, $C$ and common $\angle B$) with ratio $BN:BA=8:13$, so $[BKN]:[BAC]=8^2:13^2$.
$\triangle AMJ\sim \triangle ABC$ (right angles at $J$, $C$ and common $\angle A$), with ratio $AM:AB=1:13$, so $[AMJ]:[BAC]=1:13^2$.
$([BKN]+[AMJ] ):[ABC]=65:169=5:13$, so $[CNKJM]:[ABC]=8:13$.
Since $[ABC]=30$, $[CNKJM]=\frac{8}{13}\cdot 30=\frac{240}{13}$.
In $\triangle ABC$, $AB=15$, $AC=9$, and $BC=12$. Points $M$ and $N$ lie on $AC$ and $BC$, respectively, with $CM=CN=3$. Points $J$ and $K$ are on $AB$ so that $MJ$ and $NK$ are perpendicular to $AB$. What is the area of pentagon $CMJKN$?
$\mathrm{(A)}\ 15 \qquad$ $ \mathrm{(B)}\ \frac{81}{5} \qquad$ $ \mathrm{(C)}\ \frac{205}{12} \qquad$ $ \mathrm{(D)}\ \frac{240}{13} \qquad$ $ \mathrm{(E)}\ 20$ 
