As shown in the figure, in triangle $\mathrm{ABC}$, points $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ are on line segments $\mathrm{AZ}, \mathrm{BX}, \mathrm{CY}$ respectively, and $Y Z=2 Z C, Z X=3 X A, X Y=4 Y B$, the area of triangle $X Y Z$ is 24, find the area of triangle $\mathrm{ABC}$.
5. Answer: 59

Analysis: Connect $A Y, C X, B Z$, as shown in the figure. The area of triangle XYZ is 
$24, \mathrm{YZ}=2 \mathrm{ZZ}$, and the area of triangle $\mathrm{XZC}$ is 12.

Also, $Z X=3 X A$, and the area of triangle $\mathrm{XZC}$ is 12, so the area of triangle $\mathrm{AXC}$ is 4. The area of triangle $\mathrm{AYX}$ is 8. Note that $X Y=4 Y B$, so the area of triangle $A B Y$ is 2. The area of triangle ZBY is 6, and the area of triangle CBZ is 3.

Therefore, the area of triangle $\mathrm{ABC}$ is $=24+12+4+8+2+6+3=59$.