2. As shown, $U$ and $C$ are points on the sides of triangle $M N H$ such that $M U=s$, $U N=6, N C=20, C H=s, H M=25$. If triangle $U N C$ and quadrilateral $M U C H$ have equal areas, what is $s$ ?
Solution:
4
Using brackets to denote areas, we have $[M C H]=[U N C]+[M U C H]=2[U N C]$. On the other hand, triangles with equal altitudes have their areas in the same ratio as their bases, so
$$
2=\frac{[M N H]}{[U N C]}=\frac{[M N H]}{[M N C]} \cdot \frac{[M N C]}{[U N C]}=\frac{N H}{N C} \cdot \frac{M N}{U N}=\frac{s+20}{20} \cdot \frac{s+6}{6} .
$$

Clearing the denominator gives $(s+20)(s+6)=240$, and solving the quadratic gives $s=4$ or -30 . Since $s>0$, we must have $s=4$.