As shown, $U$ and $C$ are points on the sides of triangle MNH such that $MU = s$, $UN = 6$, $NC = 20$, $CH = s$, $HM = 25$. If triangle $UNC$ and quadrilateral $MUCH$ have equal areas, what is $s$?

[img]https://cdn.artofproblemsolving.com/attachments/3/f/52e92e47c11911c08047320d429089cba08e26.png[/img]
1. **Identify the given information and the goal:**
   - We are given a triangle \( \triangle MNH \) with points \( U \) and \( C \) on sides \( MN \) and \( NH \) respectively.
   - The lengths are \( MU = s \), \( UN = 6 \), \( NC = 20 \), \( CH = s \), and \( HM = 25 \).
   - The areas of \( \triangle UNC \) and quadrilateral \( MUCH \) are equal.
   - We need to find the value of \( s \).

2. **Calculate the area of \( \triangle MNH \):**
   - The total length of \( MN \) is \( MU + UN = s + 6 \).
   - The total length of \( NH \) is \( NC + CH = 20 + s \).
   - The area of \( \triangle MNH \) can be found using the formula for the area of a triangle:
     \[
     \text{Area}(\triangle MNH) = \frac{1}{2} \times \text{base} \times \text{height}
     \]
     However, we do not have the height directly, so we will use the given ratio of areas.

3. **Set up the ratio of the areas of \( \triangle UNC \) and \( \triangle MNH \):**
   - The area of \( \triangle UNC \) is proportional to the product of the segments \( NU \) and \( NC \):
     \[
     \text{Area}(\triangle UNC) = \frac{1}{2} \times NU \times NC = \frac{1}{2} \times 6 \times 20 = 60
     \]
   - The area of \( \triangle MNH \) is proportional to the product of the segments \( MN \) and \( NH \):
     \[
     \text{Area}(\triangle MNH) = \frac{1}{2} \times (s + 6) \times (s + 20)
     \]

4. **Use the given condition that the areas of \( \triangle UNC \) and quadrilateral \( MUCH \) are equal:**
   - The area of quadrilateral \( MUCH \) is the area of \( \triangle MNH \) minus the area of \( \triangle UNC \):
     \[
     \text{Area}(MUCH) = \text{Area}(\triangle MNH) - \text{Area}(\triangle UNC)
     \]
   - Given that the areas are equal:
     \[
     \text{Area}(\triangle UNC) = \text{Area}(MUCH)
     \]
     \[
     60 = \text{Area}(\triangle MNH) - 60
     \]
     \[
     2 \times 60 = \text{Area}(\triangle MNH)
     \]
     \[
     120 = \frac{1}{2} \times (s + 6) \times (s + 20)
     \]

5. **Solve for \( s \):**
   \[
   120 = \frac{1}{2} \times (s + 6) \times (s + 20)
   \]
   \[
   240 = (s + 6) \times (s + 20)
   \]
   \[
   240 = s^2 + 26s + 120
   \]
   \[
   s^2 + 26s + 120 - 240 = 0
   \]
   \[
   s^2 + 26s - 120 = 0
   \]

6. **Solve the quadratic equation:**
   \[
   s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   where \( a = 1 \), \( b = 26 \), and \( c = -120 \):
   \[
   s = \frac{-26 \pm \sqrt{26^2 - 4 \cdot 1 \cdot (-120)}}{2 \cdot 1}
   \]
   \[
   s = \frac{-26 \pm \sqrt{676 + 480}}{2}
   \]
   \[
   s = \frac{-26 \pm \sqrt{1156}}{2}
   \]
   \[
   s = \frac{-26 \pm 34}{2}
   \]
   \[
   s = \frac{8}{2} \quad \text{or} \quad s = \frac{-60}{2}
   \]
   \[
   s = 4 \quad \text{or} \quad s = -30
   \]

7. **Select the positive solution:**
   \[
   s = 4
   \]

The final answer is \( \boxed{ s = 4 } \)