Two convex polygons have a total of 33 sides and 243 diagonals. Find the number of diagonals in the polygon with the greater number of sides.
1. Let the two convex polygons have \( m \) and \( n \) sides respectively. We are given the following conditions:
   \[
   m + n = 33
   \]
   and the total number of diagonals is 243. The formula for the number of diagonals in a polygon with \( k \) sides is:
   \[
   \frac{k(k-3)}{2}
   \]
   Therefore, the total number of diagonals for the two polygons is:
   \[
   \frac{m(m-3)}{2} + \frac{n(n-3)}{2} = 243
   \]

2. Simplify the equation for the total number of diagonals:
   \[
   \frac{m(m-3) + n(n-3)}{2} = 243
   \]
   Multiply both sides by 2 to clear the fraction:
   \[
   m(m-3) + n(n-3) = 486
   \]
   Expand and simplify:
   \[
   m^2 - 3m + n^2 - 3n = 486
   \]

3. Substitute \( n = 33 - m \) into the equation \( m^2 - 3m + n^2 - 3n = 486 \):
   \[
   m^2 - 3m + (33 - m)^2 - 3(33 - m) = 486
   \]
   Expand and simplify:
   \[
   m^2 - 3m + (1089 - 66m + m^2) - 99 + 3m = 486
   \]
   Combine like terms:
   \[
   2m^2 - 66m + 990 = 486
   \]
   Simplify further:
   \[
   2m^2 - 66m + 504 = 0
   \]
   Divide the entire equation by 2:
   \[
   m^2 - 33m + 252 = 0
   \]

4. Solve the quadratic equation using the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   m = \frac{33 \pm \sqrt{33^2 - 4 \cdot 1 \cdot 252}}{2 \cdot 1}
   \]
   \[
   m = \frac{33 \pm \sqrt{1089 - 1008}}{2}
   \]
   \[
   m = \frac{33 \pm \sqrt{81}}{2}
   \]
   \[
   m = \frac{33 \pm 9}{2}
   \]
   Therefore, the solutions are:
   \[
   m = \frac{42}{2} = 21 \quad \text{and} \quad m = \frac{24}{2} = 12
   \]

5. Since \( m + n = 33 \), if \( m = 21 \), then \( n = 12 \), and vice versa. The polygon with the greater number of sides has 21 sides.

6. Calculate the number of diagonals in the polygon with 21 sides:
   \[
   \frac{21(21-3)}{2} = \frac{21 \cdot 18}{2} = \frac{378}{2} = 189
   \]

The final answer is \(\boxed{189}\).