The vertices of a right-angled triangle are on a circle of radius $R$ and the sides of the triangle are tangent to another circle of radius $r$ (this is the circle that is inside triangle). If the lengths of the sides about the right angles are 16 and 30, determine the value of $R+r$.
1. **Identify the sides of the right-angled triangle:**
   The problem states that the lengths of the sides about the right angle are 16 and 30. These are the legs of the right-angled triangle.

2. **Calculate the hypotenuse:**
   Using the Pythagorean theorem, we find the hypotenuse \( c \):
   \[
   c = \sqrt{16^2 + 30^2} = \sqrt{256 + 900} = \sqrt{1156} = 34
   \]

3. **Determine the circumradius \( R \):**
   For a right-angled triangle, the circumradius \( R \) is half the hypotenuse:
   \[
   R = \frac{c}{2} = \frac{34}{2} = 17
   \]

4. **Calculate the area of the triangle:**
   The area \( A \) of the triangle can be calculated using the legs:
   \[
   A = \frac{1}{2} \times 16 \times 30 = 240
   \]

5. **Determine the semi-perimeter \( s \):**
   The semi-perimeter \( s \) is half the sum of the sides:
   \[
   s = \frac{16 + 30 + 34}{2} = \frac{80}{2} = 40
   \]

6. **Calculate the inradius \( r \):**
   The inradius \( r \) can be found using the formula \( A = r \cdot s \):
   \[
   r = \frac{A}{s} = \frac{240}{40} = 6
   \]

7. **Sum the circumradius and the inradius:**
   \[
   R + r = 17 + 6 = 23
   \]

The final answer is \(\boxed{23}\).