Let a, b, and c be the lengths of the sides of the triangle ABC. 

We have a = 20 cm, b = 21 cm, and c = 29 cm. 

The semi-perimeter s is given by: s = (a + b + c) / 2 = (20 + 21 + 29) / 2 = 70 / 2 = 35 cm 

The area A of the triangle can be calculated using Heron's formula: A = √(s(s-a)(s-b)(s-c)) = √(35(35-20)(35-21)(35-29)) = √(35 * 15 * 14 * 6) = √(44100) = 210 cm² 

The circumradius R of the triangle is given by the formula: R = abc / 4A = (20 * 21 * 29) / (4 * 210) = (20 * 21 * 29) / 840 = 12180 / 840 = 14.5 cm

Use the formula: \( R = \frac{abc}{4\Delta} \)

Let sides be: 
\( a = 20 \, \text{cm}, \quad b = 29 \, \text{cm}, \quad c = 21 \, \text{cm} \)

First, find semi-perimeter: \( s = \frac{a + b + c}{2} = \frac{20 + 29 + 21}{2} = 35 \)

Use Heron’s formula for area: \( \Delta = \sqrt{s(s - a)(s - b)(s - c)} \)

\( \Delta = \sqrt{35(35 - 20)(35 - 29)(35 - 21)} = \sqrt{35 \cdot 15 \cdot 6 \cdot 14} \)

\( \Delta = \sqrt{44100} = 210 \, \text{cm}^2 \)

Now calculate radius: \( R = \frac{20 \cdot 29 \cdot 21}{4 \cdot 210} = \frac{12180}{840} = 14.5 \, \text{cm} \)

Answer:

14.5 cm

Step 1: Verify if ABC is right-angled

Check using the Pythagorean theorem:

\[ AB^2 + BC^2 = 21^2 + 20^2 = 441 + 400 = 841 \]

\[ AC^2 = 29^2 = 841 \]

Since \( AB^2 + BC^2 = AC^2 \), the triangle is right-angled at B.

Step 2: Calculate circumradius

For right-angled triangles, the hypotenuse equals the diameter of the circumcircle:

\[ \text{Circumradius} (R) = \frac{\text{Hypotenuse}}{2} = \frac{AC}{2} = \frac{29}{2} = 14.5\ \text{cm} \]

Final Answer:

The circumradius is \[ \boxed{14.5\ \text{cm}} \].