To solve this, we need to analyze the relationship between the radii of the incircle and the circumcircle of a right triangle, which are given as $r = 2 \, \text{cm}$ (radius of the incircle) and $R = 5 \, \text{cm}$ (radius of the circumcircle).

Properties and Relationships:

1. For a right triangle:

   • The circumradius $R$ is half the hypotenuse:
     $R = \frac{c}{2}$ where $c$ is the hypotenuse.
   • The inradius $r$ is given by:
     $r = \frac{a + b - c}{2}$ where $a$ and $b$ are the other two sides of the triangle.

2. Using the Pythagorean theorem:
   $c^2 = a^2 + b^2$

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Step 1: Hypotenuse from $R$

Given $R = 5 \, \text{cm}$, the hypotenuse $c$ is: $c = 2R = 2 \cdot 5 = 10 \, \text{cm}.$

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Step 2: Relation involving $r$

Using the formula for the inradius: $r = \frac{a + b - c}{2}.$ Substituting $r = 2 \, \text{cm}$ and $c = 10 \, \text{cm}$: $2 = \frac{a + b - 10}{2}.$ Simplifying: $a + b = 14.$

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Step 3: Solve for $a$ and $b$

Using the Pythagorean theorem $c^2 = a^2 + b^2$ with $c = 10$: $100 = a^2 + b^2.$ We now solve the system of equations:

1. $a + b = 14$
2. $a^2 + b^2 = 100$

Let $b = 14 - a$. Substitute into the second equation: $a^2 + (14 - a)^2 = 100.$ Expand and simplify: $a^2 + 196 - 28a + a^2 = 100,$ $2a^2 - 28a + 96 = 0.$ Divide through by 2: $a^2 - 14a + 48 = 0.$ Factorize: $(a - 8)(a - 6) = 0.$ Thus: $a = 8 \quad \text{or} \quad a = 6.$

If $a = 8$, then $b = 14 - 8 = 6$.
If $a = 6$, then $b = 14 - 6 = 8$.

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Final Answer:

The sides of the triangle are: $a = 6 \, \text{cm}, \, b = 8 \, \text{cm}, \, c = 10 \, \text{cm}.$

Would you like more details or explanations on any step?

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Follow-up Questions:

1. Why is the circumradius $R$ always half the hypotenuse in a right triangle?
2. What are some applications of the inradius and circumradius in geometry?
3. How does the relationship $a + b - c = 2r$ generalize to non-right triangles?
4. Can you derive the formula for the inradius using the area of the triangle?
5. How does the Pythagorean theorem ensure the solution is unique?

Tip: When solving systems of equations, substitution is often a powerful method for reducing variables efficiently.