$AB$ and $CD$ are two parallel chords of a circle such that $AB=10\text{cm}$ and $CD=24\text{cm}$. If the chords are on the opposite sides of the centre and the distance between them is $17\text{cm}$, then the radius of the circle is:

Let the centre of the circle be $O$ and its radius be $R$.
A perpendicular from the centre to a chord bisects the chord.

* For chord $AB=10\text{cm}$, the half-length is $\frac{10}{2}=5\text{cm}$.
* For chord $CD=24\text{cm}$, the half-length is $\frac{24}{2}=12\text{cm}$.

Let the distance from the centre to chord $AB$ be $x\text{cm}$. Since the total distance between the parallel chords on opposite sides is $17\text{cm}$, the distance from the centre to chord $CD$ is $(17-x)\text{cm}$.

Using the right-angled triangles formed with the radius:

$$R^2=5^2+x^2=25+x^2\text{— (Equation 1)}$$

$$R^2=12^2+(17-x{)}^2=144+289+x^2-34x\text{— (Equation 2)}$$

Equating Equation 1 and Equation 2:

$$25+x^2=433+x^2-34x$$

$$34x=433-25=408$$

$$x=\frac{408}{34}=12\text{cm}$$

Substitute $x=12$ back into Equation 1:

$$R^2=25+12^2=25+144=169$$

$$R=\sqrt{169}=13\text{cm}$$