Solution
The area of a triangle is $\frac{1}{2}$ base $\times$ height, so the triangle has sides of length $8 \mathrm{~cm}, 15 \mathrm{~cm}$ and $17 \mathrm{~cm}$ (using Pythagoras' theorem).
Let the radius of the circle be $r \mathrm{~cm}$.
Draw two radii, as shown. Since a tangent meets the radius at the point of contact at right angles, we have a square, with sides of length $r \mathrm{~cm}$.

Then, using the equal tangent theorem, we have
$$
(r-8)+(r-15)=17,
$$
so that $2 r=40$ and $r=20$.
Commentary
The equal tangent theorem states that the two tangents from a point outside a circle to the circle are equal in length. Here we use two of the vertices of the triangle as the points.

Hence the radius of the circle is $20 \mathrm{~cm}$.
5. The area of the right-angled triangle in the diagram alongside is $60 \mathrm{~cm}^{2}$. The triangle touches the circle, and one side of the triangle has length $15 \mathrm{~cm}$, as shown. What is the radius of the circle?