You have a valid point, Grace. If this were the exact measures of the sides of the triangle and the radius, it would be impossible to draw the triangle inside the circle.

Let's take a look at what the radius of a circle with sides $10$, $10$ and $4$ would be. We know from Heron's formula that the area of this triangle is $$\sqrt{12(12-10)(12-10)(12-4)} = \sqrt{384} = 8\sqrt{6}\approx 19.59\approx 20.$$ Using the formula $$[ABC] = \frac{abc}{4R},$$ we can solve for the radius of the circle $$8\sqrt{6} = \frac{(10)(10)(4)}{4R}\Rightarrow R = \frac{100}{8\sqrt{6}} = \frac{25}{2\sqrt{6}} \approx 5.1\approx 5$$

So, a triangle with sides $10$, $10$ and $4$, which has area close to $20$ can be inscribed in a circle of radius a little bit more than $5$. Note how the triangle in the figure on the previous post is thin, and the long sides are almost as long as the diameter of the circle.