The triangle of maximum area that can be inscribed in a given circle of radius 'r' is:

We need to find the triangle of maximum area that can be inscribed in a circle of radius $$r$$. Let the circle be centred at the origin with equation $$x^2 + y^2 = r^2$$. We parameterize the three vertices as $$A = (r\cos\theta_1, r\sin\theta_1)$$, $$B = (r\cos\theta_2, r\sin\theta_2)$$, and $$C = (r\cos\theta_3, r\sin\theta_3)$$.

Without loss of generality, fix one vertex at $$A = (r, 0)$$ (i.e., $$\theta_1 = 0$$), and let $$\theta_2 = \alpha$$ and $$\theta_3 = \alpha + \beta$$, where $$\alpha > 0$$, $$\beta > 0$$, and $$\alpha + \beta < 2\pi$$. The area of the triangle using the cross-product formula is:

$$\text{Area} = \dfrac{r^2}{2}\big|\sin(\theta_2 - \theta_1) + \sin(\theta_3 - \theta_2) + \sin(\theta_1 - \theta_3)\big| = \dfrac{r^2}{2}\big(\sin\alpha + \sin\beta + \sin(2\pi - \alpha - \beta)\big)$$

Since $$\sin(2\pi - \alpha - \beta) = -\sin(\alpha + \beta)$$, we get $$\text{Area} = \dfrac{r^2}{2}\big(\sin\alpha + \sin\beta - \sin(\alpha + \beta)\big)$$.

To maximize, we take partial derivatives. Setting $$\dfrac{\partial(\text{Area})}{\partial\alpha} = 0$$: $$\cos\alpha - \cos(\alpha + \beta) = 0$$, so $$\cos\alpha = \cos(\alpha + \beta)$$. Setting $$\dfrac{\partial(\text{Area})}{\partial\beta} = 0$$: $$\cos\beta - \cos(\alpha + \beta) = 0$$, so $$\cos\beta = \cos(\alpha + \beta)$$.

From these, $$\cos\alpha = \cos\beta$$, which gives $$\alpha = \beta$$ (since both are in $$(0, 2\pi)$$). Substituting into $$\cos\alpha = \cos 2\alpha$$: $$\cos\alpha = 2\cos^2\alpha - 1$$, i.e., $$2\cos^2\alpha - \cos\alpha - 1 = 0$$. Factoring: $$(2\cos\alpha + 1)(\cos\alpha - 1) = 0$$. Since $$\alpha \neq 0$$, we have $$\cos\alpha = -\dfrac{1}{2}$$, giving $$\alpha = \dfrac{2\pi}{3}$$. The third angle subtended is $$2\pi - 2 \cdot \dfrac{2\pi}{3} = \dfrac{2\pi}{3}$$, confirming the triangle is equilateral.

The maximum area is $$\dfrac{r^2}{2}\left(3\sin\dfrac{2\pi}{3}\right) = \dfrac{r^2}{2} \cdot \dfrac{3\sqrt{3}}{2} = \dfrac{3\sqrt{3}}{4}r^2$$.

Now we find the side length. Each side is a chord subtending an angle of $$\dfrac{2\pi}{3}$$ at the centre. Using the chord formula $$\ell = 2r\sin\dfrac{\theta}{2}$$: $$s = 2r\sin\dfrac{\pi}{3} = 2r \cdot \dfrac{\sqrt{3}}{2} = \sqrt{3}\,r$$.

Checking option 3: the height of this equilateral triangle is $$h = \dfrac{\sqrt{3}}{2} \times \sqrt{3}\,r = \dfrac{3r}{2}$$, which does not equal $$\dfrac{2r}{3}$$. Checking option 2: an isosceles triangle with base $$2r$$ inscribed in the circle has its base as a diameter, making it a right triangle with area $$\dfrac{1}{2} \times 2r \times r = r^2$$, which is less than $$\dfrac{3\sqrt{3}}{4}r^2 \approx 1.30\,r^2$$. Checking option 4: a right triangle with sides $$2r$$ and $$r$$ has hypotenuse $$\sqrt{5}\,r$$, and its circumradius would be $$\dfrac{\sqrt{5}\,r}{2} \neq r$$ in general.

Therefore, the triangle of maximum area inscribed in a circle of radius $$r$$ is an equilateral triangle having each of its sides of length $$\sqrt{3}\,r$$.