1. **State the problem:** We have an isosceles triangle inscribed in a circle of radius 4. We want to find the dimensions of the triangle when its area is maximum. 2. **Formula and important rules:** The triangle is inscribed in a circle of radius $r=4$. Let the two equal sides be $s$ and the base be $b$. The area $A$ of the triangle can be expressed as: $$A = \frac{1}{2} b h$$ where $h$ is the height. Since the triangle is inscribed in a circle, its vertices lie on the circle. The maximum area for a triangle inscribed in a circle occurs when the triangle is isosceles with the vertex angle at the center of the circle. 3. **Set up variables:** Let the vertex angle at the center be $\theta$. Then the two equal sides $s$ are the radii of the circle, so $s=4$. The base $b$ is the chord subtending angle $2\theta$ at the center: $$b = 2r \sin(\theta) = 8 \sin(\theta)$$ The height $h$ is the distance from the center to the base, which is: $$h = r \cos(\theta) = 4 \cos(\theta)$$ 4. **Area as a function of $\theta$:** $$A = \frac{1}{2} b h = \frac{1}{2} (8 \sin(\theta)) (4 \cos(\theta)) = 16 \sin(\theta) \cos(\theta)$$ Using the double angle identity: $$\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$$ So, $$A = 16 \cdot \frac{\sin(2\theta)}{2} = 8 \sin(2\theta)$$ 5. **Maximize area:** The maximum value of $\sin(2\theta)$ is 1, which occurs at $2\theta = \frac{\pi}{2}$ or $\theta = \frac{\pi}{4}$. 6. **Calculate dimensions at maximum area:** $$b = 8 \sin\left(\frac{\pi}{4}\right) = 8 \cdot \frac{\sqrt{2}}{2} = 4 \sqrt{2}$$ $$h = 4 \cos\left(\frac{\pi}{4}\right) = 4 \cdot \frac{\sqrt{2}}{2} = 2 \sqrt{2}$$ The equal sides are the radii: $$s = 4$$ 7. **Final answer:** The isosceles triangle with maximum area inscribed in a circle of radius 4 has two equal sides of length 4 and a base of length $4 \sqrt{2}$. The maximum area is: $$A_{max} = 8 \sin\left(2 \cdot \frac{\pi}{4}\right) = 8 \sin\left(\frac{\pi}{2}\right) = 8$$