1. Problem: Calculate the radius of the circumscribed circle (circumradius) of a triangle with sides $4\sqrt{2}$, $8$, and $4\sqrt{2}$. 2. Formula: The circumradius $R$ of a triangle with sides $a$, $b$, and $c$ and area $A$ is given by $$ R = \frac{abc}{4A} $$ 3. Step 1: Identify the sides: $$ a = 4\sqrt{2}, \quad b = 8, \quad c = 4\sqrt{2} $$ 4. Step 2: Calculate the semi-perimeter $s$: $$ s = \frac{a+b+c}{2} = \frac{4\sqrt{2} + 8 + 4\sqrt{2}}{2} = \frac{8\sqrt{2} + 8}{2} = 4\sqrt{2} + 4 $$ 5. Step 3: Calculate the area $A$ using Heron's formula: $$ A = \sqrt{s(s-a)(s-b)(s-c)} $$ Calculate each term: $$ s - a = (4\sqrt{2} + 4) - 4\sqrt{2} = 4 $$ $$ s - b = (4\sqrt{2} + 4) - 8 = 4\sqrt{2} - 4 $$ $$ s - c = (4\sqrt{2} + 4) - 4\sqrt{2} = 4 $$ 6. Step 4: Substitute into Heron's formula: $$ A = \sqrt{(4\sqrt{2} + 4) \times 4 \times (4\sqrt{2} - 4) \times 4} $$ Simplify inside the square root: $$ = \sqrt{16 (4\sqrt{2} + 4)(4\sqrt{2} - 4)} $$ Note that $(x+y)(x-y) = x^2 - y^2$, so: $$ (4\sqrt{2} + 4)(4\sqrt{2} - 4) = (4\sqrt{2})^2 - 4^2 = 16 \times 2 - 16 = 32 - 16 = 16 $$ Therefore: $$ A = \sqrt{16 \times 16} = \sqrt{256} = 16 $$ 7. Step 5: Calculate the circumradius $R$: $$ R = \frac{abc}{4A} = \frac{(4\sqrt{2}) \times 8 \times (4\sqrt{2})}{4 \times 16} $$ Calculate numerator: $$ (4\sqrt{2}) \times 8 = 32\sqrt{2} $$ $$ 32\sqrt{2} \times 4\sqrt{2} = 32 \times 4 \times (\sqrt{2} \times \sqrt{2}) = 128 \times 2 = 256 $$ Calculate denominator: $$ 4 \times 16 = 64 $$ Simplify fraction: $$ R = \frac{256}{64} = 4 $$ Final answer: The radius of the circumscribed circle is $\boxed{4}$.