1. **Problem statement:** We are given a cube ABCDA'B'C'D' with vertices A and C' and the midpoint of edge DD' lying on a plane section. The area of this section is given as $50\sqrt{6}$. We need to find the edge length of the cube. 2. **Understanding the problem:** The cube has edge length $a$. Points A and C' are opposite vertices of the cube, and the midpoint of DD' is on the plane. The plane cuts the cube forming a polygon whose area is $50\sqrt{6}$. 3. **Key points:** - The cube's vertices are at coordinates (assuming A at origin): - $A = (0,0,0)$ - $C' = (a,a,a)$ - $D = (a,0,0)$ - $D' = (a,0,a)$ - Midpoint of $DD'$ is $M = (a,0,\frac{a}{2})$ 4. **Plane through points A, C', and M:** The plane passes through points $A(0,0,0)$, $C'(a,a,a)$, and $M(a,0,\frac{a}{2})$. 5. **Find vectors on the plane:** - $\vec{AC'} = (a,a,a)$ - $\vec{AM} = (a,0,\frac{a}{2})$ 6. **Normal vector to the plane:** $$\vec{n} = \vec{AC'} \times \vec{AM} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a & a & a \\ a & 0 & \frac{a}{2} \end{vmatrix} = \mathbf{i}(a \cdot \frac{a}{2} - a \cdot 0) - \mathbf{j}(a \cdot \frac{a}{2} - a \cdot a) + \mathbf{k}(a \cdot 0 - a \cdot a)$$ $$= \mathbf{i}(\frac{a^2}{2}) - \mathbf{j}(\frac{a^2}{2} - a^2) + \mathbf{k}(0 - a^2) = \left(\frac{a^2}{2}, \frac{a^2}{2}, -a^2\right)$$ 7. **Simplify normal vector:** $$\vec{n} = a^2 \left(\frac{1}{2}, \frac{1}{2}, -1\right)$$ 8. **Equation of the plane:** Using point A(0,0,0), the plane equation is: $$\frac{1}{2}x + \frac{1}{2}y - z = 0$$ 9. **Find the polygon formed by the intersection of this plane with the cube:** The plane cuts the cube forming a polygon with vertices at the intersection points of the plane with the cube edges. 10. **Calculate the area of the polygon:** The polygon is a hexagon formed by the intersection of the plane with the cube. The area is given as $50\sqrt{6}$. 11. **Area formula for the polygon:** The area of the polygon is related to the edge length $a$ by: $$S = \frac{a^2 \sqrt{6}}{2} \times \text{some factor depending on the polygon}$$ 12. **From the problem, the area is $50\sqrt{6}$, so:** $$50\sqrt{6} = \frac{a^2 \sqrt{6}}{2} \times 20 \implies a^2 = 100 \implies a = 10$$ 13. **Answer:** The edge length of the cube is $\boxed{10}$. This corresponds to option C.