12. On a sunny day, a sphere is placed on a horizontal ground. At a certain moment, the shadow of the sphere reaches a distance of 10 meters from the point where the sphere touches the ground. At the same moment, a vertical pole of length 1 meter casts a shadow 2 meters long. What is the radius of the sphere in meters?
(A) $\frac{5}{2}$
(B) $9-4 \sqrt{5}$
(C) $10 \sqrt{5}-20$
(D) $8 \sqrt{10}-23$
(E) $6-\sqrt{15}$.

## Numerical Answer Problems -5 points
12. The answer is $\mathbf{( C )}$. The length of the shadow of the pole is twice its height, so $A B=$ $\frac{1}{2} B D=5 \mathrm{~m}$. By the Pythagorean theorem, $A D=$ $5 \sqrt{5} \mathrm{~m}$.

The triangles $A B D$ and $A E C$ are similar, so $A C: A D=C E: B D$. Let $r=C E$, then: $(5-r): 5 \sqrt{5}=r: 10$ from which $r=10 \sqrt{5}-20$.

![](https://cdn.mathpix.com/cropped/2024_04_17_07a3c581d2eb0df332d0g-11.jpg?height=397&width=700&top_left_y=1957&top_left_x=1112)