A cylinder has radius 12 and height 30. The top circular face of the cylinder is the base of a cone and the centre of the bottom circular base of the cylinder is the vertex of the cone. A sphere is placed inside so that it touches the cone, the base of the cylinder and the side of the cylinder as shown. Which of the following is closest to the radius of the sphere?
(A) 4.84
(B) 4.74
(C) 4.64
(D) 4.54
(E) 4.44

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-041.jpg?height=512&width=420&top_left_y=259&top_left_x=1292)
We slice the cylinder, cone and sphere using a vertical plane that passes through the centres of the top and bottom faces of the cylinder and through the centre of the sphere.

The resulting cross-sections of the cylinder, cone and sphere are a rectangle, triangle and circle, respectively.

Since the sphere is touching the cylinder and the cone, then by slicing the cylinder in this way, the resulting circle is tangent to two sides of the rectangle (at $F$ and $H$ ) and a side of the triangle (at $G$ ).

Join $O$ to $F, G$ and $H$. Since radii are perpendicular to tangents at the resulting points of tangency, then $O F$ is perpendicular to

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-081.jpg?height=542&width=477&top_left_y=168&top_left_x=1561)
$A D, O G$ is perpendicular to $A E$, and $O H$ is perpendicular to $D E$.

Let the radius of the sphere (now a circle) be $r$. Then $O F=O G=O H=r$.

Since the radius of the cylinder is 12 , then $D E=12$.

Since the height of the cylinder is 30 , then $A D=30$.

Since $F O H D$ has right angles at $F, D$ and $H$, then it must have four right angles, and so is a rectangle.

Since $O F=O H=r$, then $F O H D$ is in fact a square with $D H=D F=r$.

Since $D E=12$ and $D H=r$, then $E H=12-r$.

Since $A D=30$ and $D F=r$, then $A F=30-r$.

Since $A G$ and $A F$ are tangents to the circle from the same point, then $A G=A F=30-r$. (To see this, note that $\triangle A F O$ and $\triangle A G O$ are both right-angled, have a common side $A O$ and equal sides $F O$ and $G O$, which means that they are congruent.)

Similarly, $E G=E H=12-r$.

Finally, $A E=A G+G E$.

By the Pythagorean Theorem in $\triangle A D E, A E=\sqrt{12^{2}+30^{2}}=\sqrt{1044}$.

Thus, $\sqrt{1044}=(30-r)+(12-r)$ and so $2 r=42-\sqrt{1044}$ or $r=21-\frac{1}{2} \sqrt{1044} \approx 4.8445$.

Of the given choices, this is closest to 4.84 .

ANSWER: (A)