A cylinder has a radius of 3 cm and a height of 8 cm. What is the longest segment, in centimeters, that would fit inside the cylinder?
The longest segment stretches from the bottom to the top of the cylinder and across a diameter, and is thus the hypotenuse of a right triangle where one leg is the height $8$, and the other is a diameter of length $2(3)=6$.  Thus its length is

$$\sqrt{6^2+8^2}=\boxed{10}$$
10
Level 2
Geometry
Problem: A cylinder has a radius of 3 cm and a height of 8 cm. What is the longest segment, in centimeters, that would fit inside the cylinder?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the longest segment that can fit inside the cylinder, we need to find the length of the diagonal of the cylinder. This diagonal can be visualized as the hypotenuse of a right triangle formed by the height and the diameter of the cylinder's base.

First, calculate the diameter of the cylinder. The diameter is twice the radius, so:
Diameter = 2 * Radius = 2 * 3 cm = 6 cm.

Next, we identify the dimensions of the right triangle:
- One leg is the height of the cylinder, which is 8 cm.
- The other leg is the diameter of the base, which is 6 cm.

Now, we apply the Pythagorean theorem to find the length of the diagonal (d):
d^2 = height^2 + diameter^2.

Substituting the known values:
d^2 = (8 cm)^2 + (6 cm)^2
d^2 = 64 cm^2 + 36 cm^2
d^2 = 100 cm^2.

Taking the square root of both sides gives:
d = √(100 cm^2) = 10 cm.

Thus, the longest segment that can fit inside the cylinder is 10 cm.<|END_OF_LATENT|> The final answer is $10$.