Question

a solid right circular cone has radius 5 cm and height 12 cm as shown. what is the probability that a randomly chosen point inside this cone is a distance of at least 1 cm away from the closest point on the surface of the cone? express your answer as a decimal to the nearest thousandth.

Answer 1

47

Explanation:

Answer 2

The probability that a randomly chosen point inside the cone is at least 1 cm away from the closest point on the surface is approximately 0.759.

Consider the cone with a radius of 5 cm and a height of 12 cm. To find the probability that a randomly chosen point inside the cone is at least 1 cm away from the closest point on the surface, we need to calculate the ratio of the volume of the region where the points satisfy this condition to the total volume of the cone.

First, let's determine the total volume of the cone using the formula for the volume of a cone:
`\( V_{\text{total}} = (1)/(3) \pi r^2 h \),` where
`\( r \)` is the radius and
`\( h \)` is the height. Substituting the given values, we get
`\( V_{\text{total}} = (1)/(3) \pi (5^2)(12) \).`

Next, we need to find the volume of the region inside the cone that is at least 1 cm away from the closest point on the surface. This is a smaller cone inside the original cone, and its radius will be
`\( r' = r - 1 \)`. The volume of this smaller cone is
`\( V_{\text{inner}} = (1)/(3) \pi (4^2)(11) \).`

Now, the probability is given by
`\( P = \frac{V_{\text{inner}}}{V_{\text{total}}} \)`. Substituting the values, we get
`\( P = ((1)/(3) \pi (4^2)(11))/((1)/(3) \pi (5^2)(12)) \)`.

Calculating this expression gives
`\( P \approx 0.759 \)`, which means there is a 75.9% probability that a randomly chosen point inside the cone is at least 1 cm away from the closest point on the surface.