To calculate the volume of the cone, we use the formula for the volume of a cone:

$V = \frac{1}{3} \pi r^2 h$

Where:

• $r$ is the radius of the base,
• $h$ is the height of the cone.

From the image, the radius $r$ and height $h$ are not given directly. However, we have two values:

• $x = 19 \, \text{cm}$, which represents the radius $r$ of the base,
• $y = 24 \, \text{cm}$, which represents the slant height of the cone.

To calculate the height $h$, we use the Pythagorean theorem because the slant height forms a right triangle with the radius and the height:

$y^2 = x^2 + h^2$

Solving for $h$:

$h = \sqrt{y^2 - x^2} = \sqrt{24^2 - 19^2} = \sqrt{576 - 361} = \sqrt{215} \approx 14.66 \, \text{cm}$

Now, substitute the values of $r = 19 \, \text{cm}$ and $h \approx 14.66 \, \text{cm}$ into the volume formula:

$V = \frac{1}{3} \pi (19)^2 (14.66) = \frac{1}{3} \pi (361) (14.66) \approx \frac{1}{3} \pi (5294.26) \approx 5545.66 \, \text{cm}^3$

Rounding to the nearest whole number:

$V \approx 5546 \, \text{cm}^3$

Thus, the volume of the cone is approximately 5546 cm³.

Do you want further details or have any questions?

Here are 5 related questions you might consider:

1. How would the volume change if the radius were doubled?
2. How do you calculate the surface area of this cone?
3. What is the relationship between slant height and height in different types of cones?
4. Can the formula for volume be derived from the formula for the volume of a cylinder?
5. How would the calculation change if the cone was truncated?

Tip: For problems involving cones, it’s helpful to remember the Pythagorean theorem, especially when working with slant height.