9.04 Pyramids and cones

The diagram shows the same cylinder being filled three times, up to three different lines. If we relate the formula for the volume of a cone to the formula for the volume of a cylinder, we can see that the cone in the diagram is only filling a third of the cylinder with each pour, so it makes sense that the volume of a cone is one-third the volume of a cylinder.

If the formula for the volume of a prism or cylinder is the same, there may be another solid that is one-third the volume of a prism.

To find the density, we first want to calculate the volume of the cone. Before we can do that, we want to start by calculating the area of the circular base.

We can calculate the area of the base as follows:

So the area of the circular base is \pi \text{ in}^2.

We can now use this to find the approximate volume of the carrot:

So the approximate volume of the carrot is 4\pi \text{ in}^3.

Finally, we can approximate the density:

The density of the carrot is approximately 0.06 \, \text{lb}/\text{in}^3.

Since this is only an approximation of the density we could also express it as a decimal value, rounded to some number of decimal places. In this case, the density is approximately 0.0597 \, \text{ lb}/\text{in}^3 to four decimal places.

Note that even when approximating, it is generally good practice to leave values in exact form throughout the work and only round at the final expression.

Calculate the volume of the original cone, then take half the volume. We need the height of the cone, so use the Pythagorean theorem with the height of the slanted lateral face as the hypotenuse and the radius of the base as the length of the triangle's short leg.

First, we'll find the height of the cone:

The height of the cone is \sqrt{96 } \text{ cm}.

Now, we need the area of the base of the original cone. We know that the radius of the base is 5 \text{ cm}, so the area of the base is \pi \left(5 ^2 \right) = 25 \pi \text{ cm}^2.

We can calculate the volume of the original cone as follows:

Finally, since the cone was sawed in half, we can take half of the total cone of the original cone to be approximately 128.25 \text{ cm} ^3.

First we find the area of the base, then we can use that to find the volume. Since this solid is a square pyramid, the base is a square.

Finding the area of the base, we have:

We can now use this to calculate the volume:

So the volume of the Pyramid of Giza is 18 \, 069 \, 333 \dfrac{1}{3} cubic Egyptian royal cubits.

Subtract the volume of the smaller pyramid from the volume of the larger pyramid.

Start by calculating the volume of the larger pyramid. We have

The volume of the larger pyramid is exactly \dfrac{512}{3} \text{ cm}^3.

Now we can calculate the volume of the smaller pyramid. We have

The volume of the smaller pyramid is exactly \dfrac{64}{3} \text{ cm}^3.

The volume of the solid after removing the smaller pyramid is \dfrac{512}{3} \text{ cm}^3 - \dfrac{64}{3} \text{ cm}^3 = \dfrac{448}{3} \text{ cm}^3.

We could choose to leave the solution in exact form, since the directions do not specify rounding requirements, or we could simplify the solution and round to a measure we find appropriate.

Calculate the slant height of the cone using the Pythagorean theorem, then use the formula for the surface area of a cone to calculate the surface area.

The slant height of the cone can be found using the Pythagorean theorem as follows:

The slant height of the cone is \sqrt{90} \text{ cm}.

Now, we can calculate the surface area as follows:

The surface area of the cone is 117.7 \text{ cm}^2.

We will need the height of a lateral face of the pyramid to calculate the lateral surface area. First, find the height of one lateral face using the Pythagorean theorem, then calculate the surface area of the pyramid. The diagram that follows shows the dimensions of the Pyramid of Giza:

First, we'll use the height and half the length of the base as the legs in the Pythagorean theorem:

The height of one lateral face of the pyramid is exactly \sqrt{126800} Egyptian royal cubits.

We can use this to calculate the area of one of the faces of the Pyramid of Giza as follows:

The area of one of the faces of the Pyramid of Giza is \left(220 \right) \left(\sqrt{126 800} \right) Egyptian royal cubits^2so the area of all four lateral faces is \left(880 \right) \left(\sqrt{126800} \right) Egyptian royal cubits^2.

The area of the base of the pyramid is 440 \text{ Egyptian royal cubits} \times 440\text{ Egyptian royal cubits} = 193\,600 \text{ Egyptian royal cubits}^2

Finally, the surface area of the Pyramid of Giza is calculated as \left(880 \right) \left(\sqrt{126 800} \right) \text{ Egyptian royal cubits}^2 + 193600 \text{ Egyptian royal cubits}^2 \\= 506 \, 959.0911 \text{ Egyptian royal cubits}^2

The surface area of the Pyramid of Giza to the nearest square royal cubit is 506 \, 959 \text{ Egyptian royal cubits}^2.

The actual length of an Egyptian royal cubit is about 20 \text{ in}, consider how the surface area compares to area of a football field. Note that we did not simplify the square root until the final calculation in order to preserve the precision of the length.