### Video Transcript

In this video, we will learn how to calculate volumes of cones and solve problems including real-life situations. We will begin by looking at the properties of a cone and explain the formula that is used to calculate its volume. In this video, we will consider oblique cones and right cones, an example of which is shown in the diagram.

A cone is a three-dimensional geometric shape that has a circular base and a curved side that ends in a single vertex or apex. A right cone is a cone whose apex lies above the centroid of the circular base, whereas in an oblique cone, the apex does not lie directly above the centroid of the base. The height of a cone is the vertical or perpendicular distance from the apex to the base. The slant height of a cone is the distance from the apex to any point lying on the circumference of the base. Finally, the radius, height, and slant height form a right triangle inside the cone.

Now that we have recalled some key definitions of a cone, let’s move on and consider its volume. The volume of any cone is equal to one-third 𝜋𝑟 squared multiplied by ℎ. Let’s consider the oblique cone as shown. As the base of any cone is circular, 𝜋𝑟 squared refers to the area of the base. We know that the area of any circle is equal to 𝜋 multiplied by the radius squared. ℎ refers to the perpendicular height as shown. We recall this is the vertical distance from the apex to the base. As our cone is oblique, this will not be at the center of the circle in this example.

Let’s consider an example where this height is equal to 12 centimeters and the radius of the base is four centimeters. Substituting these values into our formula, we see that the volume of the cone is equal to one-third multiplied by 𝜋 multiplied by four squared multiplied by 12. As multiplication is commutative, we can perform these operations in any order. One-third multiplied by 12 or a third of 12 is equal to four. Four squared is equal to 16. This means that the volume is equal to four multiplied by 16 multiplied by 𝜋. This simplifies to give us 64𝜋. We measure volume in cubic units. Therefore, the volume of this cone is 64𝜋 cubic centimeters.

We will now look at some specific examples which involve finding the volume of a cone.

Work out the volume of a cone with a diameter of 10.5 and a height of 11.3. Give your solution to two decimal places.

Let’s begin by drawing a sketch of the cone. We are told that the diameter of the circular base is equal to 10.5. The height of the cone is 11.3. This is the perpendicular distance from the apex to the base. Note that the question does not tell us whether this is a right cone or an oblique one, but in fact it does not matter, since we have been given the perpendicular height. When using the perpendicular height, our formula for volume is applied in the same way to both cases.

In order to calculate the volume of any cone, we use the formula one-third 𝜋𝑟 squared multiplied by ℎ, where 𝑟 is the radius of the circular base. We know that the radius of any circle is half its diameter, and we are told the diameter is 10.5. The radius is, therefore, equal to a half multiplied by 10.5. This is equal to 5.25. We can now substitute our values for the radius and height into the formula.

The volume of the cone is equal to one-third multiplied by 𝜋 multiplied by 5.25 squared multiplied by 11.3. Typing this into the calculator gives us 326.1562 and so on. We are asked to give our solution correct to two decimal places. As the deciding number is a six, we round up, giving us an answer of 326.16. Volume is measured in cubic units. And as there were no length units given for the diameter or the height, the volume is equal to 326.16 cubic units.

We will now look at an example where we are given the height and base perimeter of a cone.

Determine to the nearest tenth the volume of a right cone having a height of 106 centimeters, given that the perimeter of its base is 318 centimeters. Use 𝜋 equal to 22 over seven.

As we are dealing with a right cone, we know that the apex lies directly above the centroid of the circular base. This means that the height and radius are at right angles or perpendicular to one another. We are told that this perpendicular distance or height is equal to 106 centimeters. We are also told that the perimeter or circumference of the base is equal to 318 centimeters. We recall that the volume of a cone 𝑉 is equal to one-third 𝜋𝑟 squared multiplied by ℎ. In order to calculate this value, we firstly need to work out the radius of the circular base.

The circumference of any circle 𝐶 is equal to two 𝜋𝑟. In this question, we know that the circumference is 318 centimeters. Therefore, 318 is equal to two 𝜋𝑟. Dividing both sides of this equation by two, we get 159 is equal to 𝜋 multiplied by 𝑟. At this stage, we could substitute 22 over seven for 𝜋. However, we will simply divide both sides of the equation by 𝜋, giving us 𝑟 is equal to 159 over 𝜋. The radius of the circular base of our cone is 159 over 𝜋 centimeters.

We can now substitute the values of ℎ and 𝑟 into our formula for the volume. 𝑉 is equal to one-third multiplied by 𝜋 multiplied by 159 over 𝜋 all squared multiplied by 106. Substituting in our value for 𝜋 and typing this into the calculator gives us an answer of 284219.727 and so on. We are asked to round our answer to the nearest tenth. This is the same as rounding to one decimal place, giving us a volume of 284219.7 cubic centimeters. Note that our units for volume will always be cubic units.

Our next example involves calculating the radius of a cone given its height and volume.

A cone has a perpendicular height of 92 inches and a volume of 420𝜋 cubic inches. Work out the radius of the cone, giving your answer to the nearest inch.

In order to answer this question, we need to recall the formula for the volume of a cone. This is equal to one-third 𝜋𝑟 squared multiplied by ℎ, where 𝑟 is the radius of the circular base and ℎ is the perpendicular distance from the apex of the cone to the base. In this question, we are given both the height and the volume and we are asked to calculate the radius. To do this, we will begin by rearranging the formula 𝑉 is equal to one-third 𝜋𝑟 squared ℎ to make 𝑟 the subject.

We begin by multiplying both sides of our equation by three such that three 𝑉 is equal to 𝜋𝑟 squared ℎ. Next, we divide both sides by 𝜋ℎ such that 𝑟 squared is equal to three 𝑉 divided by 𝜋ℎ. Finally, we take the square root of both sides of the equation. As the radius must be positive, 𝑟 is equal to the square root of three 𝑉 divided by 𝜋ℎ. We can now substitute 420𝜋 for 𝑉 and 92 for ℎ. Three multiplied by 420𝜋 is 1260𝜋, and 𝜋 multiplied by 92 is 92𝜋. Underneath the square root, we can divide the numerator and denominator by 𝜋. As 1260 and 92 are both divisible by four, 𝑟 is equal to the square root of 315 over 23. Typing this into the calculator gives us 3.7007 and so on.

As we are asked to give our answer to the nearest inch, we need to consider the digit in the tenths column. We can therefore conclude that a cone with a perpendicular height of 92 inches and a volume of 420𝜋 cubic inches will have a radius of four inches correct to the nearest inch. We could check this answer by substituting the value of 𝑟 back in to the original formula.

In our final example, we need to compare the volumes of a cone and a pyramid.

Which is greater in volume, a right cone having a base radius of 25 centimeters and a height of 56 centimeters or a right square-based pyramid having a base with perimeter of 176 centimeters and a height of 48 centimeters?

Before starting this question, it is worth recalling that the volume of a cone and the volume of a pyramid have very similar formulas. The volume of both of these shapes is equal to a third of the base area multiplied by the height. As we are dealing with a right cone and a right pyramid, the height will be the perpendicular distance from the apex to the center of the base. When dealing with a cone, the volume is equal to one-third 𝜋𝑟 squared multiplied by ℎ. This is because the base area of a cone is circular, and the area of a circle is equal to 𝜋𝑟 squared.

When dealing with a square-based pyramid, the volume is equal to one-third of 𝑙 squared multiplied by ℎ, where 𝑙 is the length of each side of the square base. In this question, the cone has a base radius of 25 centimeters and a height of 56 centimeters. The volume 𝑉 is therefore equal to one-third multiplied by 𝜋 multiplied by 25 squared multiplied by 56. Typing this into the calculator gives us 36651.91 and so on. To the nearest cubic centimeter, the volume of the cone is 36652 cubic centimeters.

Before we can calculate the volume of the pyramid, we need to work out the length of each side of the square base. We know that the perimeter of the base is equal to 176 centimeters. The side length 𝑙 is therefore equal to 176 divided by four. This is equal to 44 centimeters. The volume of the pyramid is therefore equal to one-third multiplied by 44 squared multiplied by 48, as 48 is the height of the square-based pyramid. Clearing some space and typing this into the calculator gives us 30976. The volume of the square-based pyramids is 30976 cubic centimeters. As this value is less than the volume of the cone, 36652 cubic centimeters, the shape that has the greater volume is the cone.

We will now summarize the key points from this video. A cone is a three-dimensional shape with a circular base and a curved side that ends in a single vertex or apex. The volume of a cone can be calculated using the formula one-third 𝜋𝑟 squared multiplied by ℎ. The 𝜋𝑟 squared part of the formula corresponds to the area of the circular base, where 𝑟 is the radius. ℎ is known as the perpendicular height; this is the vertical or perpendicular distance from the apex to the base.

When dealing with a right cone, this height goes from the apex to the center of the circular base. We saw that the volume of a cone is measured in cubic units, for example, cubic centimeters, cubic meters, or cubic inches. We can use the formula to not only calculate the volume of a cone, but also its radius or height if we are given the other variables. As seen in the final example, we can also compare the volumes of different three-dimensional shapes.