What is a Cone in Math? (Definition, Shape & Examples) - BYJUS

# Cone

A cone is a three-dimensional geometric structure with a smooth transition from a flat, usually circular base to the apex or vertex, a point that creates an axis to the Centre of the base. A cone can also be described as a pyramid with a circular cross section rather than a pyramid with a triangular cross section. Cones are also described as circular cones. Here, we will learn about cones, their properties, and how we can calculate their area and volume....Read MoreRead Less

## What is a Cone?

A cone is a shape created by connecting the points on a circular base to a common point, known as the apex or vertex, using a series of line segments or lines (which does not contain the apex).

The height of the cone is determined by measuring the distance between its vertex and base. The radius of the circular base is also considered. The slant height is the distance along the cone’s circumference from any point on the peak to the base.

There are formulas that may be used to calculate the cone’s surface area and volume based on these parameters. You can see a cone in the illustration, which is described by its height, base radius, and slant height.

Here, h – height

l – slant height

## Cross-Sections of Cones

Consider a plane slicing through a cone as shown in the figure. The intersection of this plane and cone is a two dimensional shape which is called the cross-section of the cone.

This intersection is a circle. So, a circular cross-section of a cone is formed by this horizontal slicing plane.

This intersection is a triangle. So, a triangular cross-section of a cone is formed by this vertical slicing plane.

## Parameters used to Calculate the Surface area and Volume of a Conen

The different parameters of a cone that are required to calculate the surface area and volume of a cone are as follows:

The height of the cone (h), the area of the circular base ‘b’ or radius ‘r’ and slant height ‘l’ are used to generate the formula for the cone’s surface area (A) and volume (V).

Surface area of Cone

The surface area (A) of a cone is equal to the sum of its lateral surface area ($$\pi rl$$) and surface area of the circular base$$\pi r^2$$.

Here, l = slant height

Also, l = $$\sqrt{h^2~+~r^2}$$

Total surface area (A) = $$\pi rl$$ + $$\pi r^2$$

Volume of a Cone

The volume of a cone (V), which has a radius of the circular base ‘r’, and height from the vertex to the base ‘h’, is given by:

Volume (V) = $$\frac{1}{3}$$Bh

(b- Area of the circular base of the cone = πr$$^2$$)

Volume (V) = $$\frac{1}{3}$$πr$$^2$$h

## Solved Cone Examples

Example 1: Jessy was looking for a conical shaped container that has a radius of 7 cm and slant height of 21 cm. Calculate the total surface area of the container.

Solution:

Slant height, l = 21 cm

Total surface area (A) = $$\pi rl$$ + $$\pi r^2$$

A = $$\frac{22}{7}$$ x 7 x 21 + $$\frac{22}{7}$$ x 7$$^2$$                  [Substitute r by 7 and l by 21]

= 462 + 154                                       [Simplify]

= 616 cm$$^2$$

Hence, the total surface area of the container is 616 cm$$^2$$.

Example 2: Find the volume of the following cones.

Solution:

(a)

Volume ( V ) = $$\frac{1}{3}$$ πr$$^2$$h cubic units

= $$\frac{1}{3}$$ x π x 3$$^2$$ x 7          [Substitute h by 7 and r by 3]

= $$\frac{1}{3}$$ x $$\frac{22}{7}$$ x 3 x 3 x 7    [Simplify]

= 22 x 3

= 66 m$$^3$$

Hence, the volume of the cone is 66 m$$^3$$.

(b)

Volume (V) = $$\frac{1}{3}$$ Bh   cubic units

= $$\frac{1}{3}$$  x 12 x 4       [Substitute B by 12 and h by 4]

= 44                    [Simplify]

= 16 m$$^3$$

Hence, the volume of the cone is 16 m$$^3$$.

Example 3: Ben has to find the ratio of the volume of one cone to the volume of another  cone. Help Ben find the ratio of the volumes of the two cones provided that the radius of the second cone is 6 times that of the first cone and the height of the second cones is $$\frac{1}{3}$$ rd that of the first cone.

Solution:

Radius of first cone = r

Height of first cone = h

Radius of second cone, r’ = 6 x radius of first cone = 6r

Height of second cone, h’ = $$\frac{1}{3}$$ rd of first cone = $$\frac{1}{3}$$ h

To find the ratio, Ben first needs to find the volume of the two cones using the cone’s formula

Volume of a cone 1 = $$\frac{1}{3}$$ πr$$^2$$h

Volume of a cone 2 = $$\frac{1}{3}$$ πr’$$^2$$h’

= $$\frac{1}{3}$$π (6r)$$^2$$ x ($$\frac{1}{3}$$  x h)   [Substitute the values of r’ and h’]

= $$\frac{1}{3}$$  x π x 36r$$^2$$ x $$\frac{h}{3}$$

= $$\frac{1}{1}$$  x π x 4r$$^2$$ x $$\frac{h}{1}$$      [Simplify]

= 4πr$$^2$$h

The ratio of the volume of cone 1 and that of cone 2 = $$\frac{1}{3}$$ πr$$^2$$h : 4πr$$^2$$h

Ratio = 1 : 12               [Simplify]

Therefore, the ratio of the volume of cone 1 to the volume of cone 2 is 1:12.