Now the base can be something other than a circle as well. However, generally, when we say “cone”, we are referring to one with a circular base - a circular cone. And more specifically a right circular cone.

In this tutorial, we'll learn how to find the volume of a cone. And we'll begin with a couple of examples of what cones look like.

So a cone has a base that tapers smoothly into a point at the other end (called vertex).

A right circular cone is a cone with a circular base whose axis - the line joining the vertex to the base's center - is perpendicular to the base.

The good thing is - the formula for the volume of a circular cone is the same, regardless of whether the cone is right or oblique.

Like with any 3-dimensional solid, the volume of a cone refers to the space it occupies.

1. For a circular cone with a radius $\hspace{0.2em} r \hspace{0.2em}$ and height $\hspace{0.2em} h \hspace{0.2em}$,

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

2. For any cone (circular or not) with a base area of $\hspace{0.2em} A_b \hspace{0.2em}$ and height $\hspace{0.2em} h \hspace{0.2em}$

$V = \frac{1}{3} A_b h$

The great Greek polymath, Archimedes, proved that the volume of a cone is one-third the volume of a cylinder with the same base radius and height.^{#}

So,

$V_{cone} = \frac{1}{3} V_{cylinder} \hspace{0.25cm} \rule[0.1cm]{1cm}{0.1em} \hspace{0.15cm} (1)$

Now, the volume of a cylinder, with a base radius $\hspace{0.2em} r \hspace{0.2em}$ and height $\hspace{0.2em} h \hspace{0.2em}$, is given by –

$V_{cylinder} = \pi r^2 h \hspace{0.25cm} \rule[0.1cm]{1cm}{0.1em} \hspace{0.15cm} (2)$

And so from $\hspace{0.2em} (1) \hspace{0.2em}$ and $\hspace{0.2em} (2) \hspace{0.2em}$, we get the volume of a cone with the same base radius and height –

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

Alright. Let's solve a couple of examples to cement our understanding of the concepts.

Example

Find the volume of a cone with a base diameter of $\hspace{0.2em} 14$ cm and height of $\hspace{0.2em} 5$ cm.

Solution

We know the volume of a cone is given by -

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

But, the question doesn't give us the radius, so let's calculate that first.

$\begin{align*} r &= \frac{d}{2} \\[1.3em] &= \frac{14}{2} = 7 \end{align*}$

Now, substituting the values of $\hspace{0.2em} r \hspace{0.2em}$ and $\hspace{0.2em} h \hspace{0.2em}$ into the formula -

$\begin{align*} V &= \frac{1}{3} \pi \cdot 7^2 \cdot 5 \\[1.3em] &= 256.56 \end{align*}$

So the volume of the cone is $\hspace{0.2em} 256.56 \text{cm}^2 \hspace{0.2em}$.

Example

A cone has a base radius of $\hspace{0.2em} 3 \hspace{0.2em}$ cm and a slant height of $\hspace{0.2em} 5 \hspace{0.2em}$ cm. Find its volume.

Solution

Here, we are given the radius $\hspace{0.2em} (r) \hspace{0.2em}$ and slant height $\hspace{0.2em} (l) \hspace{0.2em}$. But we need to be careful. Our standard formula for volume requires height $\hspace{0.2em} (h) \hspace{0.2em}$ and not slant height.

So how do we find $\hspace{0.2em} h \hspace{0.2em}$? Well, we can use the Pythagorean theorem.

Applying the theorem on the right triangle in the figure above, we have -

$\begin{align*} l^2 &= r^2 + h^2 \\[1em] 5^2 &= 3^2 + h^2 \end{align*}$

Solving for $\hspace{0.2em} h \hspace{0.2em}$, we get

$h = 4$

Great! Now, we can use our formula to find the volume of the cone.

$\begin{align*} V &= \frac{1}{3} \pi r^2 h \\[1.3em] &= \frac{1}{3} \pi 3^2 4 \\[1.3em] &= 37.7 \end{align*}$

So the volume of the come is $\hspace{0.2em} 37.7 \text{cm}^2 \hspace{0.2em}$.

And that brings us to the end of this tutorial on the volume of a cone. Until next time.

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