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Volume of Cone

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When you look at a volcanic mountain, what does it look like? Perhaps the ice cream cones are more attractive and you may be wondering how much ice cream they can contain.

In this article, you will understand what a cone and conical objects are, how to determine their volume and examples of application of this.

A **cone** is a three-dimensional solid that consists of a circular base and a continuous curved surface that tapers to a tip-top called the apex, or vertex.

Some examples of objects with conical shapes are the traffic cone, birthday cone caps, ice cream cones, carrots and so on.

There are two basic types of cones:

**The right circular cones**which have their apex perpendicular to or right above the center of their base.

**The oblique circular cones**- do not have their apex above the center of the circular base and as such are not perpendicular to the center.

An illustration of an oblique circular cone - StudySmarter Original

The volume or capacity of a cone is one-third of its corresponding cylinder. This means that when a cone and a cylinder have the same base dimension and height, the volume of the cone is one-third of the volume of the cylinder. Therefore, a cone can be obtained by dividing a cylinder into three parts. In order to visualize this, we will do a little experiment.

First, let's understand the following: The volume of a solid can also be understood as its capacity. The **capacity** of a solid is how much of a liquid – we usually think of water – it can contain.

Now to the experiment. It can be done through the following steps.

**Step 1****.** Get an empty regular cylinder with a known height and radius of the circular base.

**Step 2****.** Get an empty regular cone with its height and radius the same as that of the cylinder.

**Step 3****.** Fill the cone with water to the brim.

** Step 4.** Pour all the water in the cone into the empty cylinder and observe the level of water in the cylinder.

Note that the cylinder is yet to be filled up.

**Step 5.** Repeat steps 3 and 4 for the second time on the same cylinder which already has water in it. Notice that the water level rise but the cylinder is yet to be filled to the brim.

**Step 6.** Repeat steps 3 and 4 for the third time on the same cylinder. Notice that the cylinder is now filled to the brim.

This experiment explains that you would need 3 cones to make a cylinder.

The **volume of a cone** is the 3D space occupied by it. It is the number of unit cubes that fit in it.

Since we know that a cone is one-third of its corresponding cylinder, it is easier to calculate the volume of a cone.

We recall that the volume of a cylinder is the product of its circular base area and its height and is given by,

$Volum{e}_{Cylinder}=\pi {r}^{2}\times h$.

Thus, the volume of a cone is one-third the product of its circular base area and height.

$Volum{e}_{Cone}=\frac{Volum{e}_{Cylinder}}{3}=\frac{\pi {r}^{2}\times h}{3}$

Note that the volume of an oblique cone is calculated with the same formula as that of a regular right cone. This is because of Cavalieri's principle which explains that the volume of a solid shape equals the volume of its corresponding oblique object.

The conical cap has a base radius of 7 cm and a height of 8 cm. Find the volume of the cone. Take $\pi =3.14$.

**Solution**

We first write out the given values $r=7,h=8cm,\mathrm{\pi}=3.14.$

The volume of the cone can be calculated by the formula,

$Volum{e}_{Cone}=\frac{\pi {r}^{2}\times h}{3}=\frac{3.14\times {7}^{2}\times 8}{3}=\frac{1230.88}{3}=410.29c{m}^{3}$

Sometimes we may be asked to find the volume of cones without a known height. In this case, we need to look out for either a known slant height or a known angle. With a slant height given, you can use Pythagoras' theorem to calculate it. Therefore, let * l* be the slant height,

Pythagoras' theorem is only applied when your radius r, slant height l is given but your height is not given.

When an angle is given, SOHCAHTOA is applied to derive the value of the height. If the angle at the apex is given, then half of that angle is used to find the height. Thus;

$\mathrm{tan}\frac{\theta}{2}=\frac{r}{h}h=\frac{r}{\mathrm{tan}\frac{\theta}{2}}$where;

** θ** is the angle at the apex,

** l** is the slant height of the cone,

** h** is the height of the cone,

** r** is the base radius of the cone.

The **frustum** of a cone or a **truncated cone** has the tip-top (vertex) cut off. It is the shape of most buckets.

To calculate the volume of a frustum, proportion is used. A frustum has two radii - a radius for the bigger circular surface and another radius for the smaller circular surface.

Let * R *be the radius of the bigger circular surface, and

Therefore, while using proportion we have,

$\frac{R}{r}=\frac{H}{{h}_{c}}$

The volume of the frustum is equal to the difference between the volume of the big cone and the volume of the small cut cone, and hence we have

$Volum{e}_{frustum}=Volum{e}_{bigcone}-Volum{e}_{smallcutcone}$

But,

$Volum{e}_{bigcone}=\frac{1}{3}\pi {R}^{2}H,Volum{e}_{smallcone}=\frac{1}{3}\pi {r}^{2}{h}_{c}$

Thus, the volume of the frustum is

$Volum{e}_{frustum}=\frac{1}{3}\pi {R}^{2}H-\frac{1}{3}\pi {r}^{2}{h}_{c}$

Calculate the volume of the frustum below

**Solution**

We write out the given values,

$R=20cm\xf72=10cmr=8cm\xf72=4cm{h}_{f}=15cm$

We need to find the height of the full cone $H$, and the height of the small cone ${h}_{c}$;

$H={h}_{f}+{h}_{c}H=15+{h}_{c}$

Using proportion;

$\frac{R}{r}=\frac{H}{{h}_{c}}$

Remember that;

$H=15+{h}_{c}\frac{10}{4}=\frac{15+{h}_{c}}{{h}_{c}}10\times {h}_{c}=4(15+{h}_{c})10{h}_{c}==60+4{h}_{c}10{h}_{c}-4{h}_{c}=606{h}_{c}=60\frac{6{h}_{c}}{6}=\frac{60}{6}{h}_{c}=10cm$

And so,

$H=15+{h}_{c}H=15+10=25cmVolumeofbigcone=\frac{1}{3}\pi {R}^{2}H==\frac{1}{3}\times \frac{22}{7}\times {10}^{2}\times 25=\frac{55000}{21}=2619.05c{m}^{3}$

Using proportion;

$\frac{R}{r}=\frac{Volumeofbigcone}{Volumeofsmallcone}\frac{10}{4}=\frac{2619.05}{Volumeofsmallcone}10\times Volumeofsmallcone=4\times 2619.05Volumeofsmallcone=\frac{4\times 2619.05}{10}Volumeofsmallcone=1047.62c{m}^{3}$

Recall that;

$Volumeoffrustum=Volumeofbigcone-VolumeofsmallconeVolumeoffrustum=2619.05-1047.62Volumeoffrustum=1571.43c{m}^{3}$

A cylinder has a radius $4.2cm$ and a height $10cm$. If a conical funnel with the same circular top and height were placed into the cylinder, find the volume of the funnel.

**Solution**

$Volumeofacylinder=\pi {r}^{2}hVolumeofcylinder=\frac{22}{7}\times 4.2\times 4.2\times 10Volumeofcylinder=554.4c{m}^{3}$

We are told the cone is of the same height and circular top as that of the cylinder. This means that the volume of the cone is one-third the volume of the cylinder.

$Volumeofacone=\frac{1}{3}\times volumeofacylinderVolumeofacone=\frac{1}{3}\times 554.4Volumeofacone=184.8c{m}^{3}$

A cone is a three dimensional solid that consists of a circular base and a continuous curved surface which tapers to a tip-top called the apex.

The right circular cone and the oblique circular cones are the two types of cones.

A cone is one third of a cylinder.

The volume of a cone is one-third the circular base area multiplied by its height.

The frustum of a cone or a truncated cone has the tip-top (vertex) cut off.

The volume of a right circular cone is the same as the volume of a regular cone.

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