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Volume of Pyramid

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Do you know that the Great Pyramid of Giza measures about 146.7 m high and 230.6 m in base length? Can you imagine how many cubes of sugar measuring 1 m^{3} would be needed to fill the Great Pyramid of Giza? Herein, you shall be learning about how this can be calculated through the knowledge of the volume of pyramids.

Pyramids are 3-dimensional objects with triangular sides or surfaces that meet at a tip called an apex. The name 'pyramid' often brings to mind the Pyramids of Egypt, which is one of the seven wonders of the world.

In geometry, a **pyramid** is a polyhedron obtained connecting a polygonal base to a point, called the *apex*.

Pyramids are of various types depending on the shape of their base. A pyramid with a **triangular base** is called a **triangular pyramid,** and a **rectangular-based pyramid **is known as a **rectangular pyramid**. The sides of a pyramid are triangular and they emerge from its base. They all meet at a point called the apex.

You may be wondering how many blocks of sand can make up the Egyptian pyramids. The volume of a pyramid is the space enclosed by its faces. Generally, the volume of a pyramid is a-third of its corresponding prism. Its** corresponding prism** has the same base shape, base dimensions and height. Thus, the general formula for calculating the volume of a pyramid is,

$V=\frac{1}{3}\times bh$

where,

V is the volume of the pyramid

b is the base area of pyramid

h is the height of pyramid

Note that this is the general formula for the volume of all pyramids. Differences in the formulas are based on the shape of the base of the pyramid.

The volume of rectangular pyramids can be found by multiplying a third of the rectangular base area by the height of the pyramid. Therefore:

$Volumeofrec\mathrm{tan}gularpyramid=\frac{1}{3}\times baseArea\times height\phantom{\rule{0ex}{0ex}}Basearea=length\times breadth\phantom{\rule{0ex}{0ex}}Volume=\frac{1}{3}\times l\times b\times h$

where;

l is the length of the base

b is the breadth of the base

h is the height of the pyramid

This means that the volume of a rectangular pyramid is a third of the corresponding rectangular prism.

A square base pyramid is a pyramid whose base is a square. The volume of square-based pyramids can be gotten by multiplying one-third of the square base area by the height of the pyramid. Therefore:

$Volumeofsquarebasepyramid=\frac{1}{3}\times baseArea\times height\phantom{\rule{0ex}{0ex}}Basearea=lengt{h}^{2}\phantom{\rule{0ex}{0ex}}Volume=\frac{1}{3}\times {l}^{2}\times h$

where;

l is the length of the square base

h is the height of the pyramid

The volume of triangular base pyramids can be obtained by multiplying one-third of the triangular base area by the height of the pyramid. Therefore:

$Volumeoftriangularbasepyramid=\frac{1}{3}\times baseArea\times height\phantom{\rule{0ex}{0ex}}Basearea=\frac{1}{2}\times baselength\times heightoftriangle\phantom{\rule{0ex}{0ex}}Volume=\frac{1}{3}\times \frac{1}{2}\times b\times {h}_{triangle}\times {h}_{pyramid}\phantom{\rule{0ex}{0ex}}V=\frac{1}{6}\times b\times {h}_{triangle}\times {h}_{pyramid}$

where;

l is the length of the base

b is the triangular base length

h_{triangle} is the height of the triangular base

h_{pyramid} is the height of the pyramid

The volume of hexagonal base pyramids can be gotten by multiplying one-third of the hexagonal base area by the height of the pyramid. Therefore:

$Volumeoftriangularbasepyramid=\frac{1}{3}\times baseArea\times height\phantom{\rule{0ex}{0ex}}Basearea=\frac{3\sqrt{3}}{2}\times lengt{h}^{2}\phantom{\rule{0ex}{0ex}}Volume=\frac{1}{3}\times \frac{3\sqrt{3}}{2}\times {l}^{2}\times h\phantom{\rule{0ex}{0ex}}Volume=\frac{\sqrt{3}}{2}\times {l}^{2}\times h$

A pyramid of height 15ft has a square base of 12 ft. Determine the volume of the pyramid.

**Solution**

$Volumeofsquarebasepyramid=\frac{1}{3}\times {l}^{2}\times h\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}l=12ft\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}h=15ft\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}V=\frac{1}{3}\times {12}^{2}\times 15\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}V=5\times 144\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}V=720f{t}^{3}$

Calculate the volume of the figure below:

**Solution**

$Thevolumeofthefigure=volumeofrectangularpyramid+volumeofrectangularprism\phantom{\rule{0ex}{0ex}}Volumeofrectanglarpyramid=\frac{1}{3}\times l\times b\times h\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}l=45cm\phantom{\rule{0ex}{0ex}}b=20cm\phantom{\rule{0ex}{0ex}}h=50cm\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Volumeofrectanglarpyramid=\frac{1}{3}\times 45\times 20\times 50\phantom{\rule{0ex}{0ex}}Volumeofrectanglarpyramid=15000c{m}^{3}\phantom{\rule{0ex}{0ex}}Volumeofrectangularprism=l\times b\times h\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}l=45cm\phantom{\rule{0ex}{0ex}}b=20cm\phantom{\rule{0ex}{0ex}}h=40cm\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Volumeofrectangularprism=45\times 20\times 40\phantom{\rule{0ex}{0ex}}Volumeofrectangularprism=36000c{m}^{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Thevolumeofthefigure=volumeofrectangularpyramid+volumeofrectangularprism\phantom{\rule{0ex}{0ex}}Thevolumeofthefigure=15000+36000\phantom{\rule{0ex}{0ex}}Thevolumeofthefigure=51000c{m}^{3}$

A hexagonal pyramid and a triangular pyramid are of the same capacity. If its triangular base has a length of 6 cm and a height of 10 cm, calculate the length of each side of the hexagon when both pyramids have the same height.

**Solution**

The first step is to express the relationship in an equation.

According to the problem, the volume of the triangular pyramid equals the volume of the hexagonal pyramid.

Let b_{t} signify the base area of triangular base and b_{h} represent the base area of hexagonal base.

Then:

$Volumeoftriangularpyramid=Volumeofhexagonalpyramid\phantom{\rule{0ex}{0ex}}\frac{{b}_{t}h}{3}=\frac{{b}_{h}h}{3}$

Multiply both sides of the equation by 3 and divide by h.

$\frac{{b}_{t}h}{3}=\frac{{b}_{h}h}{3}\frac{{b}_{t}h}{3}\times \frac{3}{h}=\frac{{b}_{h}h}{3}\times \frac{3}{h}{b}_{t}={b}_{h}$

This means that the triangular base and the hexagonal base are of equal area.

Recall that we are required to find the length of each side of the hexagon.

${b}_{t}=\frac{1}{2}\times baselength\times height\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}baselengthoftriangle=6cm\phantom{\rule{0ex}{0ex}}heightoftriangle=10cm\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{b}_{h}=\frac{3\sqrt{3}}{2}\times {l}^{2}$

Where l is the length of the side of a hexagon.

Recall that b_{t} = b_{h}, then;

$\frac{1}{2}\times 6\times 10=\frac{3\sqrt{3}}{2}\times {l}^{2}\phantom{\rule{0ex}{0ex}}\frac{1}{2}\times 6\times 10\times \frac{2}{3\sqrt{3}}=\frac{3\sqrt{3}}{2}\times {l}^{2}\times \frac{2}{3\sqrt{3}}\phantom{\rule{0ex}{0ex}}\frac{20}{\sqrt{3}}={l}^{2}$

Take the roots of both sides of the equation.

$\sqrt{{l}^{2}}=\sqrt{11.547}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}l=3.398cm$

Thus each side of the hexagonal base is approximately 3.4 cm.

- A pyramid is a 3-dimensional object with triangular sides or surfaces that meet at a tip called an apex
- The various types of pyramids are based on the shape of their base
- The volume of a pyramid is one-third the base area × height

It is the capacity of a pyramid or the space it contains.

More about Volume of Pyramid

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