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# Platonic Solids

The Platonic solids are a set of five regular polyhedra - the tetrahedron, cube, octahedron, dodecahedron, and icosahedron - with equivalent faces composed of congruent convex regular polygons. Each of the five Platonic solids depends upon a two-dimensional shape: The tetrahedron, octahedron, and icosahedron's faces are based on triangles; the cube's faces are based on squares, and the dodecahedron has pentagonal faces.

## Concepts to review before learning Platonic solids

Before diving into Platonic solids, there are a few concepts to make sure you're solid on, like polyhedra, pyramids, and cross sections. It's also a good idea to review how to find the area of a polygon, as you'll need this to find the surface area and volume of a Platonic solid.

## An overview of Platonic solids

Each of the Platonic solids has faces, edges, and vertices. When finding the surface area or volume of a Platonic solid, you will need to know the measurement of the edge. Luckily, all of the edges of a Platonic solid are the same. Let's take a look at the different Platonic solids and how to find the surface area and volume of each.

## The tetrahedron

The tetrahedron has 4 faces, each of which is an equilateral triangle. It has 6 edges and 4 vertices. At each vertex, 3 edges meet.

We can find the surface area and volume of a tetrahedron if we know the length of its edge (e). For this tetrahedron, $e=6\phantom{\rule{4pt}{0ex}}\mathrm{cm}$ . To find its surface area (A), we need to plug in the length to the formula:

$A=\sqrt{3}\cdot {e}^{2}$

$A=\sqrt{3}\cdot {6}^{2}=62.354\phantom{\rule{4pt}{0ex}}{\mathrm{cm}}^{2}$

We use a slightly different formula to find the tetrahedron's volume (V):

$V=\frac{{e}^{3}}{6\sqrt{2}}$

$V=\frac{{6}^{3}}{6\sqrt{2}}=25.456\phantom{\rule{4pt}{0ex}}{\mathrm{cm}}^{3}$

## The cube

The cube has 6 faces. Each face is a square. It has 12 edges and 8 vertices. At each vertex, 3 edges meet.

We can find the surface area and volume of a cube if we know the length of its edge (e). For this cube, e = 12 in. To find its surface area (A), we will plug in the length to the formula:

$A=6{e}^{2}$

$A=6{\left(12\right)}^{2}=684\phantom{\rule{4pt}{0ex}}{\mathrm{in}}^{2}$

To find the volume (V) of a cube, we will use this formula:

$V={e}^{3}$

$V={12}^{3}=1728\phantom{\rule{4pt}{0ex}}{\mathrm{in}}^{3}$

## The octahedron

The octahedron has 8 faces. Each is an equilateral triangle. It has 12 edges and 6 vertices. At each vertex, 4 edges meet.

We can find the surface area and volume of an octahedron if we know the length of its edge (e). For this example, e = 7.5 mm. To find its surface area (A), we will plug in the length to the formula:

$A=2\sqrt{3}\cdot {e}^{2}$

$A=2\sqrt{3}{\left(7.5\right)}^{2}=194.86\phantom{\rule{4pt}{0ex}}{\mathrm{mm}}^{2}$

To find the volume (V) of an octahedron, we'll use the formula:

$V=\frac{\sqrt{2}}{3{e}^{3}}$

$V=\frac{\sqrt{2}}{3{\left(7.5\right)}^{3}}=198.87\phantom{\rule{4pt}{0ex}}{\mathrm{mm}}^{3}$

## The dodecahedron

The dodecahedron has 12 faces. Each is a regular pentagon. It has 30 edges and 20 vertices. At each vertex, 3 edges meet.

To find the surface area and volume of a dodecahedron, we need to know the length of its edge (e). For this example, $e=32$ ft. To find its surface area (A), we will plug in the length to the formula:

$A=3\sqrt{25+10\sqrt{5}}\cdot {e}^{2}$

$A=3\sqrt{25+10\sqrt{5}}\cdot {32}^{2}=21141.23\phantom{\rule{4pt}{0ex}}{\mathrm{ft}}^{2}$

To find the dodecahedron's volume, use:

$V=15+\frac{7\sqrt{5}}{4{e}^{3}}$

$V=15+\frac{7\sqrt{5}}{4\cdot {3}^{3}}=251105\phantom{\rule{4pt}{0ex}}{\mathrm{ft}}^{2}$

## The icosahedron

The icosahedron has 20 faces. Each is an equilateral triangle. It has 30 edges and 12 vertices. At each vertex, 5 edges meet.

To find the surface area and volume of an icosahedron, we need to know the length of its edge (e). For this example, $e=2\phantom{\rule{4pt}{0ex}}\mathrm{m}$ . To find its surface area (A), we will plug in the length to the formula:

$A=5\sqrt{3}{e}^{2}$

$A=5\sqrt{3}{2}^{2}=34.64\phantom{\rule{4pt}{0ex}}{\mathrm{m}}^{2}$

If you'd like to find an icosahedron's volume (V), use the formula:

$V=\frac{5\left(3+\sqrt{5}\right)}{12}{e}^{3}$

$V=\frac{5\left(3+\sqrt{5}\right)}{12}{2}^{3}=17.454\phantom{\rule{4pt}{0ex}}{\mathrm{m}}^{3}$

## Practice questions for Platonic solids

a. Find the volume (V) of a tetrahedron with an edge (e) of 14 cm.

$V=\frac{{14}^{3}}{6}\sqrt{2}=323.384\phantom{\rule{4pt}{0ex}}{\mathrm{cm}}^{3}$

b. Find the surface area (A) of a cube with an edge (e) of 3.75 in.

$A=6{\left(3.75\right)}^{2}=84.38\phantom{\rule{4pt}{0ex}}{\mathrm{in}}^{2}$

c. What is the volume (V) of an octahedron with an edge (e) of 192 mm?

$V=\frac{\sqrt{2}}{3}{192}^{3}=3.333655×{10}^{6}\phantom{\rule{4pt}{0ex}}{\mathrm{mm}}^{3}$

d. What is the surface area (A) of a dodecahedron with an edge (e) of 11 ft?

$A=3\sqrt{25+10\sqrt{5}}×{11}^{2}=2498.13\phantom{\rule{4pt}{0ex}}{\mathrm{ft}}^{2}$

e. What is the volume (V) of an icosahedron with an edge (e) of 4.33 meters?

$V=\frac{5\left(3+\sqrt{5}\right)}{12}×{4.33}^{3}=177.12\phantom{\rule{4pt}{0ex}}{\mathrm{m}}^{3}$

Pyramid

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