# 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?

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

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

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

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

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

## Topics related to the Platonic Solids

## Flashcards covering the Platonic Solids

Common Core: High School - Geometry Flashcards

## Practice tests covering the Platonic Solids

Common Core: High School - Geometry Diagnostic Tests

Advanced Geometry Diagnostic Tests

## Conquering Platonic solids with Varsity Tutors

If you or your student is struggling with Platonic solids, private tutoring is one of the most effective ways to get back on track - and stay there. Working with a tutor with experience in concepts like Platonic solids and the math behind them provides students with an individualized experience that is conducive to learning and retaining the information that's necessary for success. To find out more about math tutoring and what it can do, reach out to the Educational Directors at Varsity Tutors.

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