Tetrahedron

Tetrahedron
Regular Tetrahedron

A polyhedron with four triangular faces, or a pyramid with a triangular base.

Note: A regular tetrahedron, which has faces that are equilateral triangles, is one of the five platonic solids.

Regular Tetrahedron

a = length of an edge

Volume =

Surface Area =

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Key Formula

V = \frac{a^3}{6\sqrt{2}} \qquad \text{and} \qquad SA = a^2\sqrt{3}

Where:

$a$ = Length of an edge of the regular tetrahedron
$V$ = Volume of the regular tetrahedron
$SA$ = Total surface area of the regular tetrahedron

Worked Example

Problem: Find the volume and surface area of a regular tetrahedron with edge length 6 cm.

Step 1: Identify the edge length. Here

a = 6

cm.

Step 2: Apply the volume formula for a regular tetrahedron.

V = \frac{a^3}{6\sqrt{2}} = \frac{6^3}{6\sqrt{2}} = \frac{216}{6\sqrt{2}}

Step 3: Simplify the volume. Divide 216 by 6 to get 36, then rationalize the denominator.

V = \frac{36}{\sqrt{2}} = \frac{36\sqrt{2}}{2} = 18\sqrt{2} \approx 25.46 \text{ cm}^3

Step 4: Apply the surface area formula. A regular tetrahedron has 4 equilateral triangular faces.

SA = a^2\sqrt{3} = 6^2\sqrt{3} = 36\sqrt{3} \approx 62.35 \text{ cm}^2

Answer: The volume is

18\sqrt{2} \approx 25.46

cm³ and the surface area is

36\sqrt{3} \approx 62.35

cm².

Another Example

This example works backward from a given surface area to find the edge length, then uses it to compute volume — a common reverse-problem variation.

Problem: A regular tetrahedron has a surface area of

48\sqrt{3}

cm². Find the edge length and the volume.

Step 1: Start from the surface area formula and solve for

a

SA = a^2\sqrt{3} \quad \Rightarrow \quad 48\sqrt{3} = a^2\sqrt{3}

Step 2: Divide both sides by

\sqrt{3}

to isolate

a^2

a^2 = 48 \quad \Rightarrow \quad a = \sqrt{48} = 4\sqrt{3} \text{ cm}

Step 3: Now compute

a^3

for the volume formula.

a^3 = (4\sqrt{3})^3 = 64 \cdot 3\sqrt{3} = 192\sqrt{3}

Step 4: Substitute into the volume formula.

V = \frac{192\sqrt{3}}{6\sqrt{2}} = \frac{32\sqrt{3}}{\sqrt{2}} = \frac{32\sqrt{6}}{2} = 16\sqrt{6} \approx 39.19 \text{ cm}^3

Answer: The edge length is

4\sqrt{3} \approx 6.93

cm and the volume is

16\sqrt{6} \approx 39.19

cm³.

Frequently Asked Questions

How many faces, edges, and vertices does a tetrahedron have?

A tetrahedron has 4 triangular faces, 6 edges, and 4 vertices. You can verify this with Euler's formula for polyhedra:

V - E + F = 4 - 6 + 4 = 2

, which checks out.

What is the difference between a tetrahedron and a triangular pyramid?

They are the same shape. A tetrahedron is simply a triangular pyramid — a pyramid whose base is a triangle. The term 'tetrahedron' (from Greek, meaning 'four faces') is the more formal geometric name. Any face of a tetrahedron can serve as the base.

What is the height of a regular tetrahedron?

The height (perpendicular distance from a vertex to the opposite face) of a regular tetrahedron with edge length

a

h = a\sqrt{\frac{2}{3}}

, which simplifies to

h = \frac{a\sqrt{6}}{3}

. For example, if

a = 6

, the height is

2\sqrt{6} \approx 4.90

Regular Tetrahedron vs. Cube

	Regular Tetrahedron	Cube
Type	Platonic solid with triangular faces	Platonic solid with square faces
Faces	4 equilateral triangles	6 squares
Edges / Vertices	6 edges, 4 vertices	12 edges, 8 vertices
Volume formula	$V = \frac{a^3}{6\sqrt{2}}$	$V = a^3$
Surface area formula	$SA = a^2\sqrt{3}$	$SA = 6a^2$
Volume (a = 6)	≈ 25.46 cubic units	216 cubic units

Why It Matters

The tetrahedron appears throughout geometry courses when studying polyhedra, spatial reasoning, and the Platonic solids. In chemistry, it describes the molecular shape of compounds like methane (CH₄), where four atoms surround a central atom. Understanding its volume and surface area formulas also prepares you for more advanced topics in solid geometry and calculus involving three-dimensional integration.

Common Mistakes

Mistake: Using

\frac{1}{3} \times \text{base area} \times \text{height}

but calculating the height of the tetrahedron incorrectly — often confusing it with the height of a triangular face.

Correction: The height of a regular tetrahedron is

h = \frac{a\sqrt{6}}{3}

, which is the perpendicular distance from a vertex straight down to the center of the opposite face. The height of each equilateral triangular face is

\frac{a\sqrt{3}}{2}

, which is a different measurement.

Mistake: Forgetting that the surface area formula

a^2\sqrt{3}

already accounts for all four faces.

Correction: The area of one equilateral triangle with side

a

\frac{a^2\sqrt{3}}{4}

. Multiplying by 4 gives

a^2\sqrt{3}

for the full surface area. Do not multiply

a^2\sqrt{3}

by 4 again.

Related Terms

Polyhedron — General class of 3D solids with flat faces
Pyramid — A tetrahedron is a pyramid with a triangular base
Triangle — Each of the four faces is a triangle
Equilateral Triangle — All faces of a regular tetrahedron
Platonic Solids — The tetrahedron is the simplest Platonic solid
Face of a Polyhedron — A tetrahedron has exactly four faces
Volume — Measured using the tetrahedron volume formula
Surface Area — Sum of areas of four triangular faces