Square Pyramid Formulas (Volume, Surface Area, Slant Height)

The formulas on this page are for a right square pyramid: a square base of side b with four triangular faces meeting above the centre. Any pyramid has volume V = ⅓·B·h where B is the base area; a square base gives V = ⅓·b²·h. Its slant height is ℓ = √((b/2)² + h²), its lateral surface is 2bℓ, and its total surface area is b² + 2bℓ.

Square Pyramid formulas

Volume (any pyramid): $V = \tfrac{1}{3} B h$
Slant height: $\ell = \sqrt{(b/2)^2 + h^2}$
Volume (square base): $V = \tfrac{1}{3} b^2 h$
Lateral surface: $LSA = 2 b \ell$
Surface area: $SA = b^2 + 2 b \ell$

where B = area of the base (any pyramid), b = base edge length (square base), h = vertical height, ℓ = slant height (face), V = volume

Volume (any pyramid)

V = \tfrac{1}{3} B h

Slant height

\ell = \sqrt{(b/2)^2 + h^2}

Volume (square base)

V = \tfrac{1}{3} b^2 h

Worked example

A square pyramid with base 6 and height 4 has volume 48 cubic units.

V = \tfrac13 b^2 h = \tfrac13\cdot 36\cdot 4 = 48

Lateral surface

LSA = 2 b \ell

Surface area

SA = b^2 + 2 b \ell

Worked example

With slant height ℓ = 5, its total surface area is 96 square units.

SA = b^2 + 2b\ell = 36 + 2\cdot 6\cdot 5 = 96

Interactive square pyramid explorer

Drag a slider (or type a value, or use the arrow keys) and watch the diagram, the results, and the full step-by-step substitution update live.

Base edge b

Vertical height h

Slant height ℓ 5units

Volume V 48units³

Lateral surface LSA 60units²

Surface area SA 96units²

Step-by-step substitution

slant ℓ = √((b/2)² + h²) = √(9 + 16) = 5
V = ⅓ · b² · h = ⅓ · 6² · 4 = 48
LSA = 2 · b · ℓ = 2 · 6 · 5 = 60
SA = b² + 2bℓ = 36 + 60 = 96

Classroom use: square pyramid formulas

Who it's for: Grades 7–12 · Students learning pyramid volume and surface area, including slant-height work in high school.

Learning objectives

After using this page, students will be able to:

Compute pyramid volume using V = ⅓Bh.
Find slant height from base and height, then compute lateral surface area.
Compare a pyramid with a prism on the same base and height.

Suggested classroom activity

Run this immediately after the cone. Both carry the same ⅓, for the same reason, and asking students to predict the pyramid rule from the cone rule usually succeeds — a good moment to name the general V = ⅓Bh.

Standards alignment

Aligned to the Common Core State Standards for Mathematics. Codes are given exactly as published so they can be matched against state crosswalks.
Code	Standard
`7.G.B.6`	Solve real-world problems involving area, volume, and surface area of 2-D and 3-D objects.
`HSG.GMD.A.1`	Give an informal argument for circumference, area, and volume formulas.
`HSG.GMD.A.3`	Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems.
`8.G.B.7`	Apply the Pythagorean theorem to find unknown side lengths in right triangles.

Free to use and adapt in class under the terms on our license page — no account, no sign-up, no ads. Content last reviewed August 2026.

Common mistakes

Using slant height ℓ in the volume formula — volume uses the vertical height h.
Forgetting the ⅓ factor that makes a pyramid one third of its prism.

Square Pyramid formula card

Related & next steps

Practise square pyramid problems → Add square pyramid to your formula sheet →

Frequently asked questions

What is the volume of a pyramid?

A pyramid’s volume is V = ⅓·(base area)·h. For a square base of side b, that is V = ⅓·b²·h. A base of 6 and height 4 gives volume 48 cubic units.

How do you find the slant height of a pyramid?

For a square pyramid, the slant height is ℓ = √((b/2)² + h²), from the apex to a base-edge midpoint. With base 6 (half = 3) and height 4, ℓ = √(9 + 16) = 5 units.

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