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Right Regular Pyramid

Right Regular Pyramid

A right pyramid with a base that is a regular polygon. If the base is a triangle then the figure is called a tetrahedron.

Note: For some mathematicians regular pyramid means the same as right regular pyramid.

Additional note: The Egyptian pyramids are right regular pyramids with square bases.

Solid view of a right regular pyramid with a yellow pentagonal base and four triangular lateral faces tapering to an apex.
Solid view: right regular pyramid with a regular pentagon as base		 Solid view of a right regular pyramid with a pentagonal base, showing height h and slant height s labeled in red.
Frame view: right regular pyramid with a regular pentagon as base
Volume = (1/3)(area of base)(height)
Lateral surface area for a right regular pyramid = Formula: one-half times P times s
Total surface area for a right regular pyramid = B + (1/2)Ps, formula for total surface area of a right regular pyramid, where B is base area, P is perimeter, s is slant height.
h = height of the pyramid
B = area of the base
P = perimeter of the base
s = slant height

 

See also

Volume, area of a regular polygon

Key Formula

Lateral Surface Area=12PsTotal Surface Area=12Ps+BV=13Bh\text{Lateral Surface Area} = \tfrac{1}{2}Ps \qquad\qquad \text{Total Surface Area} = \tfrac{1}{2}Ps + B \qquad\qquad V = \tfrac{1}{3}Bh
Where:
  • PP = Perimeter of the regular polygon base
  • ss = Slant height — the distance from the apex to the midpoint of a base edge, measured along a lateral face
  • BB = Area of the regular polygon base
  • hh = Height (altitude) of the pyramid — the perpendicular distance from the apex to the base
  • VV = Volume of the pyramid

Worked Example

Problem: A right regular pyramid has a square base with side length 10 cm and a slant height of 13 cm. Find the lateral surface area, total surface area, and volume.
Step 1: Find the perimeter and area of the square base.
P=4×10=40 cmB=102=100 cm2P = 4 \times 10 = 40 \text{ cm} \qquad B = 10^2 = 100 \text{ cm}^2
Step 2: Calculate the lateral surface area using the formula.
LSA=12Ps=12(40)(13)=260 cm2\text{LSA} = \tfrac{1}{2}Ps = \tfrac{1}{2}(40)(13) = 260 \text{ cm}^2
Step 3: Calculate the total surface area by adding the base area.
TSA=260+100=360 cm2\text{TSA} = 260 + 100 = 360 \text{ cm}^2
Step 4: Find the height using the Pythagorean theorem. The slant height, the height, and half the base side form a right triangle (since the base is a square, the apothem equals half the side length, which is 5 cm).
h=s2−a2=132−52=169−25=144=12 cmh = \sqrt{s^2 - a^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12 \text{ cm}
Step 5: Calculate the volume.
V=13Bh=13(100)(12)=400 cm3V = \tfrac{1}{3}Bh = \tfrac{1}{3}(100)(12) = 400 \text{ cm}^3
Answer: Lateral surface area = 260 cm², total surface area = 360 cm², volume = 400 cm³.

Another Example

This example uses a hexagonal base instead of a square, showing how to compute the apothem and how to find the slant height from the pyramid's height rather than having it given directly.

Problem: A right regular pyramid has a regular hexagonal base with side length 6 cm and a height of 8 cm. Find the slant height, lateral surface area, and volume.
Step 1: Find the apothem of the regular hexagonal base. For a regular hexagon with side length a, the apothem is a√3/2.
apothem=632=33≈5.196 cm\text{apothem} = \frac{6\sqrt{3}}{2} = 3\sqrt{3} \approx 5.196 \text{ cm}
Step 2: Find the slant height using the Pythagorean theorem. The height, the apothem of the base, and the slant height form a right triangle.
s=h2+(33)2=64+27=91≈9.539 cms = \sqrt{h^2 + (3\sqrt{3})^2} = \sqrt{64 + 27} = \sqrt{91} \approx 9.539 \text{ cm}
Step 3: Find the perimeter and area of the hexagonal base. A regular hexagon with side a has perimeter 6a and area (3√3/2)a².
P=6×6=36 cmB=332(62)=543≈93.53 cm2P = 6 \times 6 = 36 \text{ cm} \qquad B = \frac{3\sqrt{3}}{2}(6^2) = 54\sqrt{3} \approx 93.53 \text{ cm}^2
Step 4: Calculate the lateral surface area.
LSA=12Ps=12(36)(91)=1891≈171.7 cm2\text{LSA} = \tfrac{1}{2}Ps = \tfrac{1}{2}(36)(\sqrt{91}) = 18\sqrt{91} \approx 171.7 \text{ cm}^2
Step 5: Calculate the volume.
V=13Bh=13(543)(8)=1443≈249.4 cm3V = \tfrac{1}{3}Bh = \tfrac{1}{3}(54\sqrt{3})(8) = 144\sqrt{3} \approx 249.4 \text{ cm}^3
Answer: Slant height ≈ 9.54 cm, lateral surface area ≈ 171.7 cm², volume ≈ 249.4 cm³.

Frequently Asked Questions

What is the difference between a right pyramid and a right regular pyramid?
A right pyramid has its apex directly above the centroid of its base, but the base can be any polygon. A right regular pyramid adds the requirement that the base must be a regular polygon (all sides and angles equal). So every right regular pyramid is a right pyramid, but not every right pyramid is a right regular pyramid.
How do you find the slant height of a right regular pyramid?
The slant height connects the apex to the midpoint of a base edge. It forms a right triangle with the pyramid's height and the apothem of the base. Use the Pythagorean theorem: s=h2+a2s = \sqrt{h^2 + a^2}, where hh is the height and aa is the apothem of the regular polygon base.
Why does the formula ½Ps work for lateral surface area?
Each lateral face of a right regular pyramid is a congruent isosceles triangle with base equal to one side of the polygon and height equal to the slant height. The area of one triangle is ½ × (side) × s. When you add all the triangles together, the sum of all the sides equals the perimeter P, giving ½Ps for the total lateral area.

Right Regular Pyramid vs. Oblique Pyramid

Right Regular PyramidOblique Pyramid
Apex positionDirectly above the center of the baseNot directly above the center of the base
Base requirementMust be a regular polygonCan be any polygon
Lateral facesAll congruent isosceles trianglesTriangles of different shapes and sizes
Lateral surface area formula½Ps (simple formula applies)Must calculate each face individually
Height measurementPerpendicular from apex to center of basePerpendicular from apex to base plane (not through center)

Why It Matters

Right regular pyramids appear frequently in geometry courses when studying three-dimensional solids, surface area, and volume. The Great Pyramid of Giza is a real-world example — a right regular pyramid with a square base. Understanding this shape also prepares you for more advanced topics like cross-sections of solids and calculus-based volume integration.

Common Mistakes

Mistake: Confusing the slant height with the lateral edge or the pyramid's height.
Correction: The slant height runs from the apex to the midpoint of a base edge (along a face). The lateral edge runs from the apex to a vertex of the base. The height is the perpendicular distance from the apex straight down to the base. These are three different measurements — draw and label a diagram to keep them straight.
Mistake: Using the height instead of the slant height in the lateral surface area formula.
Correction: The formula LSA = ½Ps requires the slant height s, not the vertical height h. If you are given only h, first find the apothem of the base, then compute s = √(h² + a²) before plugging into the formula.

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