Covering a pyramid-shaped cap requires the triangular face’s altitude along the surface. Substituting the whole solid’s vertical height would underestimate the material. Let us construct a square pyramid and distinguish three segments: height, apothem and lateral edge.
One apex outside the base plane
Prerequisites: polygon area, normals to a plane, sections, prisms and nets.
Take a planar convex polygon and P outside its plane. Joining P to its vertices gives a pyramid: the polygon is the base and triangles sharing apex P are lateral faces. An n-gonal pyramid has n+1 faces, n+1 vertices and 2n edges: n base and n lateral edges. A triangular pyramid is a tetrahedron; “tetrahedron” alone does not mean all edges are equal.
The height h is the length of perpendicular PH from P to the base plane. In a general pyramid H can lie outside the polygon. It is neither a lateral edge nor a triangular face’s altitude.
A pyramid is regular if its base is a regular polygon and H is its center. Then lateral edges are equal and lateral faces are congruent isosceles triangles: equal center-to-vertex distances yield equal edges by Pythagoras, followed by triangle congruence from three sides.
Three segments in one model
Define square ABCD with side 8 cm, center H, and apex P with PH=3 cm above H. Let M be AB’s midpoint. In a regular pyramid, face PAB’s altitude PM is the apothem l, or slant height. It is perpendicular to AB and lies in the lateral face. All apothems of a regular pyramid are equal.
Height h, apothem l, edge b
Section PHM at true scale
Parallel cut: t=1/2
In the square HM=4 cm and HA=4√2 cm. PHM and PHA are right triangles at H, so:
l = √((3)^(2) + (4)^(2)) = 5 cm.
The lateral edge PA=√(3²+(4√2)²)=√41≈6.40 cm. Thus h=3, l=5 and PA≈6.40 cm are different. Within face PAB, PA²=PM²+MA²=25+16=41 independently agrees.
In general, l²=h²+r² in a regular pyramid, with r the base’s inradius. For a lateral edge use circumradius R: b²=h²+R². “Base radius” alone is ambiguous.
A regular triangular pyramid has an equilateral base and apex above its center, but lateral edges need not equal base edges. Only when all six edges are equal is it a regular tetrahedron, with equilateral triangular faces.
Area: triangles plus the base
One lateral face in our pyramid has area 8·5/2=20 cm². Four give 80 cm². In general, for a regular pyramid with base perimeter Pbase:
S_(lat) = (P_(base) l)/(2).
Sum aᵢl/2 over all base sides and factor out the shared apothem. Total area adds one base: 80+8²=144 cm². A prism had two bases; carrying that habit over adds unnecessary area.
A cap without a bottom needs only the 80 cm² lateral area in our zero-thickness model. Seams, tabs and waste are unspecified. For a general pyramid calculate each face with its own altitude lᵢ, summing aᵢlᵢ/2. Do not substitute h for l or assume a shared apothem without regularity or a separate proof of equal face altitudes.
Sections change scale
A plane through P and a base diagonal gives a triangular diagonal section. A plane parallel to the base between it and P gives a polygon similar to the base. If the distance from P to that plane is fraction t of the full height, 0<t<1, lengths scale by t and area by t². This follows from similar triangles in planes through lateral edges.
Halfway down from our apex, the section is a square of side 4 cm and area 16 cm²: one quarter of 64 cm², not half. A prism would have the same section at every height; a pyramid narrows.
Optional calculationThe original pyramid and its area have already been explained.
Branch · A regular pyramid’s frustum
Discard the small top pyramid after the parallel cut t=1/2. The remainder has base sides 8 and 4 cm, vertical height 1.5 cm and apothem s=5−2.5=2.5 cm. Its lateral faces are congruent isosceles trapezoids. The small pyramid’s lateral area would be t²·80=20 cm², leaving 80−20=60 cm².
The general formula Sfrustum=(Pbig+Psmall)s/2 also follows algebraically: subtract (tP)(tl)/2 from Pl/2 to obtain P(1+t)·l(1−t)/2. This is the mean perimeter times the frustum apothem. Here (32+16)·2.5/2=60; adding bases 64+16 gives total area 140 cm². The formula requires a regular original pyramid and a parallel cut; those conditions are not automatic for an arbitrary truncated shape.
Sources and limits
Definitions of height, regular pyramid, apothem and area sum: Merzliak and coauthors, 2019, section 18, pp.109–112. Additionally, AMSI, Cones, Pyramids and Spheres explains the face altitude using Pythagoras and summing face areas. The 8 cm square, height 3 cm, cut and diagrams are our own, not a historical pyramid survey. Volume belongs to G034.