Math by First Principles

Surface Areas of Pyramids and Cones

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Learning goals

  • Derive regular-pyramid and right-cone lateral areas from their unfolded surfaces
  • Distinguish slant height from perpendicular height and calculate the needed length
  • Select the surfaces included in an open or closed boundary
  • Reconstruct a right cone and its sector pattern from linked measurements

Triangles around a regular pyramid

A regular pyramid has a convex regular polygon as its base and its apex directly above the base’s center. Its triangular side faces are congruent. Their common face altitude, measured from the apex perpendicular to a base edge within that face, is the slant height, written ℓ\ell. This is different from an edge running from the apex to a base vertex.

Its lateral area is the combined area of the triangular side faces, excluding the base. A closed pyramid’s total surface area also includes its one polygonal base.

Worked example 1 Add the unfolded faces

Problem. A regular square pyramid has base side 66 cm and slant height 55 cm. Find its lateral area and its closed total surface area.

A square-pyramid netComplete net of a regular square pyramid. The central square has side 6 cm. Four congruent triangles attach, one to each square edge. A dotted perpendicular within the top triangle labels face altitude 5 cm. Dashed square edges are folds.s = 6ℓ = 5Dimensions in cm
Cut edges are solid; the four dashed square edges fold. The dotted segment is perpendicular within one triangular face.

Solution. Each triangle has base 66 cm and perpendicular face altitude 55 cm. Its area is 1515 square centimeters. There are four triangles.

L=4(12⋅6⋅5)L=60 cm2.\begin{aligned} L&=4\left(\tfrac12\cdot6\cdot5\right)\\ L&=60\text{ cm}^2. \end{aligned}

The central square is the base. Add it once.

B=62B=36 cm2,S=60+36S=96 cm2.\begin{aligned} B&=6^2\\ B&=36\text{ cm}^2,\\ S&=60+36\\ S&=96\text{ cm}^2. \end{aligned}

The side lengths used as triangle bases add to the base perimeter. In the example that sum is 2424 cm, and half of 2424 times 55 gives the same lateral area. The argument works for any regular pyramid: every side triangle has the same face altitude, so factor half that altitude from the sum of their areas.

With PP for base perimeter and BB for base area,

L=12Pℓ,Sclosed=12Pℓ+B.\begin{aligned} L&=\tfrac12P\ell,\\ S_{\text{closed}}&=\tfrac12P\ell+B. \end{aligned}

These are the regular-pyramid shortcuts. If face altitudes differ, add the individual triangle areas with their own base lengths and altitudes. One chosen slant length cannot stand for every face.

Check your understanding

A regular pyramid has base perimeter 3232 cm and slant height 99 cm. What is its lateral area?

Find the height that lies on the surface

The solid’s perpendicular height hh runs from the apex to the base plane. It measures height through the interior. The slant height lies on a triangular face and measures that face’s area.

For a regular pyramid, join the base center to the midpoint of a base side. That segment is the base’s apothem aa. It meets the edge perpendicularly, and the apex-to-midpoint segment is the face altitude. These three segments form a right triangle with hypotenuse ℓ\ell.

Worked example 2 Use a section through a face midpoint

Problem. A regular square pyramid has perpendicular height 1212 cm and base side 1010 cm. Find its slant height and closed surface area.

Identify the perpendicular triangleTrue right triangle VOM in a regular pyramid. VO is perpendicular height 12 cm; OM is base apothem 5 cm; VM is slant height ℓ. The right angle is at O.12 cm5 cmℓVOM
A true section triangle through apex V, base center O, and side midpoint M. This is a flat section, not a perspective sketch.

Solution. The square’s center is 55 cm from a side midpoint, half its side length. In the true right triangle VOMVOM, the perpendicular height and the 55 cm base distance are legs.

ℓ2=122+52ℓ=13 cm.\begin{aligned} \ell^2&=12^2+5^2\\ \ell&=13\text{ cm}. \end{aligned}

Use the face altitude 1313 cm for lateral area, then add the square base.

L=12(40)(13)L=260 cm2,S=260+100S=360 cm2.\begin{aligned} L&=\tfrac12(40)(13)\\ L&=260\text{ cm}^2,\\ S&=260+100\\ S&=360\text{ cm}^2. \end{aligned}

Nothing depended on the particular height or side length. The same right triangle gives, for a regular pyramid,

ℓ2=h2+a2.\ell^2=h^2+a^2.

The base apothem aa goes to a side midpoint, not a vertex. For a square it is half a side; for other regular bases, use the base geometry already learned.

Check your understanding

A regular square pyramid has base side 2020 cm and perpendicular height 2424 cm. What is its slant height?

A right cone unfolds into a sector

Cut a right circular cone along one straight segment from apex to rim and flatten its curved side. Predict which length becomes the sector radius: the base radius, the perpendicular height, or the slant height.

Every straight apex-to-rim segment of a right cone has the same length ℓ\ell. When flattened, those segments become radii of a sector. The sector’s curved arc goes once around the base rim, so its length is the base circumference.

Match the cut side to a sectorA folded right cone has base radius 3 cm and slant height 6 cm. The curved side is shown below as a dashed semicircular sector of radius 6 cm. Its selected arc is labeled 6 pi cm. A separate base disk has radius 3 cm and rim length 6 pi cm. The rear half of the folded rim is dashed.36folded cone66180°sector arc 6π cm3base rim 6π cm
Dimensions are in centimeters. Compare the folded cone, its unfolded curved side, and the separate base disk.

For the illustrated cone, the slant height is 66 cm and the base radius is 33 cm. The arc length is 6π6\pi cm. A full circle of radius 66 cm has circumference 12π12\pi cm, so the needed sector is half that circle. Its area is 18π18\pi square centimeters.

For a general right cone, the sector radius is ℓ\ell and its arc length is 2πr2\pi r. The fraction of the full radius-ℓ\ell circle is therefore r/ℓr/\ell. Apply that fraction to the full circle’s area.

L=rℓ⋅πℓ2L=πrℓ.\begin{aligned} L&=\frac{r}{\ell}\cdot\pi\ell^2\\ L&=\pi r\ell. \end{aligned}

The same fraction gives the sector angle θ\theta in degrees.

θ=360∘rℓ.\theta=360^\circ\frac{r}{\ell}.

The curved surface is the cone’s lateral surface; it is not a planar face. A closed right cone includes one base disk, so

Sclosed=πrℓ+πr2.S_{\text{closed}}=\pi r\ell+\pi r^2.

These formulas need a right circular cone with positive radius and positive perpendicular height. Then ℓ>r\ell>r, and the sector angle lies strictly between zero and a full turn. An oblique cone has different apex-to-rim lengths; its curved surface is not this circular sector, and πrℓ\pi r\ell is not its surface shortcut.

Check your understanding

A sector of radius 1212 cm forms the curved side of a right cone without overlap. Its selected arc is 8π8\pi cm long. What is the base radius?

Use the cone’s right triangle and its boundary

A plane through the axis of a right cone gives an isosceles triangle. Half of it has perpendicular height hh and base radius rr as its legs. The slant height is its hypotenuse.

Worked example 3 From an axial section to curved area

Problem. A right circular cone has base radius 55 cm and perpendicular height 1212 cm. Find its slant height, lateral area, and closed total surface area exactly.

Identify the perpendicular triangleTrue right triangle VOM in a right cone. VO is perpendicular height 12 cm; OM is radius 5 cm; VM is slant height ℓ. The right angle is at O.12 cm5 cmℓVOM
Half of an axial section. The base radius and perpendicular height meet at the square mark.

Solution. In half of the axial section, apply Pythagoras to the two legs.

ℓ2=122+52ℓ=13 cm.\begin{aligned} \ell^2&=12^2+5^2\\ \ell&=13\text{ cm}. \end{aligned}

The curved side uses the slant height. The base is one disk.

L=π(5)(13)L=65π cm2,S=65π+25πS=90π cm2.\begin{aligned} L&=\pi(5)(13)\\ L&=65\pi\text{ cm}^2,\\ S&=65\pi+25\pi\\ S&=90\pi\text{ cm}^2. \end{aligned}

In general the half-section gives

ℓ2=h2+r2.\ell^2=h^2+r^2.

Surface area depends on which surfaces actually exist in the boundary. An opening has no material disk to add. The pointed apex has no extra cap.

Worked example 4 Count the included surfaces

Problem. A thin right-conical shade has radius 44 cm and slant height 99 cm. Ignore thickness, seams, and overlap. Find the material area in each case.

Case A: the shade includes its entire curved side and has an open circular rim, with no disk across it.

Case B: the same curved side also has a circular disk across that rim.

Specify which surfaces are presentTwo right-conical shade boundaries, each with base radius 4 cm and slant height 9 cm. Case A has the complete curved side and an open rim with no base disk. Case B has that curved side and one base disk, lightly shaded. Rear rim arcs are dashed. Dimensions are in centimeters.94A: disk absent94B: disk included
Case A omits the base disk; Case B includes it. Dimensions are in centimeters.

Solution. Case A includes only the sector-shaped curved side.

L=36π cm2.L=36\pi\text{ cm}^2.

Case B includes that side and one radius-44 disk.

S=36π+16πS=52π cm2.\begin{aligned} S&=36\pi+16\pi\\ S&=52\pi\text{ cm}^2. \end{aligned}

The circle around the opening is an edge of the ideal covering. Its circumference is a length, not an extra area. This model counts one layer of material, rather than separately coating its inner and outer faces.

Check your understanding

An ideal right-conical cover has radius 22 cm and slant height 55 cm. Its entire curved side is present, but its circular base is absent. What is its material area?

Reconstruct a cone and check the pattern

Area data can determine lengths, but those lengths must still fit a real right cone. Recover the base radius and slant height, check that the slant height is longer, then match the sector arc to the base circumference.

Worked example 5 Recover a cone from two surface measurements

Problem. A closed right cone has total surface area 150π150\pi square centimeters. Its curved side alone has area 125π125\pi square centimeters. Find its radius, slant height, perpendicular height, and sector angle exactly. Check the sector’s arc length.

Solution. The difference between total and lateral area is one base disk.

πr2=150π−125πr2=25r=5 cm.\begin{aligned} \pi r^2&=150\pi-125\pi\\ r^2&=25\\ r&=5\text{ cm}. \end{aligned}

Use the curved area to recover the slant height.

125π=π(5)ℓℓ=25 cm.\begin{aligned} 125\pi&=\pi(5)\ell\\ \ell&=25\text{ cm}. \end{aligned}

Since 2525 exceeds 55, the perpendicular height is positive. Subtract the radius-square from the slant-height square.

h2=252−52h=106 cm.\begin{aligned} h^2&=25^2-5^2\\ h&=10\sqrt6\text{ cm}. \end{aligned}

The sector radius is 2525 cm. Its angle occupies 5/255/25 of a full turn.

θ=72∘.\theta=72^\circ.
A sector patternA circular sector with radius 25 cm and selected angle 72 degrees. The straight radii are cut edges.25 cm72°
The reconstructed pattern shows the sector radius and selected angle.

The arc check uses the same fraction of the full circumference.

s=72360⋅2π(25)s=10π cm.\begin{aligned} s&=\tfrac{72}{360}\cdot2\pi(25)\\ s&=10\pi\text{ cm}. \end{aligned}

That is the circumference of the recovered radius-55 base. The sector area is one fifth of 625π625\pi, agreeing with the given lateral area.

Check your understanding

A right cone has base radius 88 cm and slant height 1010 cm. What angle must its sector pattern have?

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Other explanations of this lesson, if you want a second take.

A bit of history (optional)

In 1525, Albrecht Dürer published a geometry book in Nuremberg with an unusual way of showing solids. Its flat patterns laid out the faces of three-dimensional bodies on the page. Instead of trying to see through a perspective picture, a reader could inspect the polygons and their connections directly.

The first edition described fourteen solid bodies using planar nets. Dürer continued working in his own copy, adding nets for two more bodies. Those additions appeared in the 1538 edition, after his death. A printed book could preserve an artist’s revisions as well as his finished designs.

The Metropolitan Museum’s original book record identifies the 1525 edition and its printer. Vera Viana’s historical study, section 2.3 traces the nets and the later additions.

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