Sec 2 Mensuration: Pyramids, Cones & Spheres

Secondary 2 G3 Mathematics · Study guide

Separate perpendicular height from slant height, choose volume or surface area, and count only the exposed surfaces of a composite solid.

Concept → worked examples → practice14 min read · practice linked below

Decide what is being measured

Volume measures space inside a solid, in cubic units such as cm³. Surface area measures its exposed boundary, in square units such as cm². Filling a container concerns volume; painting its outside concerns surface area.

Before you begin: revise circle area $\pi r^2$ , circumference $2\pi r$ , triangle area $\frac12 bh$ , and Pythagoras’ theorem. Convert every length to the same unit before calculating. Keep $\pi$ or the full calculator value until the final rounding.

Formulae and the dimensions they need
Solid	Volume	Surface area
Pyramid	$\frac13 Ah$ , base area $A$	Base plus triangular faces
Right circular cone	$\frac13\pi r^2h$	Curved: $\pi rl$ ; with base: $\pi rl+\pi r^2$
Sphere	$\frac43\pi r^3$	$4\pi r^2$
Hemisphere	$\frac23\pi r^3$	Curved: $2\pi r^2$ ; with base: $3\pi r^2$

Here $h$ is the perpendicular height. For a cone, $l$ is the slant height, with $l^2=h^2+r^2$ . A pyramid’s triangular face also has a slant height, measured to the base edge within that face.

A pyramid has a vertical height and face heights

Worked example 1 — a square-based pyramid

A right square pyramid has base side 10 cm and perpendicular height 12 cm. Find its volume and total surface area.

Its apex is directly above the base centre. The base area is $10^2=100\text{ cm}^2$ , so the volume is

$V=\frac13(100)(12)=400\text{ cm}^3.$

For a triangular face, the perpendicular height within that face runs from the apex to the midpoint of a base edge. The horizontal distance from the base centre to that midpoint is $10\div2=5\text{ cm}$ . Hence

$l=\sqrt{12^2+5^2}=13\text{ cm}.$

Cross-section through the apex, base centre and an edge midpoint. The slant height is not an edge from the apex to a base corner.

Each of the four congruent triangular faces has area $\frac12(10)(13)=65\text{ cm}^2$ . Add all four faces and the square base:

$A=4(65)+100=360\text{ cm}^2.$

Check: the slant height 13 cm exceeds the perpendicular height 12 cm. The volume uses 12, while the face areas use 13. Their units must also differ.

A cone: distinguish its curved surface from its base

Worked example 2 — a cone with its base included

A solid right circular cone has radius 5 cm and perpendicular height 12 cm. Find its volume and total surface area, first exactly and then to 3 significant figures.

The right triangle through the axis gives $l=\sqrt{5^2+12^2}=13\text{ cm}$ . Use the perpendicular height for volume:

$V=\frac13\pi(5^2)(12)=100\pi\text{ cm}^3\approx314\text{ cm}^3.$

For surface area, use the slant height in the curved area and add the circular base:

$A=\pi(5)(13)+\pi(5^2)=65\pi+25\pi=90\pi\text{ cm}^2\approx283\text{ cm}^2.$

Check: the volume is one third of a cylinder with the same base and perpendicular height. If the question asked for the curved surface only, the answer would be $65\pi\text{ cm}^2$ ; read whether the base is included.

A cone’s curved surface can be opened into a sector of radius $l$ . Its arc length is the base circumference $2\pi r$ , so its area is $\frac12\times l\times2\pi r=\pi rl$ . This is why curved area uses the slant height.

Spheres and hemispheres: half the volume, but which area?

Worked example 3 — recasting a sphere as hemispheres

A metal sphere of radius 6 cm is melted and recast into identical solid hemispheres of radius 3 cm. No metal is lost. Find the number of hemispheres and the total surface area of one hemisphere, including its flat base.

Volume is conserved. The original sphere has volume $\frac43\pi(6^3)=288\pi\text{ cm}^3$ . One hemisphere has half the volume of a sphere of radius 3 cm:

$V_{\text{hemisphere}}=\frac23\pi(3^3)=18\pi\text{ cm}^3.$

The number is $288\pi\div18\pi=16$ . This is an exact whole number; all the metal can be used.

A hemisphere’s curved area is $2\pi r^2$ . Its flat circular base adds another $\pi r^2$ , so one hemisphere’s total area is

$A=3\pi(3^2)=27\pi\text{ cm}^2\approx84.8\text{ cm}^2\quad\text{(3 s.f.)}.$

Check: the 16 volumes sum to $288\pi\text{ cm}^3$ . Surface area is not conserved during recasting: the new solids expose more surface.

For similar solids, multiplying every length by $k$ multiplies area by $k^2$ and volume by $k^3$ . A 20% increase in a sphere’s radius gives an area factor $1.2^2=1.44$ , hence a 44% area increase, not 20%.

Composite solids: join volumes, trace exposed surfaces

Worked example 4 — a cone joined to a hemisphere

A cone of radius 3 cm and perpendicular height 4 cm is joined to a hemisphere of radius 3 cm along their circular bases. Find the volume and external surface area of the resulting solid.

An axial cross-section. The joined base disks are hidden inside the solid and do not contribute to its external area.

Find the cone’s slant height: $l=\sqrt{3^2+4^2}=5\text{ cm}$ .
Add the volumes: $V=\frac13\pi(3^2)(4)+\frac23\pi(3^3)=12\pi+18\pi=30\pi\text{ cm}^3\approx94.2\text{ cm}^3$ .
Add only the curved surfaces: $A=\pi(3)(5)+2\pi(3^2)=15\pi+18\pi=33\pi\text{ cm}^2\approx104\text{ cm}^2$ .

Both decimal answers are to 3 significant figures. Check: adding the total areas of the separate solids would count the circular join twice. Subtract $2\pi(3^2)=18\pi\text{ cm}^2$ from that sum to obtain the same exposed area.

For a hollowed-out solid, subtract the removed volume. A newly exposed inner surface may still contribute to the surface area; it is not removed just because material was removed.

A truncated cone: use subtraction and similarity

A frustum is what remains when a smaller cone is cut from the top of a larger cone by a plane parallel to its base. If the large and small radii are $R$ and $r$ , their full heights $H$ and $h$ satisfy $r/R=h/H$ . The frustum’s own height is $H-h$ , so do not substitute it for $H$ .

Subtract the cone volumes, $\frac13\pi R^2H-\frac13\pi r^2h$ . For curved area, subtract $\pi r l$ from $\pi R L$ , where $L$ and $l$ are the full cones’ slant heights. Add whichever circular end faces are exposed. Drawing the complete cone first makes the similar triangles visible.

Explore a joined solid’s volume and exposed area

Cone and hemisphere explorer

Change the shared radius and the cone’s perpendicular height. Compare the joined solid with the same two pieces separated: volume stays the same, but separating exposes two circular faces.

Axial cross-sections are drawn to scale; lengths are in centimetres. Dashed circular faces are internal; solid circular faces are exposed.

Slant height = 5 cm. Total volume = 30π cm³ ≈ 94.2 cm³. Joined external area = 33π cm² ≈ 104 cm². The two circular faces are internal.

Common mistakes and useful checks

Using diameter as radius: halve the diameter before applying a formula involving $r$ .
Mixing heights: volume uses perpendicular height; triangular faces and cone curved area use the appropriate slant height.
Halving all sphere formulae: a hemisphere has half a sphere’s curved area, but its flat base adds $\pi r^2$ when exposed.
Converting units linearly: $1\text{ m}^2=10\,000\text{ cm}^2$ and $1\text{ m}^3=1\,000\,000\text{ cm}^3$ . Also $1000\text{ cm}^3=1$ litre.
Counting hidden joins: trace the boundary that can be painted. Count each exposed face once.

Practise drawing the solid before calculating

Use the Sec 2 mensuration practice bank for pyramids, cones, spheres, frustums and composite solids. Label radius, perpendicular height and slant height, then list the volumes and exposed surfaces needed.