Surface Area of Pyramids, Cones, and Spheres

Find the slant height first. Always. Then apply the formula. Composite solids hide faces at the join — list what is exposed before calculating anything.

55–60 min MS-M1 — MEDIUM 3 MC 3 SA Lesson 8 of 22 Free
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Think First

A party hat is a cone with no base. A beach ball is a sphere. An Egyptian pyramid has a square base and triangular sides. Before learning any formula — what do you think would be the hardest surface to calculate the area of, and why?

Type your initial response below — you will revisit this at the end of the lesson.

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Come back to this at the end of the lesson.

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Surface Area Formulas — This Lesson

$\text{SA} = b^2 + 2b\ell$
Square pyramid — $b$ = base side, $\ell$ = slant height $b^2$ = square base; $2b\ell$ = four triangular faces (each $\frac{1}{2}b\ell$, four of them) Slant height: $\ell^2 = h^2 + (b/2)^2$ (Pythagoras)
$\text{SA} = \pi r\ell + \pi r^2$
Closed cone — $r$ = radius, $\ell$ = slant height Slant height: $\ell = \sqrt{r^2 + h^2}$ (Pythagoras) Open cone (no base): $\text{SA} = \pi r\ell$
$\text{SA} = 4\pi r^2$
Sphere — $r$ = radius Hemisphere: $\text{SA} = 3\pi r^2$ (curved half $2\pi r^2$ + flat circle $\pi r^2$)
Critical distinction: $\ell$ = slant height (along the face)  |  $h$ = vertical height (straight up through centre)  |  These are NEVER equal
CONE — FINDING SLANT HEIGHT SPHERE & PYRAMID SA h r ℓ slant height ℓ² = r² + h²   →   ℓ = √(r²+h²) SA(cone) = πrℓ + πr² (closed) r SA(sphere) = 4πr² hemisphere: 3πr² (curved 2πr² + flat πr²) Square pyramid (base b, slant ℓ): SA = b² + 2bℓ

Know

  • SA formulas for square pyramids, cones ($\pi r\ell + \pi r^2$), and spheres ($4\pi r^2$)
  • How slant height relates to vertical height via Pythagoras
  • How to handle composite solids with hidden faces

Understand

  • Why slant height $\ell \neq$ vertical height $h$ — and why using $h$ in the formula gives a wrong answer
  • Why a hemisphere SA = $3\pi r^2$ (not $2\pi r^2$) — the flat face must be included
  • Why hidden faces at composite joins must be excluded from SA

Can Do

  • Find slant height using Pythagoras before every cone or pyramid SA calculation
  • Calculate SA of any cone, pyramid, sphere, or hemisphere
  • List exposed faces of composite solids and calculate total SA correctly
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Key Vocabulary

Slant height ($\ell$)Distance measured along the face from the base edge to the apex — not the vertical height
Vertical height ($h$)Perpendicular distance from the base to the apex — straight up through the centre
ApexThe top point of a pyramid or cone
Lateral faceA triangular side face of a pyramid — not the base
Hidden faceA face at the join between two composite solids — not part of the outer surface

Misconceptions to Fix

✗

Wrong: Capacity and volume are completely different concepts.

✗

Right: Capacity and volume are related; 1 litre = 1000 cm³ = 1 dm³.

Core Content

01

The Critical Distinction: Slant Height vs Vertical Height

This distinction is the source of more marks lost in this topic than anything else. Before any formula — know which height you have and which you need.

Vertical height $h$
Straight line from apex to centre of base — perpendicular to base
Inside the solid — not visible on the surface
Volume formulas (L10)
Slant height $\ell$
Distance along the face — from midpoint of a base edge (pyramid) or base circle (cone) to apex
On the surface — the face you would paint
Surface area formulas — this lesson

Finding Slant Height with Pythagoras

The right-angled triangle inside the solid has: vertical height $h$ as one leg, half the base dimension as the other leg, and slant height $\ell$ as the hypotenuse.

Right triangle
$h$, $b/2$, $\ell$
$h$, $r$, $\ell$
Formula
$\ell = \sqrt{h^2 + (b/2)^2}$
$\ell = \sqrt{h^2 + r^2}$
h vertical ℓ slant r ℓ² = h² + r²
h is inside the solid; ℓ is on the surface. SA formulas always use ℓ.
Make it a mandatory first step: Write "Step 1: find $\ell$" at the top of every cone or pyramid problem where $\ell$ is not given. Skipping this step and substituting $h$ directly gives a wrong answer that can still look plausible. The Pythagoras working earns a mark even if the SA calculation has an error.
02

Surface Area of a Square Pyramid

$$\text{SA} = b^2 + 2b\ell$$

One square base ($b^2$) + four identical triangular lateral faces (each $\frac{1}{2}b\ell$, giving $4 \times \frac{1}{2}b\ell = 2b\ell$ total).

If the pyramid has no base (open at the bottom — like a tent or roof frame): $\text{SA} = 2b\ell$ (lateral faces only).

ℓ b b 4 triangular faces + 1 square base
ℓ = slant height — drawn from apex to midpoint of a base edge
Why $2b\ell$: Each of the four triangular faces has base $b$ and height $\ell$ (slant height). Area of each = $\frac{1}{2}b\ell$. Four of them = $4 \times \frac{1}{2}b\ell = 2b\ell$. This derivation is worth knowing — if you forget the formula, you can reconstruct it.
03

Surface Area of a Cone

$$\text{SA} = \pi r\ell + \pi r^2$$

Curved lateral surface ($\pi r\ell$) + circular base ($\pi r^2$).

Factored: $\text{SA} = \pi r(\ell + r)$.

Formula
Column B
$\ell$ is the slant height — not $h$. If you are given $r$ and $h$ (vertical height), calculate $\ell = \sqrt{r^2 + h^2}$ before substituting. This step is not optional.
04

Surface Area of a Sphere and Hemisphere

$$\text{SA}_\text{sphere} = 4\pi r^2$$

One of the most elegant results in geometry: the total surface area of a sphere equals exactly four circles of the same radius laid flat.

Hemisphere

A hemisphere has two surfaces:

  • Curved outer dome: $\frac{1}{2} \times 4\pi r^2 = 2\pi r^2$
  • Flat circular base: $\pi r^2$
$$\text{SA}_\text{hemisphere} = 2\pi r^2 + \pi r^2 = 3\pi r^2$$
Most common hemisphere error: Writing $\text{SA} = \frac{1}{2}(4\pi r^2) = 2\pi r^2$ — halving the sphere's SA and forgetting the flat circular base. A hemisphere has a flat bottom face that appears when you cut the sphere. Both surfaces must be included. Think physically: if you dipped a hemisphere in paint, both the dome and the flat bottom would be coated.
05

Composite Solids — Hidden Faces at Joins

When two solids are joined, the faces at the join become internal. They are hidden from the outside and are not part of the surface area.

The Rule

At any join between two solids, the joined faces are not included in the total SA. List which faces are exposed on the outer surface — then calculate only those.

Cone on Cylinder Example

Include?
✓ Yes
✓ Yes
✗ No
✓ Yes
✗ No
Why
Exposed on outside
Exposed on outside
Hidden at the join
Exposed on outside
Hidden at the join
Process: List all faces of both solids. Mark each as exposed or hidden. Calculate exposed faces only. This listing step takes 30 seconds and prevents multi-mark errors. Writing "cylinder top and cone base are hidden" at the top of your working signals clear method to the examiner.
06

Common Mistakes

Mistake 1 — Using vertical height $h$ instead of slant height $\ell$
Substituting $h$ into $\text{SA} = b^2 + 2b\ell$ or $\text{SA} = \pi r\ell + \pi r^2$. Every time you see a cone or pyramid, write "Step 1: find $\ell$" before the formula.
Mistake 2 — Hemisphere SA = half of sphere SA
The flat circular base ($\pi r^2$) appears when the sphere is cut — it must be included. SA of hemisphere = $2\pi r^2 + \pi r^2 = 3\pi r^2$, not $2\pi r^2$.
Mistake 3 — Including hidden faces in composite SA
Adding all faces of all component solids without removing the joined faces. Before calculating, list faces as exposed or hidden. Anything at the join is hidden.

Worked Examples

07
Worked Example 1
Square Pyramid SA

Problem

A square pyramid has base side 8 cm and vertical height 3 cm. Find the total surface area.

Step-by-Step Solution

1
Find slant height first
$\ell^2 = h^2 + (b/2)^2 = 3^2 + 4^2 = 9 + 16 = 25$
$\ell = 5\text{ cm}$
Right triangle inside the pyramid: $h = 3$, $b/2 = 4$, $\ell$ = hypotenuse. Write this step explicitly — it earns a mark.
08
Worked Example 2
Cone SA (Vertical Height Given)

Problem

A cone has base radius 5 cm and vertical height 12 cm. Find the total surface area correct to 2 decimal places.

Step-by-Step Solution

1
Find slant height
$\ell^2 = r^2 + h^2 = 5^2 + 12^2 = 25 + 144 = 169$
$\ell = 13\text{ cm}$
5-12-13 Pythagorean triple — a common HSC choice that gives a clean integer slant height. Recognising triples saves time.
09
Worked Example 3
Sphere and Hemisphere SA

Problem

(a) Find the SA of a sphere with radius 6 cm correct to 2 decimal places.
(b) Find the total SA of the resulting hemisphere.

Step-by-Step Solution

a
Sphere SA
$\text{SA} = 4\pi r^2 = 4\pi(36) = 144\pi$
$= 452.3893... = 452.39\text{ cm}^2$
$r = 6$, $r^2 = 36$, $4 \times 36 = 144$.
10
Worked Example 4
Composite SA — Cone on Cylinder

Problem

A cone (radius 4 cm, vertical height 3 cm) sits on top of a cylinder (radius 4 cm, height 6 cm). Find the total SA correct to 2 decimal places.

Step-by-Step Solution

1
List exposed faces
✓ Cylinder curved surface
✓ Cylinder base (bottom only)
✓ Cone curved surface
✗ Cylinder top — hidden at join
✗ Cone base — hidden at join
Write this list before calculating anything. It commits you to the correct set of faces and earns method marks.
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Surface Area of Pyramids, Cones, and Spheres

For every pyramid or cone: show Pythagoras working for slant height first. For composite solids: list exposed faces before calculating.

Section A — Square Pyramids

1 Base side 6 cm, vertical height 4 cm. Find total SA.

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2 Base side 10 m, slant height 13 m. Find total SA.

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3 Base side 8 cm, vertical height 3 cm. Find lateral SA only (no base).

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Section B — Cones

4 $r = 3$ cm, vertical height 4 cm. Find total SA to 2 decimal places.

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5 $r = 6$ m, slant height 10 m. Find total SA to 2 decimal places.

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6 $r = 5$ cm, vertical height 12 cm. Find curved surface area only to 2 decimal places.

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Section C — Spheres and Hemispheres

7 Find SA of a sphere with $r = 9$ cm to 2 decimal places.

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8 Find SA of a sphere with diameter 14 m to 2 decimal places.

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9 Find total SA of a hemisphere with $r = 7$ cm to 2 decimal places.

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Section D — Composite Solids

10 A cone ($r = 5$ cm, vertical height 12 cm) sits on top of a cylinder ($r = 5$ cm, height 8 cm). Find total SA to 2 decimal places.

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11 A hemisphere ($r = 4$ cm) sits on top of a cylinder ($r = 4$ cm, height 10 cm). Find total SA to 2 decimal places.

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Show Answers

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Revisit Your Initial Thinking

Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?

Multiple Choice

1 A cone has base radius 6 cm and slant height 10 cm. Its total SA is:

A   $60\pi$ cm²
B   $96\pi$ cm²
C   $116\pi$ cm²
D   $156\pi$ cm²

? Regarding this topic, 1 A cone has base radius 6 cm and slant height 10 cm. Its total SA is:

A     $60\pi$ cm²
B     $96\pi$ cm²
C     $116\pi$ cm²
D     $156\pi$ cm²
B - Correct!
B — $\text{SA} = \pi(6)(10) + \pi(36) = 60\pi + 36\pi = 96\pi$ cm².

2 A square pyramid has base side 12 cm and vertical height 8 cm. Its slant height is:

A   8 cm
B   10 cm
C   12 cm
D   14 cm

? Regarding this topic, 2 A square pyramid has base side 12 cm and vertical height 8 cm. Its slant height is:

A     8 cm
B     10 cm
C     12 cm
D     14 cm
B - Correct!
B — $\ell^2 = 8^2 + 6^2 = 64 + 36 = 100$. $\ell = 10$ cm. ($b/2 = 6$ cm, not 12.)

3 A hemisphere has radius 5 cm. Its total SA is:

A   $50\pi$ cm²
B   $75\pi$ cm²
C   $100\pi$ cm²
D   $25\pi$ cm²

? Regarding this topic, 3 A hemisphere has radius 5 cm. Its total SA is:

A     $50\pi$ cm²
B     $75\pi$ cm²
C     $100\pi$ cm²
D     $25\pi$ cm²
B - Correct!
B — $\text{SA} = 3\pi r^2 = 3\pi(25) = 75\pi$ cm². Option A ($50\pi$) forgets the flat base.

Short Answer

11

SA 4 3 marks A square pyramid has a base side of 10 cm and vertical height of 12 cm.

(a) Find the slant height.  (1 mark)

(b) Find the total SA.  (2 marks)

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12

SA 5 3 marks A cone has base diameter 16 cm and vertical height 15 cm.

(a) Find the slant height.  (1 mark)

(b) Find the total SA correct to 2 decimal places.  (2 marks)

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13

SA 6 4 marks A decorative ornament has a hemisphere (radius 3 cm) sitting on top of a cylinder (radius 3 cm, height 7 cm).

(a) Find the curved SA of the hemisphere.  (1 mark)

(b) Find the exposed SA of the cylinder (top hidden, base included).  (2 marks)

(c) Find the total SA correct to 2 decimal places.  (1 mark)

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