NSWMaths Standard 2Syllabus dot point

How do you find the surface area of a prism, cylinder, sphere, pyramid or cone, and when must you first find a slant height?

Calculate the surface area of right prisms, cylinders, spheres and pyramids, and of right cones using the slant height, including solids formed as a combination of these

A focused answer to the HSC Maths Standard 2 dot point on surface area. Surface area of right prisms from a net, open and closed cylinders, spheres, pyramids and cones, finding a cone's slant height with Pythagoras, and the surface area of composite solids, with worked Australian examples and the formulae you are given on the reference sheet.

Generated by Claude Opus 4.815 min answerUpdated 2026-06-21

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What this dot point is asking

NESA wants you to find the surface area of the standard solids: right prisms, cylinders, spheres, pyramids and cones, including solids built by combining these. Surface area is the total area of every face or curved surface that wraps the outside of the solid, so it is measured in square units (cm $^2$ , m $^2$ ). The reference sheet gives you the sphere, cylinder and cone formulae, so the marks are not lost on recall. They are lost on three decisions: which surfaces actually exist (is the cylinder open or closed, does a roof cover the top), whether you must first find a slant height with Pythagoras for a cone, and remembering to add the parts of a composite solid without double-counting a hidden face.

The answer

Every surface area is the sum of the areas of all the outside surfaces. For a flat-faced solid (a prism or pyramid) you imagine cutting it open and flattening it into a net, then add up the faces. For a curved solid (a cylinder, cone or sphere) you use the given formula for the curved part and add any flat ends. The whole topic is built from a few area facts you already know - the area of a rectangle, a triangle and a circle - assembled the right way.

The formulae you are given on the HSC reference sheet are:

\text{Sphere: } A = 4\pi r^2, \qquad \text{Closed cylinder: } A = 2\pi r^2 + 2\pi r h, \qquad \text{Cone: } A = \pi r l + \pi r^2.

A prism and a pyramid have no formula on the sheet - you build them from the net.

Right prisms: add up the net

A right prism has two identical end faces joined by rectangles. The fastest way to its surface area is the net: the two ends plus the band of rectangles wrapped around the sides. For any right prism,

A = 2 \times (\text{area of one end}) + (\text{perimeter of the end}) \times (\text{length of the prism}).

The second term works because the rectangles around the side all share the prism's length, and their widths add up to the perimeter of the end. For a rectangular prism (a box) of dimensions $l \times w \times h$ this gives the familiar $A = 2(lw + lh + wh)$ , but the net method handles a triangular or any other prism just as easily.

Cylinders: a rectangle wrapped into a tube

A cylinder is a prism with a circular end. Unroll its curved surface and it becomes a rectangle: its height is the cylinder's height $h$ , and its width is the circumference of the circle, $2\pi r$ . So the curved surface area is

\text{curved surface} = 2\pi r \times h = 2\pi r h.

A closed cylinder adds the two circular ends, each $\pi r^2$ , giving $A = 2\pi r h + 2\pi r^2$ . The net below shows exactly where each piece comes from.

An open cylinder is missing one or both ends, so you count only the surfaces that are really there. A pipe open at both ends is just the curved surface $2\pi r h$ ; a cup or trough open at the top is the curved surface plus one base, $2\pi r h + \pi r^2$ . Always read the wording: "open", "no lid", "tube" and "pipe" are the clues to drop an end.

Spheres: one formula

A sphere is the simplest case to recall and the easiest to over-think. It has a single curved surface and no flat faces, so its surface area is just

A = 4\pi r^2.

There is nothing to add. A useful relative is the hemisphere (half a ball): its curved surface is half of a full sphere, $\tfrac{1}{2} \times 4\pi r^2 = 2\pi r^2$ , and if the flat circular face is exposed you add $\pi r^2$ on top of that.

Pyramids: base plus triangular faces

A pyramid has a flat base and triangular faces meeting at an apex. Like a prism, there is no formula on the sheet, so build it from the net: the base plus the triangular faces. For a square-based pyramid of base side $a$ with each triangular face having slant height $s$ (the height of the triangular face, measured up the middle of the face),

A = \underbrace{a^2}_{\text{base}} + \underbrace{4 \times \tfrac{1}{2} a s}_{\text{four faces}} = a^2 + 2as.

Read carefully whether you are given the face's slant height (used directly) or the pyramid's vertical height (which would need Pythagoras first).

Cones: find the slant height first

A cone is the case examiners love, because its formula $A = \pi r l + \pi r^2$ needs the slant height $l$ , the straight distance from the apex down the sloping side to the rim, and you are usually given the perpendicular height $h$ instead. The radius, the perpendicular height and the slant height form a right-angled triangle, so Pythagoras gives

l = \sqrt{r^2 + h^2}.

The curved surface is $\pi r l$ and the circular base is $\pi r^2$ , so a solid cone is $A = \pi r l + \pi r^2$ , while an open cone (a party hat, an ice-cream cone) is the curved surface $\pi r l$ alone. The diagram shows the right-angled triangle you must spot.

How exam questions ask about surface area

The wording tells you which surfaces to count and whether a slant height is needed:

"Find the surface area / total surface area" means every outside surface. For a cone or pyramid, check whether you have the slant height or must find it first.
"Curved surface area" of a cylinder or cone means the wrapped part only ( $2\pi r h$ or $\pi r l$ ): do not add the ends.
"Open", "no lid", "tube", "pipe", "trough", "hollow" are the signals to drop one or both circular ends.
"How much material / metal / paint / wrapping ... is needed" is a surface-area question in disguise; convert to a cost or a number of tins at the end if asked.
"... made up of a cylinder and a cone / a hemisphere" is a composite solid: add the visible surfaces and subtract any face buried at the join (for example, drop the cylinder's top circle when a cone or dome sits on it).
A cone or pyramid given by its perpendicular height is the prompt to use Pythagoras for the slant height before the surface-area formula.

Surface area across four solids

Four solids, each showing a different decision: a closed cylinder (count both ends), a sphere (one formula), a cone (find the slant height first), and a composite silo (add surfaces, drop the hidden one).

Closed cylinder - count both ends

A closed cylindrical can has radius $4$ cm and height $10$ cm. Find its total surface area, to two decimal places.

Choose the surfaces. Closed, so it is the curved wall plus two ends: $A = 2\pi r h + 2\pi r^2$ .

Substitute and simplify.

A = 2\pi (4)(10) + 2\pi (4)^2 = 80\pi + 32\pi = 112\pi.

Compute. $112\pi = 351.858\ldots \approx 351.86$ cm $^2$ .

Sphere - one formula, nothing to add

A ball has radius $7$ cm. Find its surface area, to two decimal places.

Apply the formula.

A = 4\pi r^2 = 4\pi (7)^2 = 196\pi.

Compute. $196\pi = 615.752\ldots \approx 615.75$ cm $^2$ . A sphere has no flat face, so the single formula is the complete answer.

Cone - slant height before the formula

A solid cone has base radius $6$ cm and perpendicular height $8$ cm. Find its total surface area, to two decimal places.

Find the slant height with Pythagoras. The radius, height and slant height make a right-angled triangle:

l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{100} = 10 \text{ cm}.

Apply the cone formula with $l = 10$ (not the $8$ cm height):

A = \pi r l + \pi r^2 = \pi (6)(10) + \pi (6)^2 = 60\pi + 36\pi = 96\pi.

Compute. $96\pi = 301.592\ldots \approx 301.59$ cm $^2$ .

Composite silo - add the surfaces, drop the hidden face

A grain silo is a closed cylinder of radius $1.5$ m and height $4$ m topped by a hemisphere of the same radius, standing on its flat base. Find the external surface area, to two decimal places.

List the surfaces that show. The dome sits where the cylinder's top would be, so the cylinder's top circle is gone. The external surfaces are the cylinder wall, the flat base on the ground, and the hemisphere dome:

A = 2\pi r h + \pi r^2 + 2\pi r^2.

Substitute $r = 1.5$ , $h = 4$ :

A = 2\pi (1.5)(4) + \pi (1.5)^2 + 2\pi (1.5)^2 = 12\pi + 2.25\pi + 4.5\pi = 18.75\pi.

Compute. $18.75\pi = 58.904\ldots \approx 58.90$ m $^2$ .

Final answers: the can is $\approx 351.86$ cm $^2$ ; the ball is $\approx 615.75$ cm $^2$ ; the cone is $\approx 301.59$ cm $^2$ ; the silo is $\approx 58.90$ m $^2$ .

Common traps

Using the perpendicular height in the cone formula: The cone formula needs the slant height $l$ . Find it first with $l = \sqrt{r^2 + h^2}$ ; substituting the perpendicular height $h$ is the single most common error in this topic.
Counting the wrong number of ends on a cylinder: Closed has two circular ends, a trough or cup has one, a pipe has none. Read "open", "no lid" or "tube" before deciding.
Adding a hidden face in a composite solid: Where a cone or hemisphere sits on a cylinder, the circle they share is inside the solid and must not be counted. Subtract any buried face.
Rounding $\pi$ too early: Keep answers as a multiple of $\pi$ (like $96\pi$ ) until the last line, then convert. Rounding $\pi$ at the start drifts the final figure.
Confusing surface area with volume: Surface area is the outside skin in square units; volume is the space inside in cubic units. A question about paint, metal or wrapping is surface area.
Forgetting the units are squared: Surface area is always in cm $^2$ , m $^2$ and so on, never plain cm.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2021 HSC-style3 marksA closed cylindrical oil drum has radius

0.3

m and height

0.9

m. Find the total surface area of sheet steel needed to make it, correct to two decimal places.

Show worked answer →

One mark is for choosing the closed-cylinder surfaces $A = 2\pi r h + 2\pi r^2$ (curved wall plus two ends, since the drum is closed). One mark is for correct substitution, $2\pi(0.3)(0.9) + 2\pi(0.3)^2 = 0.54\pi + 0.18\pi = 0.72\pi$ . The final mark is for $0.72\pi \approx 2.26$ m $^2$ , with the squared unit shown. A student who counts only one end, or who works in centimetres without converting, loses the accuracy mark; markers accept any answer rounded from the exact $0.72\pi$ .

2022 HSC-style4 marksA conical party hat is open at the base. It has base radius

5

cm and a perpendicular height of

12

cm. (a) Find the slant height of the hat. (b) Find the area of card used to make it, correct to two decimal places.

Show worked answer →

Part (a), one mark, is the Pythagoras step $l = \sqrt{5^2 + 12^2} = \sqrt{169} = 13$ cm. Part (b) awards one mark for recognising that an open cone is the curved surface only, $A = \pi r l$ (no base circle), one mark for $\pi(5)(13) = 65\pi$ , and one mark for $65\pi \approx 204.20$ cm $^2$ . The classic lost mark is using the $12$ cm perpendicular height in place of the slant height, or adding a base the open hat does not have.

2020 HSC-style4 marksA vitamin capsule is modelled as a cylinder of radius

0.4

cm and length

1.2

cm with a hemisphere on each end (so the two hemispheres together form a full sphere of radius

0.4

cm). Find the total surface area of the capsule, correct to two decimal places.

Show worked answer →

One mark is for seeing that the two hemispheres form one whole sphere, so there are no flat circular ends to count. One mark is for the combined surface $A = 2\pi r h + 4\pi r^2$ (cylinder curved surface plus a full sphere). One mark is for substitution, $2\pi(0.4)(1.2) + 4\pi(0.4)^2 = 0.96\pi + 0.64\pi = 1.6\pi$ , and one for $1.6\pi \approx 5.03$ cm $^2$ . A student who adds two circular ends to the cylinder has double-counted faces that are sealed inside the rounded ends and loses a mark.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksA closed cylindrical tin has radius

4

cm and height

10

cm. Find its total surface area, correct to two decimal places. (Use

A = 2\pi r^2 + 2\pi r h

Show worked solution →

Write the formula and substitute. A closed cylinder has a curved surface plus two circular ends:

A = 2\pi r^2 + 2\pi r h = 2\pi (4)^2 + 2\pi (4)(10).

Evaluate each part. The two ends give $2\pi (16) = 32\pi$ and the curved surface gives $2\pi (40) = 80\pi$ :

A = 32\pi + 80\pi = 112\pi.

Compute and round. $112\pi = 351.858\ldots$ , so

A \approx 351.86 \text{ cm}^2.

(Keep $\pi$ in the working until the last line; rounding $\pi$ early is where marks slip.)

foundation2 marksA solid sphere has radius

7

cm. Find its surface area, correct to two decimal places. (Use

A = 4\pi r^2

Show worked solution →

Write the formula and substitute. A sphere has the single formula:

A = 4\pi r^2 = 4\pi (7)^2.

Evaluate. Since $7^2 = 49$ ,

A = 4\pi (49) = 196\pi.

Compute and round. $196\pi = 615.752\ldots$ , so

A \approx 615.75 \text{ cm}^2.

(A sphere has no flat faces, so there is nothing to add - the one formula is the whole answer.)

foundation3 marksAn open cylindrical water trough (no lid) has radius

0.5

m and height

1.2

m. Find the area of metal needed to make it (the curved wall plus the one circular base), correct to two decimal places.

Show worked solution →

Decide which surfaces exist. Open means no top, so the trough is one curved surface plus one circular base:

A = 2\pi r h + \pi r^2.

Substitute $r = 0.5$ and $h = 1.2$ :

A = 2\pi (0.5)(1.2) + \pi (0.5)^2 = 1.2\pi + 0.25\pi = 1.45\pi.

Compute and round. $1.45\pi = 4.555\ldots$ , so

A \approx 4.56 \text{ m}^2.

(The whole question turns on counting one end, not two. A closed cylinder would add another $0.25\pi$ .)

core3 marksA solid cone has base radius

6

cm and perpendicular height

8

cm. (a) Find the slant height

l

. (b) Find the total surface area, correct to two decimal places. (Use

A = \pi r l + \pi r^2

Show worked solution →

Part (a) - find the slant height with Pythagoras. The radius, the perpendicular height and the slant height form a right-angled triangle with $l$ as the hypotenuse:

l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ cm}.

Part (b) - total surface area. A solid cone is the curved surface ( $\pi r l$ ) plus the circular base ( $\pi r^2$ ):

A = \pi r l + \pi r^2 = \pi (6)(10) + \pi (6)^2 = 60\pi + 36\pi = 96\pi.

Compute and round. $96\pi = 301.592\ldots$ , so

A \approx 301.59 \text{ cm}^2.

(You must find $l$ first: the formula uses the slant height, never the perpendicular height.)

core3 marksA square-based pyramid has a base of side

10

cm. Each triangular face has a perpendicular (slant) height of

13

cm. Find the total surface area.

Show worked solution →

Find the base area. The base is a square of side $10$ :

\text{base} = 10 \times 10 = 100 \text{ cm}^2.

Find the area of the four triangular faces. Each face is a triangle with base $10$ and height $13$ :

\text{one face} = \tfrac{1}{2} \times 10 \times 13 = 65 \text{ cm}^2,

and there are four identical faces, so $4 \times 65 = 260$ cm $^2$ .

Add the parts.

A = 100 + 260 = 360 \text{ cm}^2.

(A pyramid has no single formula on the sheet - build it from the net: one base plus the triangular faces.)

exam5 marksA grain silo is a closed cylinder of radius

1.5

m and height

4

m, topped by a hemisphere of the same radius. The flat circular base sits on the ground. (a) Find the total external surface area (curved wall, flat base and hemisphere dome), correct to two decimal places. (b) Paint covers

1

^2

per $6. Find the cost of one coat, correct to the nearest cent.

Show worked solution →

Part (a) - identify the three surfaces. The dome replaces the flat top, so the external surface is the cylinder wall, the flat circular base on the ground, and the curved surface of the hemisphere:

A = \underbrace{2\pi r h}_{\text{wall}} + \underbrace{\pi r^2}_{\text{base}} + \underbrace{2\pi r^2}_{\text{hemisphere}}.

Substitute $r = 1.5$ and $h = 4$ :

A = 2\pi (1.5)(4) + \pi (1.5)^2 + 2\pi (1.5)^2 = 12\pi + 2.25\pi + 4.5\pi = 18.75\pi.

Compute and round. $18.75\pi = 58.904\ldots$ , so

A \approx 58.90 \text{ m}^2.

Part (b) - cost of one coat. At $6 per square metre:

18.75\pi \times 6 = 353.43,

so one coat costs about $353.43. (A hemisphere's curved surface is $2\pi r^2$ , exactly half a full sphere's $4\pi r^2$ ; the silo's flat top is gone, so do not add it.)

exam6 marksA storage hopper is a closed cylinder of radius

2

m and height

3

m, with a cone of base radius

2

m and perpendicular height

1.5

m fixed on top as a roof. The flat circular base is included. (a) Find the slant height of the cone. (b) Find the total external surface area (cylinder wall, flat base, and cone roof), correct to two decimal places. (c) The maker quotes sheet steel at $48 per square metre. Estimate the steel cost, correct to the nearest dollar.

Show worked solution →

Part (a) - slant height of the cone. The radius and perpendicular height give the slant height by Pythagoras:

l = \sqrt{r^2 + h^2} = \sqrt{2^2 + 1.5^2} = \sqrt{4 + 2.25} = \sqrt{6.25} = 2.5 \text{ m}.

Part (b) - add the three external surfaces. The cone roof covers the cylinder's top, so there is no top circle: the surfaces are the cylinder wall ( $2\pi r h$ ), the flat base ( $\pi r^2$ ) and the cone's curved surface ( $\pi r l$ ):

A = 2\pi r h + \pi r^2 + \pi r l.

Substitute $r = 2$ , $h = 3$ , $l = 2.5$ :

A = 2\pi (2)(3) + \pi (2)^2 + \pi (2)(2.5) = 12\pi + 4\pi + 5\pi = 21\pi.

Compute and round. $21\pi = 65.973\ldots$ , so

A \approx 65.97 \text{ m}^2.

Part (c) - cost. At $48 per square metre:

65.97 \times 48 = 3166.56,

so the steel costs about $3167. (Two traps avoided here: use the cone's slant height $2.5$ , not its $1.5$ m perpendicular height, and drop the cylinder's top circle because the roof sits on it.)

What this dot point is asking

The answer

Right prisms: add up the net

Cylinders: a rectangle wrapped into a tube

Spheres: one formula

Pyramids: base plus triangular faces

Cones: find the slant height first

How exam questions ask about surface area

Exam-style practice questions

Practice questions

Related dot points