Measurement (Grade 10) — Mathematics Grades 8–11

Calculate the volume and surface area of prisms, cylinders, spheres, cones and pyramids.

Gr 10 Term 3 - MeasurementTerm 3 - 1 week (CAPS)

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What gets asked

Calculate the volume or surface area of a prism, cylinder, sphere, cone or pyramid.
Find an unknown dimension from a given volume or surface area.
Work out how volume or surface area change when a dimension is scaled by k.
Calculate the volume or surface area of a composite solid.

You must be able to

Use the perpendicular height, never the slant height, in a volume formula.
Use Pythagoras to find a slant height from the perpendicular height and the base.
Multiply volume by k³ when every dimension is scaled by k, not just by k.
Work out which faces show and which are hidden in a composite solid.

Traps that cost marks

Using the slant height instead of the perpendicular height in a volume formula. Volume always needs the perpendicular, straight-up height, not the slanted edge.
Using a full base side as the triangle's leg when finding a square-based pyramid's slant height. For a square base, that leg is half a side: from the centre to the middle of an edge.
Swapping the powers: scaling volume by k squared and surface area by k cubed. Area scales by k squared. Volume scales by k cubed. Area has 2 dimensions, volume has 3.

Worked example

A cylinder has radius 4 cm and height 10 cm. Calculate its volume, to the nearest cm³ (use

π \approx 3, 14

$V = π r^{2} h$
$V = 3, 14 \times 4^{2} \times 10$
$V = 502, 4$ , so $V \approx 502$ cm³.

Prisms and cylinders

How much a solid holds, how much you would have to paint, and why those two never share a unit.

right prism: A solid with two identical parallel end faces, joined by rectangles that stand square to them.
cross-section: The shape of a prism's end face. Every slice parallel to it looks exactly the same.
volume: The space a solid fills, counted in cubes. Its unit always carries a small $3$ .
surface area: The total area of every face you could paint. Its unit always carries a small $2$ .
perpendicular height: A height measured at a right angle to the base, never along a slanted edge.

A right prism is the same shape all the way along. So its volume is the area of the cross-section times the length. Work the cross-section out completely first, halving included if it is a triangle.

Surface area is a different job. Add up every face separately. The two end faces are identical, so the cross-section area is counted twice, not once. The answer is in square units, never cubic ones.

A cylinder is a prism whose cross-section is a circle. That gives $V = π r^{2} h$ . Its curved side unrolls into a rectangle $2 π r$ long and $h$ high, and a closed tin adds two circles on top of that.

Questions often hand you the diameter instead. Halve it before it goes anywhere near a formula, because the radius is what every circle formula wants.

Working backwards, put the numbers you have into the formula and solve. Divide by the whole base area, not by one side of it. A length can never be negative, so throw a negative answer away.

Rules to remember

right prism: $V = A h$ , with $A$ the cross-section area
cylinder: $V = π r^{2} h$
closed cylinder: $S = 2 π r^{2} + 2 π r h$

Examples

A chocolate box is a triangular prism

20

cm long. Its cross-section has base

6

cm and perpendicular height

4

cm. Find its volume.

Worked answer

The cross-section is a triangle, so halve the base times the height.
$\frac{1}{2} \times 6 \times 4 = 12$ square centimetres.
Now multiply by the length: $12 \times 20 = 240$ .

Answer: $240$ cm $^{3}$

Multiplying $6 \times 4 \times 20$ leaves the half out and doubles the answer.

A closed tin of beans has radius

4

cm and height

10

cm. Find its surface area in terms of

π

Worked answer

A closed tin is two circles plus one curved surface.
Circles: $2 \times π \times 4^{2} = 32 π$ .
Curved part: $2 \times π \times 4 \times 10 = 80 π$ .
Add them: $32 π + 80 π = 112 π$ .

Answer: $112 π$ cm $^{2}$

The tin has a lid and a base, so the circle area is counted twice.

A rectangular water tank has a base

2

m by

1, 5

m. Full, it holds

6

^{3}

. How deep is it?

Worked answer

Base area first: $2 \times 1, 5 = 3$ .
Volume is base area times depth, so $3 h = 6$ .
Divide by the whole base area: $h = \frac{6}{3} = 2$ .

Answer: $2$ m

Dividing by the $2$ alone would give $3$ m, using only one side of the base.

Traps

Multiplying all three given lengths of a triangular prism together. The cross-section is a triangle, so it needs halving first. Only then multiply by the length.
Counting the two identical end faces once instead of twice in a surface area. Both ends are painted. Add the cross-section area in twice.
Putting the diameter into a cylinder formula without halving it. Every circle formula wants the radius. A diameter of $8$ cm gives a radius of $4$ cm.
Dividing the volume by one side of the base to find the height. Divide by the whole base area. For a $2$ m by $1, 5$ m base that is $3$ , not $2$ .

Learn and practise “Prisms and cylinders” in the app →

Effect of a factor k

Stretch a solid and its size changes, but the answer depends on how many of its dimensions moved.

factor: The number a length is multiplied by. Doubling a length means the factor is $2$ .
enlargement: A new solid made by multiplying every dimension by the same factor. The shape is unchanged.
square root: The number that multiplies by itself to give it. Since $5 \times 5 = 25$ , $25$ has square root $5$ .
cube root: The number that gives it when used three times as a factor. Since $2 \times 2 \times 2 = 8$ , $8$ has cube root $2$ .

Volume is length times breadth times height. Change only the height and just one of those numbers moves. So the volume is multiplied by the factor once. Triple the height and the volume triples, not $27$ times.

Surface area has no such shortcut when only one dimension moves. The top and bottom faces do not contain the height, so they never change, while the four side faces do. Rebuild it face by face.

An enlargement is different: every dimension moves together. Then the surface area is multiplied by $k^{2}$ and the volume by $k^{3}$ . Area is two dimensions wide, volume three, and the powers follow.

Working backwards from an enlargement, undo the power. If the volume grew $64$ times then $k^{3} = 64$ , so the cube root gives $k = 4$ . If the surface area grew $9$ times then $k^{2} = 9$ , so $k = 3$ .

Check first that every dimension really was scaled. A cube root on a solid whose height alone was stretched means nothing.

Rules to remember

one dimension multiplied by $k$ : the volume is multiplied by $k$
every dimension multiplied by $k$ : surface area by $k^{2}$ , volume by $k^{3}$

Examples

A matchbox holds

30

^{3}

. Only its height is tripled. Find the new volume.

Worked answer

The base is untouched, so the base area does not change.
Only one dimension moved, so multiply the volume by $3$ once.
$30 \times 3 = 90$ .

Answer: $90$ cm $^{3}$

Nothing was cubed here, because two of the three dimensions never moved.

A cube has surface area

24

^{2}

and volume

8

^{3}

. Every edge is doubled. Find the new surface area and volume.

Worked answer

Every dimension moved, so this is an enlargement with $k = 2$ .
Surface area is multiplied by $k^{2} = 4$ , giving $24 \times 4 = 96$ .
Volume is multiplied by $k^{3} = 8$ , giving $8 \times 8 = 64$ .

Answer: $96$ cm $^{2}$ and $64$ cm $^{3}$

The same $k$ gave two different powers, one for area and one for volume.

A plastic bottle is enlarged. Its volume grows from

500

ml to

4000

ml. Find the factor

k

Worked answer

Compare the volumes: $\frac{4 000}{500} = 8$ .
Every dimension was scaled, so that $8$ is $k^{3}$ .
$k^{3} = 8$ , so $k = 38 = 2$ .
Every length on the bottle has doubled.

Answer: $k = 2$

The cube root was allowed only because the whole bottle was enlarged.

Traps

Cubing the factor when only one dimension of the solid was changed. One dimension moved, so multiply the volume by $k$ once. The cube is for a full enlargement.
Multiplying the whole surface area by $k$ when only one dimension was changed. Faces without that dimension do not grow at all. Work the new surface area out face by face.
Swapping the powers: scaling volume by k squared and surface area by k cubed. Area scales by k squared. Volume scales by k cubed. Area has 2 dimensions, volume has 3.
Taking a cube root of the volume ratio when only the height was stretched. Check what moved. With one dimension changed the ratio is $k$ itself, with no root at all.

Learn and practise “Effect of a factor k” in the app →

Spheres, pyramids and cones

Solids that come to a point or curve all over, and the two different heights each of them has.

apex: The single point at the top of a pyramid or a cone, where the sloping faces meet.
slant height: The distance down a sloping face, from the apex to the base edge. It is longer than the height.
curved surface: The sloping outside of a cone, not counting the circle it stands on.
hemisphere: Exactly half a ball, cut through the centre. It has one flat circular face.

A sphere has two formulas and they look alike. Volume cubes the radius and carries $\frac{4}{3}$ . Surface area squares it and carries $4$ . Read the question, then pick. A hemisphere halves both, and then its flat circle must be added back.

A pyramid or a cone holds exactly one third of the prism or cylinder standing on the same base with the same perpendicular height. That one third is the piece most often left out.

Every volume formula wants the perpendicular height: straight up from the middle of the base to the apex. The slant height runs down the outside instead, so it is always the longer of the two.

Get the slant height from the Theorem of Pythagoras. For a cone the two short sides are the height and the radius. For a square-based pyramid they are the height and half a base edge.

Surface area of a pyramid is the base plus its triangular faces, and those triangles use the slant height, not the pyramid's own height. A cone is $π r^{2}$ for the base plus $π r ℓ$ for the curved surface. An open cone drops the base.

Rules to remember

sphere: $V = \frac{4}{3} π r^{3}$ and $S = 4 π r^{2}$
cone: $V = \frac{1}{3} π r^{2} h$ and $S = π r^{2} + π r ℓ$
square pyramid: $V = \frac{1}{3} b^{2} h$ , with $b$ the base edge

Examples

A wooden ball has a radius of

6

cm. Find its volume in terms of

π

Worked answer

Volume was asked for, so use the formula with the cube in it.
$6^{3} = 216$ .
$V = \frac{4}{3} \times π \times 216 = 288 π$ .

Answer: $288 π$ cm $^{3}$

The surface area formula squares the radius instead, and would have given $144 π$ .

A square-based pyramid has base edges of

10

cm and a perpendicular height of

12

cm. Find its slant height.

Worked answer

The right-angled triangle uses half a base edge: $\frac{1}{2} \times 10 = 5$ .
Its other short side is the perpendicular height, $12$ cm.
$ℓ^{2} = 1 2^{2} + 5^{2} = 144 + 25 = 169$ .
$ℓ = 169 = 13$ .

Answer: $13$ cm

Using the whole $10$ cm edge would give $244$ , which is not even a whole number.

A party hat is a cone with radius

6

cm and slant height

10

cm. Find its volume in terms of

π

Worked answer

Volume needs the perpendicular height, and only the slant height is given.
The slant height is the longest side: $h^{2} = 1 0^{2} - 6^{2} = 64$ , so $h = 8$ .
$V = \frac{1}{3} \times π \times 6^{2} \times 8$ .
$\frac{1}{3} \times 36 \times 8 = 96$ , so $V = 96 π$ .

Answer: $96 π$ cm $^{3}$

Feeding the $10$ straight in would give $120 π$ , a quarter too big.

Traps

Using the slant height instead of the perpendicular height in a volume formula. Volume always needs the perpendicular, straight-up height, not the slanted edge.
Using a full base side as the triangle's leg when finding a square-based pyramid's slant height. For a square base, that leg is half a side: from the centre to the middle of an edge.
Reaching for the surface area formula when the sphere question asked for volume. Both hold $r$ and $π$ , so check the power. Volume cubes the radius, surface area squares it.
Adding the base circle to the surface area of an open cone. An open cone has no base to paint. Use the curved surface $π r ℓ$ on its own.

Learn and practise “Spheres, pyramids and cones” in the app →

Composite solids

Two solids stuck together, and the faces that vanish where they meet.

composite solid: A solid built by joining two simpler solids, or by cutting one out of another.
hidden face: A face sealed inside where two solids meet. Nobody can paint it, so it is left out.
capacity: How much liquid a container holds. A volume of $1000$ cm $^{3}$ is a capacity of $1$ litre.

For volume, do each piece on its own and then add them. If a hole has been cut out, subtract instead. Keep the full number on the calculator the whole way, and round once, at the very end.

Surface area is not the two surface areas added. Where the solids meet, a face on each of them becomes a hidden face. So take that shape off twice, once from each solid.

If a small solid stands on a big flat face, the rest of that big face is still out in the open. Subtract only the part actually covered, not the whole face.

Read what the context wants. Painting, wrapping or tiling means surface area. Filling, holding or pouring means volume. A volume of $1000$ cm $^{3}$ is $1$ litre, and $1$ m $^{3}$ is $1000$ litres.

To find a missing length from a known volume, undo each operation in turn. Multiply away the one third, divide by $π$ , then take the root last. Dividing by $π$ and stopping leaves a cube, not a length.

Rules to remember

$1000$ cm $^{3}$ is $1$ litre, and $1$ m $^{3}$ is $1000$ litres
composite solid: add the volumes, but never add the two full surface areas

Examples

A spherical gas tank has a volume of

36 π

^{3}

. Find its radius.

Worked answer

Substitute into the sphere formula: $\frac{4}{3} π r^{3} = 36 π$ .
Divide both sides by $π$ : $\frac{4}{3} r^{3} = 36$ .
Multiply by $3$ and divide by $4$ : $r^{3} = 27$ .
Take the cube root: $r = 327 = 3$ .

Answer: $3$ m

Stopping at $r^{3} = 27$ would answer $27$ m, a tank nine times too wide.

A grain silo is a cylinder of radius

3

m and height

10

m, with a hemisphere on top. Find its volume, to the nearest cubic metre. Take

π

3, 14

Worked answer

Cylinder: $π \times 3^{2} \times 10 = 90 π$ .
Hemisphere is half a sphere: $\frac{1}{2} \times \frac{4}{3} π \times 3^{3} = 18 π$ .
Add them: $90 π + 18 π = 108 π$ .
Round only now: $108 \times 3, 14 = 339, 12$ .

Answer: $339$ m $^{3}$

Rounding each piece first gives $283 + 57$ , which is $340$ and a whole cubic metre out.

A toy is a wooden cube of side

8

cm with a

3

cm cube glued on top. It is painted all over. Find the area painted.

Worked answer

Big cube alone: $6 \times 8^{2} = 384$ .
Small cube alone: $6 \times 3^{2} = 54$ .
Two squares vanish where they join, each $3^{2} = 9$ .
$384 + 54 - 9 - 9 = 420$ .

Answer: $420$ cm $^{2}$

One square came off the small cube's bottom and one off the big cube's top.

Traps

Adding the two solids' surface areas in full. The joining faces are sealed inside. Subtract that shape twice, once from each solid.
Rounding each part of a composite solid before adding them together. Rounding twice can push the total out. Carry full accuracy and round the final answer only.
Working out a volume when the question asks how much paint or covering is needed. Anything to be covered is a surface area. Only filling and holding are volume.
Dividing by $π$ and stopping, when solving for the radius of a sphere. That leaves you with $r^{3}$ . The cube root is still to come.

Learn and practise “Composite solids” in the app →

Grade 10 Measurement

What gets asked

You must be able to

Traps that cost marks

Worked example

Prisms and cylinders

Rules to remember

Examples

Traps

Effect of a factor k

Rules to remember

Examples

Traps

Spheres, pyramids and cones

Rules to remember

Examples

Traps

Composite solids

Rules to remember

Examples

Traps

About this material