Volume of Cones, Spheres and Pyramids | IEB Grade 11 Maths

Pyramids, cones and spheres use formulae that are usually given in the question. The skill lies in finding the right measurements, such as slant heights, and in understanding what happens when the dimensions are scaled.

Right pyramids, cones and spheres

Pyramid: $V = \frac{1}{3} \times area of base \times H$ ; $S = base + area of the triangular faces$
Cone: $V = \frac{1}{3} π r^{2} H$ ; $S = π r^{2} + π r s$ , where $s$ is the slant height
Sphere: $V = \frac{4}{3} π r^{3}$ ; $S = 4 π r^{2}$

$H$ is the perpendicular height. The slant height $s$ comes from Pythagoras.

Example. A cone with $r = 5$ cm and $H = 12$ cm has $s = 13$ cm, $V = 100 π \approx 314, 16$ cm³ and $S = 25 π + 65 π = 90 π \approx 282, 74$ cm².

Example. A square pyramid with base 10 cm and $H = 12$ cm. The slant height of each face is $1 2^{2} + 5^{2} = 13$ cm. $V = \frac{1}{3} (100) (12) = 400$ cm³ and $S = 100 + 4 \times \frac{1}{2} (10) (13) = 360$ cm².

Multiplying a dimension by a factor k

If every dimension is multiplied by $k$ :

surface area is multiplied by $k^{2}$
volume is multiplied by $k^{3}$

Example. If all dimensions of a cone are tripled, the surface area is 9 times larger and the volume is 27 times larger.

If only one dimension changes, use the formula. Doubling only the radius of a cone multiplies its volume by 4, because $r$ is squared and $H$ stays the same.

Composite objects of at most two shapes

Add or subtract the parts, and count only the outside surfaces.

Example. An ice-cream cone is a cone with a hemisphere on top, both with radius 3 cm. The cone is 10 cm high.

$V = \frac{1}{3} π (3)^{2} (10) + \frac{1}{2} \cdot \frac{4}{3} π (3)^{3} = 30 π + 18 π = 48 π \approx 150, 80 cm^{3}$

A hemisphere is half a sphere: its curved surface area is $2 π r^{2}$ .

Key terms

Perpendicular height: The height measured at right angles to the base.
Slant height: The height of a sloping face, measured along the face.
Hemisphere: Half of a sphere.
Scale factor: The number $k$ by which every dimension is multiplied.

Common mistakes

Using the slant height in the volume formula instead of the perpendicular height.
Forgetting the $\frac{1}{3}$ for cones and pyramids.
Multiplying volume by $k$ instead of $k^{3}$ when all dimensions are scaled.
Including the flat face of a hemisphere when it is joined to another shape.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

3 marks total

knowledge

A sphere has a radius of 6 cm. Calculate its volume and surface area. Use $V = \frac{4}{3} π r^{3}$ and $S = 4 π r^{2}$ .

Worked solution

Hidden - try the question first

Solution

$V = \frac{4}{3} π (6)^{3} = 288 π$
$\approx 904, 78$ cm³
$S = 4 π (6)^{2} = 144 π \approx 452, 39$ cm²

Question 2

5 marks total

routine

A cone has a base radius of 5 cm and a perpendicular height of 12 cm. Calculate its volume and total surface area. Use $V = \frac{1}{3} π r^{2} H$ and $S = π r^{2} + π r s$ .

Worked solution

Hidden - try the question first

Solution

$V = \frac{4}{3} π (6)^{3} = 288 π$
$\approx 314, 16$ cm³
$s = 5^{2} + 1 2^{2} = 13$ cm
$S = 25 π + 65 π = 90 π$
$\approx 282, 74$ cm²

Question 3

6 marks total

complex

A right pyramid has a square base with sides of 10 cm and a perpendicular height of 12 cm. Use $V = \frac{1}{3} (area of base) \times H$ and $S = (area of base) + \frac{1}{2} (perimeter of base) \times (slant height)$ .

(a) Calculate the volume. (2)

(b) Calculate the slant height of each triangular face. (2)

Worked solution

Hidden - try the question first

Solution

(a) $V = \frac{1}{3} (10 \times 10) (12)$
$= 400$ cm³
(b) From the centre of the base to the middle of a side is 5 cm
Slant height $= 1 2^{2} + 5^{2} = 13$ cm
(c) $S = 100 + 4 \times \frac{1}{2} (10) (13)$
$= 360$ cm²

Question 4

9 marks total

problem-solving

(a) Every dimension of a cone is multiplied by 3. By what factor does its volume change, and by what factor does its surface area change? (3)

(b) An ice-cream is made of a cone with a hemisphere on top. Both have a radius of 3 cm, and the cone is 10 cm high. Calculate the total volume. Use $V_{cone} = \frac{1}{3} π r^{2} H$ and $V_{sphere} = \frac{4}{3} π r^{3}$ . (4)

Worked solution

Hidden - try the question first

Solution

(a) Volume: $V = \frac{1}{3} (10 \times 10) (12)$
$= 27$ times larger
Surface area: $3^{2} = 9$ times larger
(b) Cone: $\frac{1}{3} π (3)^{2} (10) = 30 π$
Hemisphere: $\frac{1}{2} \cdot \frac{4}{3} π (3)^{3} = 18 π$
Total $= 48 π$
$\approx 150, 80$ cm³
(c) $V = \frac{1}{3} π (2 r)^{2} H = 4 \times \frac{1}{3} π r^{2} H$
4 times larger

FAQs

Are the formulae for cones and spheres given?

They are not on the information sheet. Our questions print the formulae you need, and many school papers do the same, but check with your teacher and learn them to be safe. You must still know when to use the slant height and when to use the perpendicular height.

What happens to volume when dimensions are doubled?

If every dimension is doubled, the volume is multiplied by $2^{3} = 8$ and the surface area by $2^{2} = 4$ .

Is measurement examined in Grade 12?

Measurement can appear in Grade 12 trigonometry questions about 2D and 3D shapes.

Surface area and volume of cones, spheres and pyramids

Right pyramids, cones and spheres

Multiplying a dimension by a factor k

Composite objects of at most two shapes

Key terms

Common mistakes

Practice questions

FAQs

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