Sec 3 G1 Maths Surface area and volume worksheet with answers

A free Sec 3 G1 Maths worksheet on surface area and volume: 8 questions, 30 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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Sec 3 G1 Maths worksheet: Surface area and volume

Total: 30 marksSuggested time: 50 minutes

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

The diagram shows a solid pyramid with a rectangular base

4

m by

7

m. Its apex is

9

m vertically above the centre of the base. Find the volume of the pyramid.

Answerm³[2]

The diagram shows a traffic cone with base radius

4

cm and slant height

13

cm. Find its curved surface area, correct to 3 significant figures.

Answercm²[2]

The diagram shows a paperweight in the shape of a pyramid with a square base of side

6

cm. Its apex is

6

cm vertically above the centre of the base. Find the volume of the pyramid.

Answercm³[2]

The diagram shows a globe of diameter

4

cm. Find its volume, correct to 3 significant figures.

Answercm³[3]

The diagram shows a bullet-shaped solid: a cylinder with a hemisphere on top. The cylinder has height

12

cm and the common radius is

3

cm. The solid stands on its flat circular base. Find the total surface area of the solid, correct to 3 significant figures.

Answercm²[3]

The diagram shows a conical party hat of height

8

cm and base diameter

12

cm. Find its volume, correct to 3 significant figures.

Answercm³[3]

Section B

(a)The diagram shows a pyramid with a square base of side

6

cm. Its apex is

12

cm vertically above the centre of the base. Find the volume of the pyramid.

Answer (a)cm³[2]

(b)The diagram shows a ball of radius

6

cm. Find its volume, correct to 3 significant figures.

Answer (b)cm³[2]

(c)The diagram shows a model of a grain silo: a cylinder topped by a hemisphere. The cylinder has height

5

cm and the common radius is

2.5

cm. The solid stands on its flat circular base. Find the total surface area of the solid, correct to 3 significant figures.

Answer (c)cm²[3]

(a)The diagram shows a pyramid with a square base of side

6

m. Its volume is

96

m³. Find its height,

h

Answer (a)m[2]

(b)The diagram shows a capsule-shaped bottle: a cylinder with a hemispherical top. The cylinder has height

10

cm and the common radius is

5

cm. The solid stands on its flat circular base. Find the total surface area of the solid, correct to 3 significant figures.

Answer (b)cm²[3]

(c)The diagram shows a conical party hat with base radius

4.5

cm and height

6

cm. Find its slant height

l

cm, and hence its curved surface area, correct to 3 significant figures.

Answer (c)cm²[3]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

= 84

m³[2]

Base area

= 4 \times 7

= 28

m²

V = \dfrac{1}{3} \times 28 \times 9

= 84

m³

Question 2

Answer:

= 163

cm² (3 s.f.)[2]

Curved surface area

= \pi r l

Curved surface area

= \pi \times 4 \times 13

= 52\pi

= 163.3628\ldots

= 163

cm² (3 s.f.)

Question 3

Answer:

= 72

cm³[2]

Base area

= 6 \times 6

= 36

cm²

V = \dfrac{1}{3} \times 36 \times 6

= 72

cm³

Question 4

Answer:

= 33.5

cm³ (3 s.f.)[3]

r = \dfrac{4}{2}

= 2

V = \dfrac{4}{3}\pi r^3

V = \dfrac{4}{3} \times \pi \times 2^3

= 33.5103\ldots

= 33.5

cm³ (3 s.f.)

Question 5

Answer:

= 311

cm² (3 s.f.)[3]

The circle where the cylinder meets the hemisphere is hidden inside, so it is not surface; the flat base is.Surface area

=

base

+

curved side of cylinder

+

curved surface of hemisphereBase

= \pi \times 3^2

= 28.2743\ldots

cm²Curved side

= 2 \times \pi \times 3 \times 12

= 226.1946\ldots

cm²Hemisphere

= 2 \times \pi \times 3^2

= 56.5486\ldots

cm²Total surface area

= 28.2743\ldots + 226.1946\ldots + 56.5486\ldots

= 311.0176\ldots

= 311

cm² (3 s.f.)

Question 6

Answer:

= 302

cm³ (3 s.f.)[3]

r = \dfrac{12}{2}

= 6

V = \dfrac{1}{3}\pi r^2 h

V = \dfrac{1}{3} \times \pi \times 6^2 \times 8

= 96\pi

= 301.5928\ldots

= 302

cm³ (3 s.f.)

Question 7

(a)Answer:

= 144

cm³[2]

Base area

= 6 \times 6

= 36

cm²

V = \dfrac{1}{3} \times 36 \times 12

= 144

cm³

(b)Answer:

= 905

cm³ (3 s.f.)[2]

V = \dfrac{4}{3}\pi r^3

V = \dfrac{4}{3} \times \pi \times 6^3

= 904.7786\ldots

= 905

cm³ (3 s.f.)

(c)Answer:

= 137

cm² (3 s.f.)[3]

The circle where the cylinder meets the hemisphere is hidden inside, so it is not surface; the flat base is.Surface area

=

base

+

curved side of cylinder

+

curved surface of hemisphereBase

= \pi \times 2.5^2

= 19.6349\ldots

cm²Curved side

= 2 \times \pi \times 2.5 \times 5

= 78.5398\ldots

cm²Hemisphere

= 2 \times \pi \times 2.5^2

= 39.2699\ldots

cm²Total surface area

= 19.6349\ldots + 78.5398\ldots + 39.2699\ldots

= 137.4446\ldots

= 137

cm² (3 s.f.)

Question 8

(a)Answer:

= 8

m[2]

Base area

= 6 \times 6

= 36

m²

\dfrac{1}{3} \times 36 \times h = 96

h = \dfrac{96 \times 3}{36}

= 8

(b)Answer:

= 550

cm² (3 s.f.)[3]

The circle where the cylinder meets the hemisphere is hidden inside, so it is not surface; the flat base is.Surface area

=

base

+

curved side of cylinder

+

curved surface of hemisphereBase

= \pi \times 5^2

= 78.5398\ldots

cm²Curved side

= 2 \times \pi \times 5 \times 10

= 314.1592\ldots

cm²Hemisphere

= 2 \times \pi \times 5^2

= 157.0796\ldots

cm²Total surface area

= 78.5398\ldots + 314.1592\ldots + 157.0796\ldots

= 549.7787\ldots

= 550

cm² (3 s.f.)

(c)Answer:

= 106

cm² (3 s.f.)[3]

l^2 = 4.5^2 + 6^2

= 56.25

l = \sqrt{56.25}

= 7.5000\ldots

Curved surface area

= \pi r l

Curved surface area

= \pi \times 4.5 \times 7.5000\ldots

= 106.0287\ldots

= 106

cm² (3 s.f.)

Sec 3 G1 Maths Surface area and volume worksheet with answers

In the print window, choose Save as PDF to download it.

Or practise it online with step-by-step help when you get one wrong.

Section B

(a)The diagram shows a pyramid with a square base of side

6

cm. Its apex is

12

cm vertically above the centre of the base. Find the volume of the pyramid.

Answer (a)cm³[2]

(b)The diagram shows a ball of radius

6

cm. Find its volume, correct to 3 significant figures.

Answer (b)cm³[2]

(c)The diagram shows a model of a grain silo: a cylinder topped by a hemisphere. The cylinder has height

5

cm and the common radius is

2.5

cm. The solid stands on its flat circular base. Find the total surface area of the solid, correct to 3 significant figures.

Answer (c)cm²[3]

Answers and worked solutions

Question 1

Answer:

= 84

m³[2]

Base area

= 4 \times 7

= 28

m²

V = \dfrac{1}{3} \times 28 \times 9

= 84

m³

Question 2

Answer:

= 163

cm² (3 s.f.)[2]

Curved surface area

= \pi r l

Curved surface area

= \pi \times 4 \times 13

= 52\pi

= 163.3628\ldots

= 163

cm² (3 s.f.)

Question 3

Answer:

= 72

cm³[2]

Base area

= 6 \times 6

= 36

cm²

V = \dfrac{1}{3} \times 36 \times 6

= 72

cm³

Question 4

Answer:

= 33.5

cm³ (3 s.f.)[3]

r = \dfrac{4}{2}

= 2

V = \dfrac{4}{3}\pi r^3

V = \dfrac{4}{3} \times \pi \times 2^3

= 33.5103\ldots

= 33.5

cm³ (3 s.f.)

Question 5

Answer:

= 311

cm² (3 s.f.)[3]

The circle where the cylinder meets the hemisphere is hidden inside, so it is not surface; the flat base is.Surface area

=

base

+

curved side of cylinder

+

curved surface of hemisphereBase

= \pi \times 3^2

= 28.2743\ldots

cm²Curved side

= 2 \times \pi \times 3 \times 12

= 226.1946\ldots

cm²Hemisphere

= 2 \times \pi \times 3^2

= 56.5486\ldots

cm²Total surface area

= 28.2743\ldots + 226.1946\ldots + 56.5486\ldots

= 311.0176\ldots

= 311

cm² (3 s.f.)

Question 6

Answer:

= 302

cm³ (3 s.f.)[3]

r = \dfrac{12}{2}

= 6

V = \dfrac{1}{3}\pi r^2 h

V = \dfrac{1}{3} \times \pi \times 6^2 \times 8

= 96\pi

= 301.5928\ldots

= 302

cm³ (3 s.f.)

Question 7

(a)Answer:

= 144

cm³[2]

Base area

= 6 \times 6

= 36

cm²

V = \dfrac{1}{3} \times 36 \times 12

= 144

cm³

(b)Answer:

= 905

cm³ (3 s.f.)[2]

V = \dfrac{4}{3}\pi r^3

V = \dfrac{4}{3} \times \pi \times 6^3

= 904.7786\ldots

= 905

cm³ (3 s.f.)

(c)Answer:

= 137

cm² (3 s.f.)[3]

The circle where the cylinder meets the hemisphere is hidden inside, so it is not surface; the flat base is.Surface area

=

base

+

curved side of cylinder

+

curved surface of hemisphereBase

= \pi \times 2.5^2

= 19.6349\ldots

cm²Curved side

= 2 \times \pi \times 2.5 \times 5

= 78.5398\ldots

cm²Hemisphere

= 2 \times \pi \times 2.5^2

= 39.2699\ldots

cm²Total surface area

= 19.6349\ldots + 78.5398\ldots + 39.2699\ldots

= 137.4446\ldots

= 137

cm² (3 s.f.)

Question 8

(a)Answer:

= 8

m[2]

Base area

= 6 \times 6

= 36

m²

\dfrac{1}{3} \times 36 \times h = 96

h = \dfrac{96 \times 3}{36}

= 8

(b)Answer:

= 550

cm² (3 s.f.)[3]

The circle where the cylinder meets the hemisphere is hidden inside, so it is not surface; the flat base is.Surface area

=

base

+

curved side of cylinder

+

curved surface of hemisphereBase

= \pi \times 5^2

= 78.5398\ldots

cm²Curved side

= 2 \times \pi \times 5 \times 10

= 314.1592\ldots

cm²Hemisphere

= 2 \times \pi \times 5^2

= 157.0796\ldots

cm²Total surface area

= 78.5398\ldots + 314.1592\ldots + 157.0796\ldots

= 549.7787\ldots

= 550

cm² (3 s.f.)

(c)Answer:

= 106

cm² (3 s.f.)[3]

l^2 = 4.5^2 + 6^2

= 56.25

l = \sqrt{56.25}

= 7.5000\ldots

Curved surface area

= \pi r l

Curved surface area

= \pi \times 4.5 \times 7.5000\ldots

= 106.0287\ldots

= 106

cm² (3 s.f.)