Sec 2 G2 Maths Mensuration worksheet with answers

A free Sec 2 G2 Maths worksheet on mensuration: 7 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 2 G2 Maths worksheet: Mensuration

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 sphere of radius

2.5

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

Answercm²[2]

The diagram shows a solid made of a cylinder with a hemisphere on top. The cylinder has height

3

cm and the common radius is

4

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 pyramid with a square base of side

16

cm. Its apex is

9

cm vertically above the centre of the base. Find the slant height,

l

cm, of a triangular face, and hence the total surface area of the pyramid, 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 ball of radius

6

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

Answer (a)cm³[2]

(b)

2

identical ball bearings, each of radius

3

cm, are stacked one on top of another in a closed cylinder. They fit exactly: they touch its curved side, and the end ones touch its top and bottom. Find the volume of the empty space in the cylinder, correct to 3 significant figures.

Answer (b)cm³[3]

(a)The diagram shows a globe of radius

3.5

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

Answer (a)cm²[2]

(b)The diagram shows a conical funnel of height

10

cm and base radius

2

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

Answer (b)cm³[2]

(c)The diagram shows a glass building in the shape of a pyramid with a rectangular base

6

m by

15

m. The slant heights of the triangular faces are

8.5

m (on the

6

m sides) and

5

m (on the

15

m sides). Find the total surface area of the pyramid.

Answer (c)m²[3]

(a)The diagram shows a glass building in the shape of a pyramid with a square base of side

4

m. Its volume is

64

m³. Find its height,

h

Answer (a)m[2]

(b)The surface area of a solid sphere is

324\pi

cm². Find its radius,

r

cm.

Answer (b)cm[2]

(c)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. Find the volume of the solid, correct to 3 significant figures.

Answer (c)cm³[3]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

= 78.5

cm² (3 s.f.)[2]

Surface area

= 4\pi r^2

Surface area

= 4 \times \pi \times 2.5^2

= 25\pi

= 78.5398\ldots

= 78.5

cm² (3 s.f.)

Question 2

Answer:

= 226

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 4^2

= 50.2654\ldots

cm²Curved side

= 2 \times \pi \times 4 \times 3

= 75.3982\ldots

cm²Hemisphere

= 2 \times \pi \times 4^2

= 100.5309\ldots

cm²Total surface area

= 50.2654\ldots + 75.3982\ldots + 100.5309\ldots

= 226.1946\ldots

= 226

cm² (3 s.f.)

Question 3

Answer:

= 641

cm² (3 s.f.)[3]

l^2 = 9^2 + 8^2

= 145

l = \sqrt{145}

= 12.0415\ldots

Base area

= 16 \times 16

= 256

cm²Area of one face

= \dfrac{1}{2} \times 16 \times 12.0415\ldots

= 96.3327\ldots

Total surface area

= 256 + 4 \times 96.3327\ldots

= 641.3310\ldots

= 641

cm² (3 s.f.)

Question 4

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 5

(a)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.)

(b)Answer:

= 113

cm³ (3 s.f.)[3]

Height

= 2 \times 2 \times 3

= 12

cmVolume of cylinder

= \pi \times 3^2 \times 12

= 339.2920\ldots

cm³Volume of the balls

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

= 226.1946\ldots

cm³Empty space

= 339.2920\ldots - 226.1946\ldots

= 113.0973\ldots

= 113

cm³ (3 s.f.)

Question 6

(a)Answer:

= 154

cm² (3 s.f.)[2]

Surface area

= 4\pi r^2

Surface area

= 4 \times \pi \times 3.5^2

= 49\pi

= 153.9380\ldots

= 154

cm² (3 s.f.)

(b)Answer:

= 41.9

cm³ (3 s.f.)[2]

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

V = \dfrac{1}{3} \times \pi \times 2^2 \times 10

= 41.8879\ldots

= 41.9

cm³ (3 s.f.)

(c)Answer:

= 216

m²[3]

Base area

= 6 \times 15

= 90

m²Each face on a

6

m side

= \dfrac{1}{2} \times 6 \times 8.5

= 25.5

m²Each face on a

15

m side

= \dfrac{1}{2} \times 15 \times 5

= 37.5

m²Total surface area

= 90 + 2 \times 25.5 + 2 \times 37.5

= 216

m²

Question 7

(a)Answer:

= 12

m[2]

Base area

= 4 \times 4

= 16

m²

\dfrac{1}{3} \times 16 \times h = 64

h = \dfrac{64 \times 3}{16}

= 12

(b)Answer:

= 9

cm[2]

4\pi r^2 = 324\pi

r^2 = \dfrac{324}{4}

= 81

r = \sqrt{81}

= 9

(c)Answer:

= 396

cm³ (3 s.f.)[3]

Volume of cylinder

= \pi \times 3^2 \times 12

= 339.2920\ldots

cm³Volume of hemisphere

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

= 56.5486\ldots

cm³Total volume

= 339.2920\ldots + 56.5486\ldots

= 395.8406\ldots

= 396

cm³ (3 s.f.)

Sec 2 G2 Maths Mensuration 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 ball of radius

6

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

Answer (a)cm³[2]

(b)

2

identical ball bearings, each of radius

3

Answer (b)cm³[3]

Answers and worked solutions

Question 1

Answer:

= 78.5

cm² (3 s.f.)[2]

Surface area

= 4\pi r^2

Surface area

= 4 \times \pi \times 2.5^2

= 25\pi

= 78.5398\ldots

= 78.5

cm² (3 s.f.)

Question 2

Answer:

= 226

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 4^2

= 50.2654\ldots

cm²Curved side

= 2 \times \pi \times 4 \times 3

= 75.3982\ldots

cm²Hemisphere

= 2 \times \pi \times 4^2

= 100.5309\ldots

cm²Total surface area

= 50.2654\ldots + 75.3982\ldots + 100.5309\ldots

= 226.1946\ldots

= 226

cm² (3 s.f.)

Question 3

Answer:

= 641

cm² (3 s.f.)[3]

l^2 = 9^2 + 8^2

= 145

l = \sqrt{145}

= 12.0415\ldots

Base area

= 16 \times 16

= 256

cm²Area of one face

= \dfrac{1}{2} \times 16 \times 12.0415\ldots

= 96.3327\ldots

Total surface area

= 256 + 4 \times 96.3327\ldots

= 641.3310\ldots

= 641

cm² (3 s.f.)

Question 4

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 5

(a)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.)

(b)Answer:

= 113

cm³ (3 s.f.)[3]

Height

= 2 \times 2 \times 3

= 12

cmVolume of cylinder

= \pi \times 3^2 \times 12

= 339.2920\ldots

cm³Volume of the balls

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

= 226.1946\ldots

cm³Empty space

= 339.2920\ldots - 226.1946\ldots

= 113.0973\ldots

= 113

cm³ (3 s.f.)

Question 6

(a)Answer:

= 154

cm² (3 s.f.)[2]

Surface area

= 4\pi r^2

Surface area

= 4 \times \pi \times 3.5^2

= 49\pi

= 153.9380\ldots

= 154

cm² (3 s.f.)

(b)Answer:

= 41.9

cm³ (3 s.f.)[2]

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

V = \dfrac{1}{3} \times \pi \times 2^2 \times 10

= 41.8879\ldots

= 41.9

cm³ (3 s.f.)

(c)Answer:

= 216

m²[3]

Base area

= 6 \times 15

= 90

m²Each face on a

6

m side

= \dfrac{1}{2} \times 6 \times 8.5

= 25.5

m²Each face on a

15

m side

= \dfrac{1}{2} \times 15 \times 5

= 37.5

m²Total surface area

= 90 + 2 \times 25.5 + 2 \times 37.5

= 216

m²

Question 7

(a)Answer:

= 12

m[2]

Base area

= 4 \times 4

= 16

m²

\dfrac{1}{3} \times 16 \times h = 64

h = \dfrac{64 \times 3}{16}

= 12

(b)Answer:

= 9

cm[2]

4\pi r^2 = 324\pi

r^2 = \dfrac{324}{4}

= 81

r = \sqrt{81}

= 9

(c)Answer:

= 396

cm³ (3 s.f.)[3]

Volume of cylinder

= \pi \times 3^2 \times 12

= 339.2920\ldots

cm³Volume of hemisphere

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

= 56.5486\ldots

cm³Total volume

= 339.2920\ldots + 56.5486\ldots

= 395.8406\ldots

= 396

cm³ (3 s.f.)