Sec 2 Maths Pyramids, cones and spheres worksheet with answers

A free Sec 2 Maths worksheet on pyramids, cones and spheres: 6 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 Maths worksheet: Pyramids, cones and spheres

Total: 30 marksSuggested time: 45 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 pyramid with a square base of side

6

cm. The slant height of each triangular face is

5

cm. Find the total surface area of the pyramid.

Answercm²[3]

6

identical solid metal spheres, each of radius

2.5

cm, are melted down and recast into one solid sphere, with no metal lost. Find the radius of this sphere, correct to 3 significant figures.

Answercm[3]

A solid cone has base radius

r

cm and height

h

cm. Its volume is equal to the volume of a solid cylinder of radius

r

cm and height

6r

cm. Find

h

in terms of

r

. Hence show that the curved surface area of the cone is

k\pi r^2

cm², and find

k

, leaving your answer in surd form.

Answer[3]

Section B

(a)A solid cylinder of radius

r

cm and height

r

cm is melted down and recast into a solid cylinder of radius

2r

cm and height

h

cm, with no metal lost. Express

h

in terms of

r

Answer (a)[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]

(a)The diagram shows a pyramid with a rectangular base

9

cm by

6

cm. Its apex is

9

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

Answer (a)cm³[2]

(b)The diagram shows a solid made of a cylinder of radius

6

cm and height

10

cm, with a cone of base radius

4

cm and height

3

cm standing on the centre of its top face. Find the total surface area of the solid, correct to 3 significant figures.

Answer (b)cm²[3]

(c)A solid metal cylinder of radius

3

cm and height

7

cm is melted down and recast into a prism of length

15

cm, with no metal lost. The cross-section of the prism is a sector of a circle with radius

R

cm and angle

120^\circ

, as shown. Find

R

, correct to 3 significant figures.

Answer (c)cm[3]

(a)The diagram shows a conical funnel with base radius

2

cm and slant height

12

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

Answer (a)cm²[2]

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

12

cm. Its apex

V

is vertically above the centre

O

of the base, and each of its four slant edges is

12

cm long. Find the height of the pyramid, and hence its volume, correct to 3 significant figures.

Answer (b)cm³[3]

(c)A solid metal hemisphere of radius

4

cm is melted down and recast into a prism of length

5

cm, with no metal lost. The cross-section of the prism is a sector of a circle with radius

R

cm and angle

45^\circ

, as shown. Find

R

, correct to 3 significant figures.

Answer (c)cm[3]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

= 96

cm²[3]

Base area

= 6 \times 6

= 36

cm²Area of one face

= \dfrac{1}{2} \times 6 \times 5

= 15

cm²Total surface area

= 36 + 4 \times 15

= 96

cm²

Question 2

Answer:

= 4.54

cm (3 s.f.)[3]

Total volume

= 6 \times \dfrac{4}{3} \times \pi \times 2.5^3

= 392.6990\ldots

cm³

\dfrac{4}{3}\pi R^3 = 392.6990\ldots

R^3 = 93.7500\ldots

R = \sqrt[3]{93.7500\ldots}

= 4.5428\ldots

= 4.54

cm (3 s.f.)

Question 3

Answer:

\pi r \times 5\sqrt{13}\,r = 5\sqrt{13}\,\pi r^2

[3]

\dfrac{1}{3}\pi r^2 h = 6\pi r^3

h = 18r

l^2 = r^2 + (18r)^2

\phantom{l^2} = 325r^2

l = \sqrt{325}\,r

\phantom{l} = 5\sqrt{13}\,r

\pi r \times 5\sqrt{13}\,r = 5\sqrt{13}\,\pi r^2

Question 4

(a)Answer:

h = \dfrac{1}{4}r

[2]

Volume

= \pi r^3

Volume

= \pi \left(2r\right)^2 h

= 4\pi r^2 h

4\pi r^2 h = \pi r^3

h = \dfrac{1}{4}r

(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.)

Question 5

(a)Answer:

= 162

cm³[2]

Base area

= 9 \times 6

= 54

cm²

V = \dfrac{1}{3} \times 54 \times 9

= 162

cm³

(b)Answer:

= 616

cm² (3 s.f.)[3]

l = \sqrt{4^2 + 3^2}

\phantom{l} = 5

Base:

\pi \times 6^2 = 36\pi

Curved side:

2\pi \times 6 \times 10 = 120\pi

Ring:

\pi \times 6^2 - \pi \times 4^2 = 20\pi

Cone:

\pi \times 4 \times 5 = 20\pi

Total

= 196\pi

= 615.7521\ldots

= 616

cm² (3 s.f.)

(c)Answer:

= 3.55

cm (3 s.f.)[3]

V = \pi \times 3^2 \times 7

\phantom{V} = 63\pi

cm³

\dfrac{120}{360} \times \pi R^2 \times 15 = 63\pi

R^2 = \dfrac{63 \times 360}{120 \times 15}

\phantom{R^2} = 12.6

= 3.5496\ldots

= 3.55

cm (3 s.f.)

Question 6

(a)Answer:

= 75.4

cm² (3 s.f.)[2]

Curved surface area

= \pi r l

Curved surface area

= \pi \times 2 \times 12

= 24\pi

= 75.3982\ldots

= 75.4

cm² (3 s.f.)

(b)Answer:

= 407

cm³ (3 s.f.)[3]

OB^2 = 6^2 + 6^2

= 72

h^2 = 12^2 - 72

= 72

h = \sqrt{72}

= 8.4852\ldots

V = \dfrac{1}{3} \times 12^2 \times 8.4852\ldots

= 407.2935\ldots

= 407

cm³ (3 s.f.)

(c)Answer:

= 8.26

cm (3 s.f.)[3]

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

\phantom{V} = \dfrac{128}{3}\pi

cm³

\dfrac{45}{360} \times \pi R^2 \times 5 = \dfrac{128}{3}\pi

R^2 = \dfrac{42.666667 \times 360}{45 \times 5}

\phantom{R^2} = 68.2666\ldots

= 8.2623\ldots

= 8.26

cm (3 s.f.)

Practice questions by skill

Sec 2 Maths Pyramids, cones and spheres worksheet with answers

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

Need the formulas first? Read the Pyramids, cones and spheres notes. Or practise it online with step-by-step help when you get one wrong.

Section B

(a)A solid cylinder of radius

r

cm and height

r

cm is melted down and recast into a solid cylinder of radius

2r

cm and height

h

cm, with no metal lost. Express

h

in terms of

r

Answer (a)[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]

Answers and worked solutions

Question 1

Answer:

= 96

cm²[3]

Base area

= 6 \times 6

= 36

cm²Area of one face

= \dfrac{1}{2} \times 6 \times 5

= 15

cm²Total surface area

= 36 + 4 \times 15

= 96

cm²

Question 2

Answer:

= 4.54

cm (3 s.f.)[3]

Total volume

= 6 \times \dfrac{4}{3} \times \pi \times 2.5^3

= 392.6990\ldots

cm³

\dfrac{4}{3}\pi R^3 = 392.6990\ldots

R^3 = 93.7500\ldots

R = \sqrt[3]{93.7500\ldots}

= 4.5428\ldots

= 4.54

cm (3 s.f.)

Question 3

Answer:

\pi r \times 5\sqrt{13}\,r = 5\sqrt{13}\,\pi r^2

[3]

\dfrac{1}{3}\pi r^2 h = 6\pi r^3

h = 18r

l^2 = r^2 + (18r)^2

\phantom{l^2} = 325r^2

l = \sqrt{325}\,r

\phantom{l} = 5\sqrt{13}\,r

\pi r \times 5\sqrt{13}\,r = 5\sqrt{13}\,\pi r^2

Question 4

(a)Answer:

h = \dfrac{1}{4}r

[2]

Volume

= \pi r^3

Volume

= \pi \left(2r\right)^2 h

= 4\pi r^2 h

4\pi r^2 h = \pi r^3

h = \dfrac{1}{4}r

(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.)

Question 5

(a)Answer:

= 162

cm³[2]

Base area

= 9 \times 6

= 54

cm²

V = \dfrac{1}{3} \times 54 \times 9

= 162

cm³

(b)Answer:

= 616

cm² (3 s.f.)[3]

l = \sqrt{4^2 + 3^2}

\phantom{l} = 5

Base:

\pi \times 6^2 = 36\pi

Curved side:

2\pi \times 6 \times 10 = 120\pi

Ring:

\pi \times 6^2 - \pi \times 4^2 = 20\pi

Cone:

\pi \times 4 \times 5 = 20\pi

Total

= 196\pi

= 615.7521\ldots

= 616

cm² (3 s.f.)

(c)Answer:

= 3.55

cm (3 s.f.)[3]

V = \pi \times 3^2 \times 7

\phantom{V} = 63\pi

cm³

\dfrac{120}{360} \times \pi R^2 \times 15 = 63\pi

R^2 = \dfrac{63 \times 360}{120 \times 15}

\phantom{R^2} = 12.6

= 3.5496\ldots

= 3.55

cm (3 s.f.)

Question 6

(a)Answer:

= 75.4

cm² (3 s.f.)[2]

Curved surface area

= \pi r l

Curved surface area

= \pi \times 2 \times 12

= 24\pi

= 75.3982\ldots

= 75.4

cm² (3 s.f.)

(b)Answer:

= 407

cm³ (3 s.f.)[3]

OB^2 = 6^2 + 6^2

= 72

h^2 = 12^2 - 72

= 72

h = \sqrt{72}

= 8.4852\ldots

V = \dfrac{1}{3} \times 12^2 \times 8.4852\ldots

= 407.2935\ldots

= 407

cm³ (3 s.f.)

(c)Answer:

= 8.26

cm (3 s.f.)[3]

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

\phantom{V} = \dfrac{128}{3}\pi

cm³

\dfrac{45}{360} \times \pi R^2 \times 5 = \dfrac{128}{3}\pi

R^2 = \dfrac{42.666667 \times 360}{45 \times 5}

\phantom{R^2} = 68.2666\ldots

= 8.2623\ldots

= 8.26

cm (3 s.f.)