Sec 3 E-Math Area and volume of similar figures worksheet with answers

A free Sec 3 E-Math worksheet on area and volume of similar figures: 9 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 E-Math worksheet: Area and volume of similar figures

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

Triangles

ABC

and

PQR

are similar, with areas

8

cm² and

24.5

cm² respectively.

PQ = 7

cm. Find

AB

Answercm[2]

Two similar cylinders have total surface areas

144

cm² and

441

cm². The volume of the smaller cylinder is

128

cm³. Find the volume of the larger cylinder.

Answercm³[2]

Two similar cylinders have heights

3

cm and

4

cm. The volume of the smaller cylinder is

27

cm³. Find the volume of the larger cylinder.

Answercm³[2]

Two similar cylinders have total surface areas

588

cm² and

1323

cm². The volume of the larger cylinder is

3375

cm³. Find the volume of the smaller cylinder.

Answercm³[2]

A map is drawn to a scale of

1 : 500\,000

. A park has an area of

6

cm² on the map. Find the actual area, in km², of the park.

Answerkm²[3]

Section B

In the diagram,

D

lies on

AB

and

E

lies on

AC

, with

DE

parallel to

BC

AD = 3

cm,

DB = 4

cm and the area of triangle

ABC

98

cm².

(a)Find the area of triangle

ADE

Answer (a)cm²[2]

(b)Find the area of the quadrilateral

DBCE

Answer (b)cm²[1]

In the diagram,

D

lies on

AB

and

E

lies on

AC

, with

DE

parallel to

BC

AD = 2

cm,

DB = 1

cm and the area of triangle

ABC

36

cm².

(a)Find the area of triangle

ADE

Answer (a)cm²[2]

(b)Find the area of the quadrilateral

DBCE

Answer (b)cm²[1]

Two similar containers are made of the same material. The smaller container is

2

cm tall and the larger one is

5

cm tall.

(a)The capacity of the smaller container is

128

cm³. Find the capacity of the larger container.

Answer (a)cm³[2]

(b)The surface area of the larger container is

735

cm². Find the surface area of the smaller container.

Answer (b)cm²[2]

(c)The mass of the smaller container is

160

g. Find the mass of the larger container.

Answer (c)g[1]

ABCD

is a trapezium in which

AB

is parallel to

DC

. The diagonals

AC

and

BD

meet at

X

AB = 20

cm,

DC = 8

cm and the area of triangle

ABX

125

cm².

(a)Show that triangle

ABX

is similar to triangle

CDX

[2]

(b)Find the area of triangle

CDX

Answer (b)cm²[2]

(c)Find the area of triangle

ADX

Answer (c)cm²[2]

(d)Find the area of the trapezium

ABCD

Answer (d)cm²[2]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

AB = 4

cm[2]

\dfrac{l_1}{l_2} = \sqrt{\dfrac{A_1}{A_2}}

\mathstrut \dfrac{AB}{7} = \sqrt{\dfrac{8}{24.5}}

\dfrac{AB}{7} = \dfrac{4}{7}

AB = \dfrac{4}{7} \times 7

AB = 4

Question 2

Answer:

V = 686

cm³[2]

\mathstrut \dfrac{l_1}{l_2} = \sqrt{\dfrac{144}{441}}

\dfrac{l_1}{l_2} = \dfrac{4}{7}

\mathstrut \dfrac{V}{128} = \left(\dfrac{7}{4}\right)^3

V = \left(\dfrac{7}{4}\right)^3 \times 128

V = 686

cm³

Question 3

Answer:

V = 64

cm³[2]

\dfrac{V_1}{V_2} = \left(\dfrac{l_1}{l_2}\right)^3

\mathstrut \dfrac{V}{27} = \left(\dfrac{4}{3}\right)^3

V = \left(\dfrac{4}{3}\right)^3 \times 27

V = 64

cm³

Question 4

Answer:

V = 1000

cm³[2]

\mathstrut \dfrac{l_1}{l_2} = \sqrt{\dfrac{588}{1323}}

\dfrac{l_1}{l_2} = \dfrac{2}{3}

\mathstrut \dfrac{V}{3375} = \left(\dfrac{2}{3}\right)^3

V = \left(\dfrac{2}{3}\right)^3 \times 3375

V = 1000

cm³

Question 5

Answer:

6\text{ cm}^2 : 150\text{ km}^2

km²[3]

\mathstrut 1 : 500\,000

\mathstrut 1\text{ cm} : 500\,000\text{ cm}

1\text{ cm} : 5\text{ km}

1^2\text{ cm}^2 : 5^2\text{ km}^2

1\text{ cm}^2 : 25\text{ km}^2

\mathstrut 6\text{ cm}^2 : 6 \times 25\text{ km}^2

6\text{ cm}^2 : 150\text{ km}^2

Question 6

(a)Answer:

18

cm²[2]

AB = 3 + 4

\phantom{AB} = 7

\dfrac{\text{Area of } \triangle ADE}{98} = \left(\dfrac{3}{7}\right)^2

\text{Area of } \triangle ADE = 18

cm²

(b)Answer:

80

cm²[1]

\text{Area of } DBCE = 98 - 18

\phantom{\text{Area of } DBCE} = 80

cm²

Question 7

(a)Answer:

16

cm²[2]

AB = 2 + 1

\phantom{AB} = 3

\dfrac{\text{Area of } \triangle ADE}{36} = \left(\dfrac{2}{3}\right)^2

\text{Area of } \triangle ADE = 16

cm²

(b)Answer:

20

cm²[1]

\text{Area of } DBCE = 36 - 16

\phantom{\text{Area of } DBCE} = 20

cm²

Question 8

(a)Answer:

2000

cm³[2]

\dfrac{V}{128} = \left(\dfrac{5}{2}\right)^3

V = 2000

cm³

(b)Answer:

117.6

cm²[2]

\dfrac{A}{735} = \left(\dfrac{2}{5}\right)^2

A = 117.6

cm²

(c)Answer:

2500

g[1]

\text{Same material, so mass} \propto \text{volume}

\dfrac{m}{160} = \left(\dfrac{5}{2}\right)^3

m = 2500

Question 9

(a)[2]

\angle XAB = \angle XCD

(alt. ∠s,

AB \parallel DC

)

\angle AXB = \angle CXD

(vert. opp. ∠s)

\triangle ABX

is similar to

\triangle CDX

(AA)

(b)Answer:

20

cm²[2]

\dfrac{\text{Area of } \triangle CDX}{125} = \left(\dfrac{2}{5}\right)^2

\text{Area of } \triangle CDX = 20

cm²

(c)Answer:

50

cm²[2]

\dfrac{DX}{XB} = \dfrac{DC}{AB}

\phantom{\dfrac{DX}{XB}} = \dfrac{2}{5}

(similar △s)

\triangle ADX

and

\triangle ABX

have the same height from

A

\text{Area of } \triangle ADX = \dfrac{2}{5} \times 125

\phantom{\text{Area of } \triangle ADX} = 50

cm²

(d)Answer:

245

cm²[2]

\triangle ABD

and

\triangle ABC

have the same base

AB

and the same height

\text{Area of } \triangle BCX = \text{area of } \triangle ADX

\phantom{\text{Area of } \triangle BCX} = 50

\text{Area of } ABCD = 125 + 20 + 50 + 50

\phantom{\text{Area of } ABCD} = 245

cm²

Practice questions by skill

Sec 3 E-Math Area and volume of similar figures worksheet with answers

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

Section B

In the diagram,

D

lies on

AB

and

E

lies on

AC

, with

DE

parallel to

BC

AD = 3

cm,

DB = 4

cm and the area of triangle

ABC

98

cm².

(a)Find the area of triangle

ADE

Answer (a)cm²[2]

(b)Find the area of the quadrilateral

DBCE

Answer (b)cm²[1]

Answers and worked solutions

Question 1

Answer:

AB = 4

cm[2]

\dfrac{l_1}{l_2} = \sqrt{\dfrac{A_1}{A_2}}

\mathstrut \dfrac{AB}{7} = \sqrt{\dfrac{8}{24.5}}

\dfrac{AB}{7} = \dfrac{4}{7}

AB = \dfrac{4}{7} \times 7

AB = 4

Question 2

Answer:

V = 686

cm³[2]

\mathstrut \dfrac{l_1}{l_2} = \sqrt{\dfrac{144}{441}}

\dfrac{l_1}{l_2} = \dfrac{4}{7}

\mathstrut \dfrac{V}{128} = \left(\dfrac{7}{4}\right)^3

V = \left(\dfrac{7}{4}\right)^3 \times 128

V = 686

cm³

Question 3

Answer:

V = 64

cm³[2]

\dfrac{V_1}{V_2} = \left(\dfrac{l_1}{l_2}\right)^3

\mathstrut \dfrac{V}{27} = \left(\dfrac{4}{3}\right)^3

V = \left(\dfrac{4}{3}\right)^3 \times 27

V = 64

cm³

Question 4

Answer:

V = 1000

cm³[2]

\mathstrut \dfrac{l_1}{l_2} = \sqrt{\dfrac{588}{1323}}

\dfrac{l_1}{l_2} = \dfrac{2}{3}

\mathstrut \dfrac{V}{3375} = \left(\dfrac{2}{3}\right)^3

V = \left(\dfrac{2}{3}\right)^3 \times 3375

V = 1000

cm³

Question 5

Answer:

6\text{ cm}^2 : 150\text{ km}^2

km²[3]

\mathstrut 1 : 500\,000

\mathstrut 1\text{ cm} : 500\,000\text{ cm}

1\text{ cm} : 5\text{ km}

1^2\text{ cm}^2 : 5^2\text{ km}^2

1\text{ cm}^2 : 25\text{ km}^2

\mathstrut 6\text{ cm}^2 : 6 \times 25\text{ km}^2

6\text{ cm}^2 : 150\text{ km}^2

Question 6

(a)Answer:

18

cm²[2]

AB = 3 + 4

\phantom{AB} = 7

\dfrac{\text{Area of } \triangle ADE}{98} = \left(\dfrac{3}{7}\right)^2

\text{Area of } \triangle ADE = 18

cm²

(b)Answer:

80

cm²[1]

\text{Area of } DBCE = 98 - 18

\phantom{\text{Area of } DBCE} = 80

cm²

Question 7

(a)Answer:

16

cm²[2]

AB = 2 + 1

\phantom{AB} = 3

\dfrac{\text{Area of } \triangle ADE}{36} = \left(\dfrac{2}{3}\right)^2

\text{Area of } \triangle ADE = 16

cm²

(b)Answer:

20

cm²[1]

\text{Area of } DBCE = 36 - 16

\phantom{\text{Area of } DBCE} = 20

cm²

Question 8

(a)Answer:

2000

cm³[2]

\dfrac{V}{128} = \left(\dfrac{5}{2}\right)^3

V = 2000

cm³

(b)Answer:

117.6

cm²[2]

\dfrac{A}{735} = \left(\dfrac{2}{5}\right)^2

A = 117.6

cm²

(c)Answer:

2500

g[1]

\text{Same material, so mass} \propto \text{volume}

\dfrac{m}{160} = \left(\dfrac{5}{2}\right)^3

m = 2500

Question 9

(a)[2]

\angle XAB = \angle XCD

(alt. ∠s,

AB \parallel DC

)

\angle AXB = \angle CXD

(vert. opp. ∠s)

\triangle ABX

is similar to

\triangle CDX

(AA)

(b)Answer:

20

cm²[2]

\dfrac{\text{Area of } \triangle CDX}{125} = \left(\dfrac{2}{5}\right)^2

\text{Area of } \triangle CDX = 20

cm²

(c)Answer:

50

cm²[2]

\dfrac{DX}{XB} = \dfrac{DC}{AB}

\phantom{\dfrac{DX}{XB}} = \dfrac{2}{5}

(similar △s)

\triangle ADX

and

\triangle ABX

have the same height from

A

\text{Area of } \triangle ADX = \dfrac{2}{5} \times 125

\phantom{\text{Area of } \triangle ADX} = 50

cm²

(d)Answer:

245

cm²[2]

\triangle ABD

and

\triangle ABC

have the same base

AB

and the same height

\text{Area of } \triangle BCX = \text{area of } \triangle ADX

\phantom{\text{Area of } \triangle BCX} = 50

\text{Area of } ABCD = 125 + 20 + 50 + 50

\phantom{\text{Area of } ABCD} = 245

cm²