Sec 3 E-Math Congruence and similarity worksheet with answers

A free Sec 3 E-Math worksheet on congruence and similarity: 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 E-Math worksheet: Congruence and similarity

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. Using the parts marked, state the similarity test: AA, SSS or SAS.

Answer[2]

The plan of a house is drawn to a scale of

1 : 500

. A room is

7.5

cm long on the plan. Find the actual length, in metres, of the room.

Answerm[2]

In the diagram, triangle

ABC

is congruent to triangle

QPR

. Find

x

Answercm[2]

In the diagram, triangle

ABC

is congruent to triangle

PRQ

. Find

x

Answer°[2]

In the diagram, the lines

AD

and

BC

meet at

X

, and

AB

is parallel to

DC

. Find

DC

, correct to 3 significant figures.

Answercm[3]

Section B

A map is drawn to a scale of

1 : 50\,000

(a)Two towns are

12.4

cm apart on the map. Find the actual distance between them, in kilometres.

Answer (a)km[1]

(b)A road is

4.5

km long. Find its length on the map, in centimetres.

Answer (b)cm[1]

(c)On a second map, the two towns in part (a) are

5

cm apart. Find the scale of the second map in the form

1 : n

Answer (c)[2]

In the diagram,

D

lies on

AB

and

E

lies on

AC

, with

DE

parallel to

BC

AD = 6

cm,

DB = 2

cm,

BC = 9

cm and

AC = 12

cm.

(a)Complete the statement: triangle

ABC

is similar to triangle

\ldots

Answer (a)[1]

(b)Find

DE

Answer (b)cm[2]

(c)Find

EC

Answer (c)cm[2]

ABCD

is a square of side

12

cm.

E

lies on

BC

and

F

lies on

CD

such that

BE = CF = 5

cm.

AE

and

BF

meet at

G

(a)Show that triangle

ABE

is congruent to triangle

BCF

, giving a reason for each statement.

[3]

(b)Find the length of

AE

Answer (b)cm[2]

(c)Hence write down the length of

BF

Answer (c)cm[1]

(d)Find angle

BAE

, correct to 1 decimal place.

Answer (d)°[2]

(e)Using your answer to part (a), find angle

AGB

Answer (e)°[2]

END OF PAPER

Answers and worked solutions

Question 1

Answer: SSS[2]

\mathstrut \dfrac{PR}{AB} = \dfrac{15}{10}

\mathstrut \dfrac{RQ}{BC} = \dfrac{10.5}{7}

\mathstrut \dfrac{QP}{CA} = \dfrac{12}{8}

All three pairs of sides are in the same ratio,

\dfrac{3}{2}

: SSS.

Question 2

Answer:

7.5\text{ cm} : 37.5\text{ m}

[2]

\mathstrut 1 : 500

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

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

\mathstrut 7.5\text{ cm} : 7.5 \times 5\text{ m}

7.5\text{ cm} : 37.5\text{ m}

Question 3

Answer:

x = 9

cm[2]

\triangle ABC \equiv \triangle QPR

, so

AB

matches

QP

QP = AB

x = 9

Question 4

Answer:

x = 97

°[2]

\mathstrut \angle C = 180^\circ - 48^\circ - 35^\circ

\angle C = 97^\circ

\triangle ABC \equiv \triangle PRQ

, so

\angle C

matches

\angle Q

x = 97

Question 5

Answer:

\phantom{DC} = 11.7

cm (3 s.f.)[3]

\angle XAB = \angle XDC

(alternate angles)

\angle AXB = \angle DXC

(vertically opposite angles)

\dfrac{DC}{AB} = \dfrac{XD}{XA}

\mathstrut \dfrac{DC}{7} = \dfrac{10}{6}

DC = \dfrac{10}{6} \times 7

DC = 11.6666\ldots

\phantom{DC} = 11.7

cm (3 s.f.)

Question 6

(a)Answer:

6.2

km[1]

1 : 50\,000

1 \text{ cm} : 50\,000 \text{ cm}

1 \text{ cm} : 0.5 \text{ km}

12.4 \text{ cm} : 6.2 \text{ km}

(b)Answer:

9

cm[1]

1 \text{ cm} : 0.5 \text{ km}

\dfrac{4.5}{0.5} \text{ cm} : 4.5 \text{ km}

9 \text{ cm} : 4.5 \text{ km}

(c)Answer:

1 : 124\,000

[2]

5 \text{ cm} : 6.2 \text{ km}

5 \text{ cm} : 620\,000 \text{ cm}

5 : 620\,000

1 : 124\,000

Question 7

(a)Answer:

ADE

[1]

\angle ADE = \angle ABC

(corresponding angles)

\angle A

is common

\triangle ABC

is similar to

\triangle ADE

(AA)

(b)Answer:

6.75

cm[2]

AB = 6 + 2

\phantom{AB} = 8

\dfrac{DE}{BC} = \dfrac{AD}{AB}

\dfrac{DE}{9} = \dfrac{6}{8}

DE = \dfrac{6}{8} \times 9

\phantom{DE} = 6.75

(c)Answer:

3

cm[2]

\dfrac{AE}{AC} = \dfrac{AD}{AB}

AE = \dfrac{6}{8} \times 12

\phantom{AE} = 9

EC = 12 - 9

\phantom{EC} = 3

Question 8

(a)[3]

AB = BC

(sides of a square)

\angle ABE = 90^\circ

and

\angle BCF = 90^\circ

(angles of a square)

BE = CF

(given)

\triangle ABE \equiv \triangle BCF

(SAS)

(b)Answer:

13

cm[2]

AE^2 = AB^2 + BE^2

(Pythagoras' theorem)

AE^2 = 12^2 + 5^2

\phantom{AE^2} = 169

AE = \sqrt{169}

\phantom{AE} = 13

(c)Answer:

13

cm[1]

BF = AE

\phantom{BF} = 13

cm (corr. sides of congruent △s)

(d)Answer:

22.6^\circ

°[2]

\tan \angle BAE = \dfrac{BE}{AB}

\phantom{\tan \angle BAE} = \dfrac{5}{12}

\angle BAE = 22.6198\ldots^\circ

\angle BAE = 22.6^\circ

(e)Answer:

90^\circ

°[2]

\angle CBF = \angle BAE

(corr. ∠s of congruent △s)

\angle ABG = 90^\circ - \angle CBF

\angle ABG = 90^\circ - \angle BAE

\angle AGB = 180^\circ - \angle BAE - (90^\circ - \angle BAE)

(∠ sum of △)

\angle AGB = 90^\circ

Practice questions by skill

Sec 3 E-Math Congruence and similarity worksheet with answers

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

Need the formulas first? Read the Congruence and similarity notes. Or practise it online with step-by-step help when you get one wrong.

Section B

A map is drawn to a scale of

1 : 50\,000

(a)Two towns are

12.4

cm apart on the map. Find the actual distance between them, in kilometres.

Answer (a)km[1]

(b)A road is

4.5

km long. Find its length on the map, in centimetres.

Answer (b)cm[1]

(c)On a second map, the two towns in part (a) are

5

cm apart. Find the scale of the second map in the form

1 : n

Answer (c)[2]

Answers and worked solutions

Question 1

Answer: SSS[2]

\mathstrut \dfrac{PR}{AB} = \dfrac{15}{10}

\mathstrut \dfrac{RQ}{BC} = \dfrac{10.5}{7}

\mathstrut \dfrac{QP}{CA} = \dfrac{12}{8}

All three pairs of sides are in the same ratio,

\dfrac{3}{2}

: SSS.

Question 2

Answer:

7.5\text{ cm} : 37.5\text{ m}

[2]

\mathstrut 1 : 500

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

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

\mathstrut 7.5\text{ cm} : 7.5 \times 5\text{ m}

7.5\text{ cm} : 37.5\text{ m}

Question 3

Answer:

x = 9

cm[2]

\triangle ABC \equiv \triangle QPR

, so

AB

matches

QP

QP = AB

x = 9

Question 4

Answer:

x = 97

°[2]

\mathstrut \angle C = 180^\circ - 48^\circ - 35^\circ

\angle C = 97^\circ

\triangle ABC \equiv \triangle PRQ

, so

\angle C

matches

\angle Q

x = 97

Question 5

Answer:

\phantom{DC} = 11.7

cm (3 s.f.)[3]

\angle XAB = \angle XDC

(alternate angles)

\angle AXB = \angle DXC

(vertically opposite angles)

\dfrac{DC}{AB} = \dfrac{XD}{XA}

\mathstrut \dfrac{DC}{7} = \dfrac{10}{6}

DC = \dfrac{10}{6} \times 7

DC = 11.6666\ldots

\phantom{DC} = 11.7

cm (3 s.f.)

Question 6

(a)Answer:

6.2

km[1]

1 : 50\,000

1 \text{ cm} : 50\,000 \text{ cm}

1 \text{ cm} : 0.5 \text{ km}

12.4 \text{ cm} : 6.2 \text{ km}

(b)Answer:

9

cm[1]

1 \text{ cm} : 0.5 \text{ km}

\dfrac{4.5}{0.5} \text{ cm} : 4.5 \text{ km}

9 \text{ cm} : 4.5 \text{ km}

(c)Answer:

1 : 124\,000

[2]

5 \text{ cm} : 6.2 \text{ km}

5 \text{ cm} : 620\,000 \text{ cm}

5 : 620\,000

1 : 124\,000

Question 7

(a)Answer:

ADE

[1]

\angle ADE = \angle ABC

(corresponding angles)

\angle A

is common

\triangle ABC

is similar to

\triangle ADE

(AA)

(b)Answer:

6.75

cm[2]

AB = 6 + 2

\phantom{AB} = 8

\dfrac{DE}{BC} = \dfrac{AD}{AB}

\dfrac{DE}{9} = \dfrac{6}{8}

DE = \dfrac{6}{8} \times 9

\phantom{DE} = 6.75

(c)Answer:

3

cm[2]

\dfrac{AE}{AC} = \dfrac{AD}{AB}

AE = \dfrac{6}{8} \times 12

\phantom{AE} = 9

EC = 12 - 9

\phantom{EC} = 3

Question 8

(a)[3]

AB = BC

(sides of a square)

\angle ABE = 90^\circ

and

\angle BCF = 90^\circ

(angles of a square)

BE = CF

(given)

\triangle ABE \equiv \triangle BCF

(SAS)

(b)Answer:

13

cm[2]

AE^2 = AB^2 + BE^2

(Pythagoras' theorem)

AE^2 = 12^2 + 5^2

\phantom{AE^2} = 169

AE = \sqrt{169}

\phantom{AE} = 13

(c)Answer:

13

cm[1]

BF = AE

\phantom{BF} = 13

cm (corr. sides of congruent △s)

(d)Answer:

22.6^\circ

°[2]

\tan \angle BAE = \dfrac{BE}{AB}

\phantom{\tan \angle BAE} = \dfrac{5}{12}

\angle BAE = 22.6198\ldots^\circ

\angle BAE = 22.6^\circ

(e)Answer:

90^\circ

°[2]

\angle CBF = \angle BAE

(corr. ∠s of congruent △s)

\angle ABG = 90^\circ - \angle CBF

\angle ABG = 90^\circ - \angle BAE

\angle AGB = 180^\circ - \angle BAE - (90^\circ - \angle BAE)

(∠ sum of △)

\angle AGB = 90^\circ