Sec 2 G2 Maths Congruence and similarity worksheet with answers

A free Sec 2 G2 Maths 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 2 G2 Maths worksheet: Congruence and similarity

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

In the diagram, triangle

ABC

is similar to triangle

PRQ

. Find

x

Answercm[2]

In the diagram, triangle

ABC

is congruent to triangle

RQP

. Find

x

Answer°[2]

In the diagram, the perpendicular bisector of

AB

has been constructed with compasses. It meets

AB

M

, and

P

is a point on it. Find

x

Answer°[2]

On the squared paper, shape

B

is an enlargement of shape

A

. Find the scale factor of the enlargement.

Answer[2]

Triangles

ABC

and

PQR

are congruent, with the equal parts marked. Complete the statement

\triangle ABC \equiv \triangle \ldots

Answer[2]

Section B

(a)In the diagram, triangle

ABC

is congruent to triangle

QPR

. Find

x

Answer (a)°[2]

(b)In the diagram,

D

lies on

AB

and

E

lies on

AC

, with

DE

parallel to

BC

. Find

DB

Answer (b)cm[3]

(a)On the squared paper, shape

B

is an enlargement of shape

A

. Find the scale factor of the enlargement.

Answer (a)[2]

(b)In the diagram, the perpendicular bisector of

AB

has been constructed with compasses. It meets

AB

M

, and

P

is a point on it.

AB = 8

cm. Find the perimeter of triangle

PAB

Answer (b)cm[2]

(c)In the diagram,

D

lies on

AB

and

E

lies on

AC

, with

DE

parallel to

BC

. Find

DE

Answer (c)cm[3]

(a)In the diagram, triangle

ABC

is similar to triangle

QPR

. Find

x

Answer (a)cm[2]

(b)Ali is constructing the bisector of

\angle ABC

with a pair of compasses and a ruler. The diagram shows the construction so far. State the next step.

Answer (b)[2]

(c)On the squared paper, shape

B

is an enlargement of shape

A

. Find the scale factor of the enlargement.

Answer (c)[2]

(d)In the diagram, triangle

ABC

is congruent to triangle

PRQ

. Find

x

Answer (d)°[2]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

x = 10

cm[2]

CA

matches

QP

and

AB

matches

PR

\dfrac{AB}{PR} = \dfrac{CA}{QP}

\mathstrut \dfrac{x}{15} = \dfrac{9}{13.5}

x = \dfrac{9}{13.5} \times 15

x = 10

Question 2

Answer:

x = 88

°[2]

\mathstrut \angle C = 180^\circ - 57^\circ - 35^\circ

\angle C = 88^\circ

\triangle ABC \equiv \triangle RQP

, so

\angle C

matches

\angle P

x = 88

Question 3

Answer:

x = 52

(base ∠s of isos. △) °[2]

PA = PB

(

P

is on the perpendicular bisector of

AB

)So triangle

PAB

is isosceles.

x = 52

(base ∠s of isos. △)

Question 4

Answer: Scale factor

= 2

[2]

Bottom side of shape

A

6

units; of shape

B

12

unitsScale factor

= \dfrac{12}{6}

Scale factor

= 2

Question 5

Answer: QRP[2]

AB

(

11

cm) matches

\text{QR}

CA

(

8

cm) matches

\text{PQ}

\angle A

(

58^\circ

) matches

\angle \text{Q}

A

matches

Q

B

matches

R

C

matches

P

\triangle ABC \equiv \triangle \text{QRP}

Question 6

(a)Answer:

x = 97

°[2]

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

\angle C = 97^\circ

\triangle ABC \equiv \triangle QPR

, so

\angle C

matches

\angle R

x = 97

(b)Answer:

DB = 4

cm[3]

\angle ADE = \angle ABC

(corresponding angles)

\angle A

is common, so the triangles are similar (AA).

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

\mathstrut \dfrac{AB}{6} = \dfrac{10}{6}

AB = \dfrac{10}{6} \times 6

AB = 10

\mathstrut DB = 10 - 6

DB = 4

Question 7

(a)Answer: Scale factor

= \dfrac{1}{2}

(less than 1: the image is smaller)[2]

Bottom side of shape

A

6

units; of shape

B

3

unitsScale factor

= \dfrac{3}{6}

Scale factor

= \dfrac{1}{2}

(less than 1: the image is smaller)

(b)Answer:

\text{Perimeter} = 20

cm[2]

PB = PA

(

P

is on the perpendicular bisector of

AB

)

PB = 6

\text{Perimeter} = 6 + 6 + 8

\text{Perimeter} = 20

(c)Answer:

DE = 4.8

cm[3]

\angle ADE = \angle ABC

(corresponding angles)

\angle A

is common, so the triangles are similar (AA).

\mathstrut AB = 6 + 4

AB = 10

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

\mathstrut \dfrac{DE}{8} = \dfrac{6}{10}

DE = \dfrac{6}{10} \times 8

DE = 4.8

Question 8

(a)Answer:

x = 8

cm[2]

AB

matches

QP

and

BC

matches

PR

\dfrac{PR}{BC} = \dfrac{QP}{AB}

\mathstrut \dfrac{x}{4} = \dfrac{12}{6}

x = \dfrac{12}{6} \times 4

x = 8

(b)Answer: arcs from X and Y[2]

The arc about

B

has cut the arms at

X

and

Y

.1. With centre

B

, draw an arc that cuts

BA

X

and

BC

Y

. (done)2. With centres

X

and

Y

and the same radius, draw two arcs that meet at

Z

. (next)3. Join

B

and

Z

with a straight line.

(c)Answer: Scale factor

= 2

[2]

Bottom side of shape

A

4

units; of shape

B

8

unitsScale factor

= \dfrac{8}{4}

Scale factor

= 2

(d)Answer:

x = 79

°[2]

\mathstrut \angle C = 180^\circ - 57^\circ - 44^\circ

\angle C = 79^\circ

\triangle ABC \equiv \triangle PRQ

, so

\angle C

matches

\angle Q

x = 79

Sec 2 G2 Maths Congruence and similarity 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.

Answers and worked solutions

Question 1

Answer:

x = 10

cm[2]

CA

matches

QP

and

AB

matches

PR

\dfrac{AB}{PR} = \dfrac{CA}{QP}

\mathstrut \dfrac{x}{15} = \dfrac{9}{13.5}

x = \dfrac{9}{13.5} \times 15

x = 10

Question 2

Answer:

x = 88

°[2]

\mathstrut \angle C = 180^\circ - 57^\circ - 35^\circ

\angle C = 88^\circ

\triangle ABC \equiv \triangle RQP

, so

\angle C

matches

\angle P

x = 88

Question 3

Answer:

x = 52

(base ∠s of isos. △) °[2]

PA = PB

(

P

is on the perpendicular bisector of

AB

)So triangle

PAB

is isosceles.

x = 52

(base ∠s of isos. △)

Question 4

Answer: Scale factor

= 2

[2]

Bottom side of shape

A

6

units; of shape

B

12

unitsScale factor

= \dfrac{12}{6}

Scale factor

= 2

Question 5

Answer: QRP[2]

AB

(

11

cm) matches

\text{QR}

CA

(

8

cm) matches

\text{PQ}

\angle A

(

58^\circ

) matches

\angle \text{Q}

A

matches

Q

B

matches

R

C

matches

P

\triangle ABC \equiv \triangle \text{QRP}

Question 6

(a)Answer:

x = 97

°[2]

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

\angle C = 97^\circ

\triangle ABC \equiv \triangle QPR

, so

\angle C

matches

\angle R

x = 97

(b)Answer:

DB = 4

cm[3]

\angle ADE = \angle ABC

(corresponding angles)

\angle A

is common, so the triangles are similar (AA).

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

\mathstrut \dfrac{AB}{6} = \dfrac{10}{6}

AB = \dfrac{10}{6} \times 6

AB = 10

\mathstrut DB = 10 - 6

DB = 4

Question 7

(a)Answer: Scale factor

= \dfrac{1}{2}

(less than 1: the image is smaller)[2]

Bottom side of shape

A

6

units; of shape

B

3

unitsScale factor

= \dfrac{3}{6}

Scale factor

= \dfrac{1}{2}

(less than 1: the image is smaller)

(b)Answer:

\text{Perimeter} = 20

cm[2]

PB = PA

(

P

is on the perpendicular bisector of

AB

)

PB = 6

\text{Perimeter} = 6 + 6 + 8

\text{Perimeter} = 20

(c)Answer:

DE = 4.8

cm[3]

\angle ADE = \angle ABC

(corresponding angles)

\angle A

is common, so the triangles are similar (AA).

\mathstrut AB = 6 + 4

AB = 10

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

\mathstrut \dfrac{DE}{8} = \dfrac{6}{10}

DE = \dfrac{6}{10} \times 8

DE = 4.8

Question 8

(a)Answer:

x = 8

cm[2]

AB

matches

QP

and

BC

matches

PR

\dfrac{PR}{BC} = \dfrac{QP}{AB}

\mathstrut \dfrac{x}{4} = \dfrac{12}{6}

x = \dfrac{12}{6} \times 4

x = 8

(b)Answer: arcs from X and Y[2]

The arc about

B

has cut the arms at

X

and

Y

.1. With centre

B

, draw an arc that cuts

BA

X

and

BC

Y

. (done)2. With centres

X

and

Y

and the same radius, draw two arcs that meet at

Z

. (next)3. Join

B

and

Z

with a straight line.

(c)Answer: Scale factor

= 2

[2]

Bottom side of shape

A

4

units; of shape

B

8

unitsScale factor

= \dfrac{8}{4}

Scale factor

= 2

(d)Answer:

x = 79

°[2]

\mathstrut \angle C = 180^\circ - 57^\circ - 44^\circ

\angle C = 79^\circ

\triangle ABC \equiv \triangle PRQ

, so

\angle C

matches

\angle Q

x = 79