Sec 1 Maths Triangles, quadrilaterals and polygons worksheet with answers

A free Sec 1 Maths worksheet on triangles, quadrilaterals and polygons: 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 1 Maths worksheet: Triangles, quadrilaterals and polygons

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 an equilateral triangle. State whether the dashed line is a line of symmetry of the shape.

Answer[1]

Property	Has it?
All four sides are equal.	✗
Both pairs of opposite sides are equal.	✓
All four angles are right angles.	✗

Hui Min describes a quadrilateral. It has every property ticked (✓) in the table, and none of the properties crossed (✗). Write down the special name of the quadrilateral.

Answer[1]

In triangle

ABC

AB = AC

. Find

x

Answer[2]

ABCD

is a rhombus. Find

x

Answer[2]

The diagram shows a regular pentagon

ABCDE

and a square

CDQP

drawn inside it on the side

CD

B

is joined to

P

. Find

x

Answer[3]

Section B

(a)The diagram shows a square. Find the smallest angle through which it can be rotated about its centre to look exactly the same.

Answer (a)°[2]

(b)The diagram shows triangle

LMN

. Find

x

Answer (b)[2]

Property	Has it?
Both pairs of opposite sides are equal.	✓
The diagonals are equal in length.	✗
The diagonals cross at right angles.	✗

(a)Devi describes a quadrilateral. It has every property ticked (✓) in the table, and none of the properties crossed (✗). Write down the special name of the quadrilateral.

Answer (a)[1]

(b)The diagram shows a pinwheel. Find the smallest angle through which it can be rotated about its centre to look exactly the same.

Answer (b)°[2]

(c)(i)State whether a triangle can be constructed with sides of 4 cm, 9 cm and 5 cm.

Answer (c)(i)[1]

(c)(ii)Which set of data fixes exactly one triangle

ABC

Answer (c)(ii)[1]

(a)The diagram shows a quadrilateral

ABCD

. Find

x

Answer (a)[2]

(b)(i)State whether a triangle can be constructed with sides of 2 cm, 7 cm and 3 cm.

Answer (b)(i)[1]

(b)(ii)Construct triangle

ABC

with

AB = 10

cm,

BC = 5

cm and

CA = 9

cm. Measure

\angle ABC

, to the nearest degree.

Answer (b)(ii)°[2]

(a)

ABCD

is a rectangle. Its diagonals meet at

O

AC = 8

cm. Find

OB

Answer (a)cm[2]

(b)The diagram shows a five-pointed star. Find the smallest angle through which it can be rotated about its centre to look exactly the same.

Answer (b)°[2]

(c)(i)State whether a triangle can be constructed with sides of 3 cm, 4 cm and 9 cm.

Answer (c)(i)[1]

(c)(ii)State whether triangle

ABC

can be constructed with

AB = 7

cm,

\angle BAC = 35^\circ

and

\angle ABC = 95^\circ

Answer (c)(ii)[2]

END OF PAPER

Answers and worked solutions

Question 1

Answer: YES[1]

Each corner on one side must land exactly on a corner on the other side, and each side on a side.Yes: the two halves fold exactly onto each other, so the dashed line is a line of symmetry.

Question 2

Answer: parallelogram[1]

✗ Left: rectangle, parallelogram, kite, trapezium, isosceles trapezium.✓ Left: rectangle, parallelogram.✗ Left: parallelogram.It is a parallelogram.

Question 3

Answer:

x = 62

(∠ sum of △)[2]

\angle ACB = x^\circ

(base ∠s of isos. △)

x^\circ + x^\circ + 56^\circ = 180^\circ

2x = 124

x = 62

(∠ sum of △)

Question 4

Answer:

x = 28

(∠ sum of △)[2]

\angle BCA = x^\circ

(base ∠s of isos. △)

x^\circ + x^\circ + 124^\circ = 180^\circ

2x = 56

x = 28

(∠ sum of △)

Question 5

Answer:

\phantom{x} = 81

(base ∠s of isos. △,

CB = CP

)[3]

\angle BCD = \dfrac{(5 - 2) \times 180^\circ}{5}

\phantom{\angle BCD} = 108^\circ

\angle DCP = 90^\circ

(angle of a square)

\angle BCP = 108^\circ - 90^\circ

\phantom{\angle BCP} = 18^\circ

x = \dfrac{180 - 18}{2}

\phantom{x} = 81

(base ∠s of isos. △,

CB = CP

)

Question 6

(a)Answer:

\text{Angle} = 90^\circ

°[2]

It looks the same every quarter turn.

\text{Order} = 4

\text{Angle} = \dfrac{360^\circ}{4}

\text{Angle} = 90^\circ

(b)Answer:

x = 95

(∠ sum of △)[2]

x^\circ + 46^\circ + 39^\circ = 180^\circ

x = 180 - 46 - 39

x = 95

(∠ sum of △)

Question 7

(a)Answer: parallelogram[1]

✓ Left: square, rectangle, rhombus, parallelogram.✗ Left: rhombus, parallelogram.✗ Left: parallelogram.It is a parallelogram.

(b)Answer:

\text{Angle} = 90^\circ

°[2]

It looks the same every quarter turn.

\text{Order} = 4

\text{Angle} = \dfrac{360^\circ}{4}

\text{Angle} = 90^\circ

(c)(i)Answer: NO[1]

4 + 5 = 9

9

is equal to

9

.No: the two shorter sides only reach along the longest one, so the triangle is flat.

(c)(ii)Answer: B[1]

A: 3 + 4 is not more than 9, so the sides cannot meet.C: three angles and no length. Triangles of any size have these angles.D:

\angle BAC

is not between

AB

and

BC

. The arc about

B

can cut the ray from

A

twice: two different triangles.B: two sides and the angle between them fix one triangle.

Question 8

(a)Answer:

x = 52

[2]

(x + 10)^\circ + 2x^\circ + 80^\circ + (2x + 10)^\circ = 360^\circ

(∠ sum of quad.)

5x + 100 = 360

5x = 260

x = 52

(b)(i)Answer: NO[1]

2 + 3 = 5

5

is less than

7

.No: the arcs about the two ends never meet.

(b)(ii)Answer:

\angle ABC = 64^\circ

(to the nearest degree) °[2]

Draw

AB = 10

cm.With centre

A

and radius

AC = 9

cm, draw an arc.With centre

B

and radius

BC = 5

cm, draw an arc to cut the first one at

C

. Join

AC

and

BC

\angle ABC = 64^\circ

(to the nearest degree)

Question 9

(a)Answer:

OB = 4

cm[2]

The diagonals of a rectangle are equal and bisect each other.

OA = OB

\phantom{OA} = OC

\phantom{OA} = OD

OB = \dfrac{8}{2}

OB = 4

(b)Answer:

\text{Angle} = 72^\circ

°[2]

It looks the same every fifth of a turn.

\text{Order} = 5

\text{Angle} = \dfrac{360^\circ}{5}

\text{Angle} = 72^\circ

(c)(i)Answer: NO[1]

3 + 4 = 7

7

is less than

9

.No: the arcs about the two ends never meet.

(c)(ii)Answer: YES[2]

35 + 95 = 130

130^\circ

is less than

180^\circ

.Yes: the third angle is

50^\circ

and the rays from

A

and

B

cross.

Practice questions by skill

Sec 1 Maths Triangles, quadrilaterals and polygons worksheet with answers

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

Property

Has it?

All four sides are equal.

✗

Both pairs of opposite sides are equal.

✓

All four angles are right angles.

✗

Property

Has it?

Both pairs of opposite sides are equal.

✓

The diagonals are equal in length.

✗

The diagonals cross at right angles.

✗

Answers and worked solutions

Question 1

Answer: YES[1]

Each corner on one side must land exactly on a corner on the other side, and each side on a side.Yes: the two halves fold exactly onto each other, so the dashed line is a line of symmetry.

Question 2

Answer: parallelogram[1]

✗ Left: rectangle, parallelogram, kite, trapezium, isosceles trapezium.✓ Left: rectangle, parallelogram.✗ Left: parallelogram.It is a parallelogram.

Question 3

Answer:

x = 62

(∠ sum of △)[2]

\angle ACB = x^\circ

(base ∠s of isos. △)

x^\circ + x^\circ + 56^\circ = 180^\circ

2x = 124

x = 62

(∠ sum of △)

Question 4

Answer:

x = 28

(∠ sum of △)[2]

\angle BCA = x^\circ

(base ∠s of isos. △)

x^\circ + x^\circ + 124^\circ = 180^\circ

2x = 56

x = 28

(∠ sum of △)

Question 5

Answer:

\phantom{x} = 81

(base ∠s of isos. △,

CB = CP

)[3]

\angle BCD = \dfrac{(5 - 2) \times 180^\circ}{5}

\phantom{\angle BCD} = 108^\circ

\angle DCP = 90^\circ

(angle of a square)

\angle BCP = 108^\circ - 90^\circ

\phantom{\angle BCP} = 18^\circ

x = \dfrac{180 - 18}{2}

\phantom{x} = 81

(base ∠s of isos. △,

CB = CP

)

Question 6

(a)Answer:

\text{Angle} = 90^\circ

°[2]

It looks the same every quarter turn.

\text{Order} = 4

\text{Angle} = \dfrac{360^\circ}{4}

\text{Angle} = 90^\circ

(b)Answer:

x = 95

(∠ sum of △)[2]

x^\circ + 46^\circ + 39^\circ = 180^\circ

x = 180 - 46 - 39

x = 95

(∠ sum of △)

Question 7

(a)Answer: parallelogram[1]

✓ Left: square, rectangle, rhombus, parallelogram.✗ Left: rhombus, parallelogram.✗ Left: parallelogram.It is a parallelogram.

(b)Answer:

\text{Angle} = 90^\circ

°[2]

It looks the same every quarter turn.

\text{Order} = 4

\text{Angle} = \dfrac{360^\circ}{4}

\text{Angle} = 90^\circ

(c)(i)Answer: NO[1]

4 + 5 = 9

9

is equal to

9

.No: the two shorter sides only reach along the longest one, so the triangle is flat.

(c)(ii)Answer: B[1]

A: 3 + 4 is not more than 9, so the sides cannot meet.C: three angles and no length. Triangles of any size have these angles.D:

\angle BAC

is not between

AB

and

BC

. The arc about

B

can cut the ray from

A

twice: two different triangles.B: two sides and the angle between them fix one triangle.

Question 8

(a)Answer:

x = 52

[2]

(x + 10)^\circ + 2x^\circ + 80^\circ + (2x + 10)^\circ = 360^\circ

(∠ sum of quad.)

5x + 100 = 360

5x = 260

x = 52

(b)(i)Answer: NO[1]

2 + 3 = 5

5

is less than

7

.No: the arcs about the two ends never meet.

(b)(ii)Answer:

\angle ABC = 64^\circ

(to the nearest degree) °[2]

Draw

AB = 10

cm.With centre

A

and radius

AC = 9

cm, draw an arc.With centre

B

and radius

BC = 5

cm, draw an arc to cut the first one at

C

. Join

AC

and

BC

\angle ABC = 64^\circ

(to the nearest degree)

Question 9

(a)Answer:

OB = 4

cm[2]

The diagonals of a rectangle are equal and bisect each other.

OA = OB

\phantom{OA} = OC

\phantom{OA} = OD

OB = \dfrac{8}{2}

OB = 4

(b)Answer:

\text{Angle} = 72^\circ

°[2]

It looks the same every fifth of a turn.

\text{Order} = 5

\text{Angle} = \dfrac{360^\circ}{5}

\text{Angle} = 72^\circ

(c)(i)Answer: NO[1]

3 + 4 = 7

7

is less than

9

.No: the arcs about the two ends never meet.

(c)(ii)Answer: YES[2]

35 + 95 = 130

130^\circ

is less than

180^\circ

.Yes: the third angle is

50^\circ

and the rays from

A

and

B

cross.