Sec 4 G1 Maths Pythagoras' theorem worksheet with answers

A free Sec 4 G1 Maths worksheet on pythagoras' theorem: 7 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 4 G1 Maths worksheet: Pythagoras' theorem

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

A ship travels

2.5

km due north from

P

Q

, then

12

km due east from

Q

R

. Find the distance

PR

x

km, correct to 3 significant figures.

Answerkm[3]

In the diagram,

\angle KLM = 90^\circ

. Find

x

Answerm[3]

In the diagram,

\angle LMN = 90^\circ

. Find

x

Answercm[3]

The diagram shows a rectangular door

ABCD

21

cm long and

10

cm wide. Find the length of the diagonal

AC

x

cm, correct to 3 significant figures.

Answercm[3]

(a)In triangle

PQR

PQ = 1.5

cm,

QR = 0.9

cm and

PR = 1.2

cm. Use the converse of Pythagoras' theorem to find which angle of triangle

PQR

is a right angle.

Answer (a)[3]

(b)In triangle

XYZ

XY = 12

cm,

YZ = 12.5

cm and

XZ = 3.5

cm. Use the converse of Pythagoras' theorem to find which angle of triangle

XYZ

is a right angle.

Answer (b)[3]

Section B

(a)A hiker travels

16

km due west from

P

Q

, then

6

km due north from

Q

R

. Find the distance

PR

x

km, correct to 3 significant figures.

Answer (a)km[3]

(b)In the diagram,

\angle XYZ = 90^\circ

. Find

x

Answer (b)cm[3]

(a)In the diagram,

\angle XYZ = 90^\circ

. Find

x

Answer (a)cm[3]

(b)In the diagram,

\angle LMN = 90^\circ

. Find

x

Answer (b)m[3]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

= 12.3

km (3 s.f.)[3]

\angle PQR = 90^\circ

, so

PR

is the hypotenuse.

x^2 = 2.5^2 + 12^2

= 150.25

x = \sqrt{150.25}

= 12.2576\ldots

= 12.3

km (3 s.f.)

Question 2

Answer:

= 37

m[3]

The hypotenuse is

KM

, opposite the right angle at

L

KM^2 = KL^2 + LM^2

x^2 = 12^2 + 35^2

= 144 + 1225

= 1369

x = \sqrt{1369}

= 37

Question 3

Answer: Check:

11.1 < 22

, shorter than the hypotenuse. cm[3]

The hypotenuse is

LN

, opposite the right angle at

M

LN^2 = LM^2 + MN^2

x^2 = 22^2 - 19^2

= 484 - 361

= 123

x = \sqrt{123}

= 11.0905\ldots

= 11.1

cm (3 s.f.)Check:

11.1 < 22

, shorter than the hypotenuse.

Question 4

Answer:

= 23.3

cm (3 s.f.)[3]

In triangle

ABC

\angle ABC = 90^\circ

, so the hypotenuse is

AC

x^2 = 21^2 + 10^2

= 541

x = \sqrt{541}

= 23.2594\ldots

= 23.3

cm (3 s.f.)

Question 5

(a)Answer: ∠PRQ[3]

PQ^2 = 1.5^2

= 2.25

QR^2 = 0.9^2

= 0.81

PR^2 = 1.2^2

= 1.44

The longest side is

PQ

0.81 + 1.44 = 2.25

PQ^2 = QR^2 + PR^2

, so by the converse of Pythagoras' theorem the angle opposite

PQ

90^\circ

\angle PRQ = 90^\circ

(b)Answer: ∠YXZ[3]

XY^2 = 12^2

= 144

YZ^2 = 12.5^2

= 156.25

XZ^2 = 3.5^2

= 12.25

The longest side is

YZ

12.25 + 144 = 156.25

YZ^2 = XZ^2 + XY^2

, so by the converse of Pythagoras' theorem the angle opposite

YZ

90^\circ

\angle YXZ = 90^\circ

Question 6

(a)Answer:

= 17.1

km (3 s.f.)[3]

\angle PQR = 90^\circ

, so

PR

is the hypotenuse.

x^2 = 16^2 + 6^2

= 292

x = \sqrt{292}

= 17.0880\ldots

= 17.1

km (3 s.f.)

(b)Answer: Check:

15.9 < 17

, shorter than the hypotenuse. cm[3]

The hypotenuse is

XZ

, opposite the right angle at

Y

XZ^2 = XY^2 + YZ^2

x^2 = 17^2 - 6^2

= 289 - 36

= 253

x = \sqrt{253}

= 15.9059\ldots

= 15.9

cm (3 s.f.)Check:

15.9 < 17

, shorter than the hypotenuse.

Question 7

(a)Answer:

= 29

cm[3]

The hypotenuse is

XZ

, opposite the right angle at

Y

XZ^2 = XY^2 + YZ^2

x^2 = 21^2 + 20^2

= 441 + 400

= 841

x = \sqrt{841}

= 29

(b)Answer: Check:

16 < 20

, shorter than the hypotenuse. m[3]

The hypotenuse is

LN

, opposite the right angle at

M

LN^2 = LM^2 + MN^2

x^2 = 20^2 - 12^2

= 400 - 144

= 256

x = \sqrt{256}

= 16

mCheck:

16 < 20

, shorter than the hypotenuse.

Sec 4 G1 Maths Pythagoras' theorem 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.

Section B

(a)A hiker travels

16

km due west from

P

Q

, then

6

km due north from

Q

R

. Find the distance

PR

x

km, correct to 3 significant figures.

Answer (a)km[3]

(b)In the diagram,

\angle XYZ = 90^\circ

. Find

x

Answer (b)cm[3]

Answers and worked solutions

Question 1

Answer:

= 12.3

km (3 s.f.)[3]

\angle PQR = 90^\circ

, so

PR

is the hypotenuse.

x^2 = 2.5^2 + 12^2

= 150.25

x = \sqrt{150.25}

= 12.2576\ldots

= 12.3

km (3 s.f.)

Question 2

Answer:

= 37

m[3]

The hypotenuse is

KM

, opposite the right angle at

L

KM^2 = KL^2 + LM^2

x^2 = 12^2 + 35^2

= 144 + 1225

= 1369

x = \sqrt{1369}

= 37

Question 3

Answer: Check:

11.1 < 22

, shorter than the hypotenuse. cm[3]

The hypotenuse is

LN

, opposite the right angle at

M

LN^2 = LM^2 + MN^2

x^2 = 22^2 - 19^2

= 484 - 361

= 123

x = \sqrt{123}

= 11.0905\ldots

= 11.1

cm (3 s.f.)Check:

11.1 < 22

, shorter than the hypotenuse.

Question 4

Answer:

= 23.3

cm (3 s.f.)[3]

In triangle

ABC

\angle ABC = 90^\circ

, so the hypotenuse is

AC

x^2 = 21^2 + 10^2

= 541

x = \sqrt{541}

= 23.2594\ldots

= 23.3

cm (3 s.f.)

Question 5

(a)Answer: ∠PRQ[3]

PQ^2 = 1.5^2

= 2.25

QR^2 = 0.9^2

= 0.81

PR^2 = 1.2^2

= 1.44

The longest side is

PQ

0.81 + 1.44 = 2.25

PQ^2 = QR^2 + PR^2

, so by the converse of Pythagoras' theorem the angle opposite

PQ

90^\circ

\angle PRQ = 90^\circ

(b)Answer: ∠YXZ[3]

XY^2 = 12^2

= 144

YZ^2 = 12.5^2

= 156.25

XZ^2 = 3.5^2

= 12.25

The longest side is

YZ

12.25 + 144 = 156.25

YZ^2 = XZ^2 + XY^2

, so by the converse of Pythagoras' theorem the angle opposite

YZ

90^\circ

\angle YXZ = 90^\circ

Question 6

(a)Answer:

= 17.1

km (3 s.f.)[3]

\angle PQR = 90^\circ

, so

PR

is the hypotenuse.

x^2 = 16^2 + 6^2

= 292

x = \sqrt{292}

= 17.0880\ldots

= 17.1

km (3 s.f.)

(b)Answer: Check:

15.9 < 17

, shorter than the hypotenuse. cm[3]

The hypotenuse is

XZ

, opposite the right angle at

Y

XZ^2 = XY^2 + YZ^2

x^2 = 17^2 - 6^2

= 289 - 36

= 253

x = \sqrt{253}

= 15.9059\ldots

= 15.9

cm (3 s.f.)Check:

15.9 < 17

, shorter than the hypotenuse.

Question 7

(a)Answer:

= 29

cm[3]

The hypotenuse is

XZ

, opposite the right angle at

Y

XZ^2 = XY^2 + YZ^2

x^2 = 21^2 + 20^2

= 441 + 400

= 841

x = \sqrt{841}

= 29

(b)Answer: Check:

16 < 20

, shorter than the hypotenuse. m[3]

The hypotenuse is

LN

, opposite the right angle at

M

LN^2 = LM^2 + MN^2

x^2 = 20^2 - 12^2

= 400 - 144

= 256

x = \sqrt{256}

= 16

mCheck:

16 < 20

, shorter than the hypotenuse.