Sec 2 Maths Pythagoras' theorem worksheet with answers

A free Sec 2 Maths worksheet on pythagoras' theorem: 7 questions, 31 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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

Need the formulas first? Read the Pythagoras' theorem notes. Or practise it online with step-by-step help when you get one wrong.

Try a question: Pythagoras' theorem

Free · no sign-up

Answer a real practice question. Get it wrong and MugCorner shows you where it went wrong, with hints and a step-by-step guide.

MugCornerT-Leng Tuition · mugcornersg.com

Sec 2 Maths worksheet: Pythagoras' theorem

Total: 31 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 triangle

PQR

PQ = 15

cm,

QR = x

cm and

PR = y

cm, where

QR

is the longest side. It is given that

(x + y)(x - y) = 225

. Determine whether triangle

PQR

is right-angled. Choose the answer with its reason.

Answer[2]

In the diagram,

\angle KLM = 90^\circ

. Find

x

Answerm[3]

In the diagram,

\angle ABC = 90^\circ

and

D

lies on

BC

AD = 13

cm,

BD = 5

cm and

DC = 4

cm. Find

x

Answercm[3]

In the diagram,

\angle DEF = 90^\circ

. Find

x

Answercm[3]

Section B

(a)In the diagram,

\angle ABC = 90^\circ

. Find

x

Answer (a)cm[3]

(b)A boat travels

2.5

km due north from

P

Q

, then

4

km due east from

Q

R

. Find the distance

PR

x

km, correct to 3 significant figures.

Answer (b)km[3]

(a)In the diagram,

\angle XYZ = 90^\circ

. Find

x

Answer (a)cm[3]

(b)In the diagram,

\angle DEF = 90^\circ

. Find

x

Answer (b)m[3]

(a)In the diagram,

\angle ABC = 90^\circ

AB = 20

cm and

BC = 21

cm.

D

is the point on

AC

such that

BD

is perpendicular to

AC

. Find the shortest distance from

B

AC

h

cm, correct to 3 significant figures.

Answer (a)cm[3]

(b)(i)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)(i)[3]

(b)(ii)In triangle

LMN

LM = 5

cm,

MN = x

cm and

LN = y

cm, where

MN

is the longest side. It is given that

(x + y)(x - y) = 23

. Determine whether triangle

LMN

is right-angled. Choose the answer with its reason.

Answer (b)(ii)[2]

END OF PAPER

Answers and worked solutions

Question 1

Answer: Yes[2]

(x + y)(x - y) = x^2 - y^2

, so

x^2 - y^2 = 225

15^2 = 225

x^2 - y^2 = 225

\phantom{x^2 - y^2} = 15^2

, so

x^2 = y^2 + 15^2

: triangle

PQR

is right-angled, with the right angle opposite

QR

, at

P

Question 2

Answer: Check:

10.2 < 11

, shorter than the hypotenuse. m[3]

The hypotenuse is

KM

, opposite the right angle at

L

KM^2 = KL^2 + LM^2

x^2 = 11^2 - 4^2

= 121 - 16

= 105

x = \sqrt{105}

= 10.2469\ldots

= 10.2

m (3 s.f.)Check:

10.2 < 11

, shorter than the hypotenuse.

Question 3

Answer:

= 15

cm[3]

AB^2 = 13^2 - 5^2

= 144

AB = \sqrt{144}

= 12

BC = 5 + 4

= 9

x^2 = 12^2 + 9^2

= 225

x = \sqrt{225}

= 15

Question 4

Answer:

= 14.3

cm (3 s.f.)[3]

The hypotenuse is

DF

, opposite the right angle at

E

DF^2 = DE^2 + EF^2

x^2 = 6^2 + 13^2

= 36 + 169

= 205

x = \sqrt{205}

= 14.3178\ldots

= 14.3

cm (3 s.f.)

Question 5

(a)Answer:

= 26

cm[3]

The hypotenuse is

AC

, opposite the right angle at

B

AC^2 = AB^2 + BC^2

x^2 = 24^2 + 10^2

= 576 + 100

= 676

x = \sqrt{676}

= 26

(b)Answer:

= 4.72

km (3 s.f.)[3]

\angle PQR = 90^\circ

, so

PR

is the hypotenuse.

x^2 = 2.5^2 + 4^2

= 22.25

x = \sqrt{22.25}

= 4.7169\ldots

= 4.72

km (3 s.f.)

Question 6

(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:

15 < 25

, shorter than the hypotenuse. m[3]

The hypotenuse is

DF

, opposite the right angle at

E

DF^2 = DE^2 + EF^2

x^2 = 25^2 - 20^2

= 625 - 400

= 225

x = \sqrt{225}

= 15

mCheck:

15 < 25

, shorter than the hypotenuse.

Question 7

(a)Answer:

= 14.5

cm (3 s.f.)[3]

AC^2 = 20^2 + 21^2

= 841

AC = \sqrt{841}

= 29

Area

= \dfrac{1}{2} \times 20 \times 21

= 210

cm²

\dfrac{1}{2} \times 29 \times h = 210

h = \dfrac{2 \times 210}{29}

= 14.4827\ldots

= 14.5

cm (3 s.f.)

(b)(i)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

(b)(ii)Answer: No[2]

(x + y)(x - y) = x^2 - y^2

, so

x^2 - y^2 = 23

5^2 = 25

x^2 - y^2 = 23

, but

5^2 = 25

, so

x^2 \neq y^2 + 5^2

: triangle

LMN

is not right-angled.

Practice questions by skill

Sec 2 Maths Pythagoras' theorem worksheet with answers

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

Need the formulas first? Read the Pythagoras' theorem notes. Or practise it online with step-by-step help when you get one wrong.

Section B

(a)In the diagram,

\angle ABC = 90^\circ

. Find

x

Answer (a)cm[3]

(b)A boat travels

2.5

km due north from

P

Q

, then

4

km due east from

Q

R

. Find the distance

PR

x

km, correct to 3 significant figures.

Answer (b)km[3]

Answers and worked solutions

Question 1

Answer: Yes[2]

(x + y)(x - y) = x^2 - y^2

, so

x^2 - y^2 = 225

15^2 = 225

x^2 - y^2 = 225

\phantom{x^2 - y^2} = 15^2

, so

x^2 = y^2 + 15^2

: triangle

PQR

is right-angled, with the right angle opposite

QR

, at

P

Question 2

Answer: Check:

10.2 < 11

, shorter than the hypotenuse. m[3]

The hypotenuse is

KM

, opposite the right angle at

L

KM^2 = KL^2 + LM^2

x^2 = 11^2 - 4^2

= 121 - 16

= 105

x = \sqrt{105}

= 10.2469\ldots

= 10.2

m (3 s.f.)Check:

10.2 < 11

, shorter than the hypotenuse.

Question 3

Answer:

= 15

cm[3]

AB^2 = 13^2 - 5^2

= 144

AB = \sqrt{144}

= 12

BC = 5 + 4

= 9

x^2 = 12^2 + 9^2

= 225

x = \sqrt{225}

= 15

Question 4

Answer:

= 14.3

cm (3 s.f.)[3]

The hypotenuse is

DF

, opposite the right angle at

E

DF^2 = DE^2 + EF^2

x^2 = 6^2 + 13^2

= 36 + 169

= 205

x = \sqrt{205}

= 14.3178\ldots

= 14.3

cm (3 s.f.)

Question 5

(a)Answer:

= 26

cm[3]

The hypotenuse is

AC

, opposite the right angle at

B

AC^2 = AB^2 + BC^2

x^2 = 24^2 + 10^2

= 576 + 100

= 676

x = \sqrt{676}

= 26

(b)Answer:

= 4.72

km (3 s.f.)[3]

\angle PQR = 90^\circ

, so

PR

is the hypotenuse.

x^2 = 2.5^2 + 4^2

= 22.25

x = \sqrt{22.25}

= 4.7169\ldots

= 4.72

km (3 s.f.)

Question 6

(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:

15 < 25

, shorter than the hypotenuse. m[3]

The hypotenuse is

DF

, opposite the right angle at

E

DF^2 = DE^2 + EF^2

x^2 = 25^2 - 20^2

= 625 - 400

= 225

x = \sqrt{225}

= 15

mCheck:

15 < 25

, shorter than the hypotenuse.

Question 7

(a)Answer:

= 14.5

cm (3 s.f.)[3]

AC^2 = 20^2 + 21^2

= 841

AC = \sqrt{841}

= 29

Area

= \dfrac{1}{2} \times 20 \times 21

= 210

cm²

\dfrac{1}{2} \times 29 \times h = 210

h = \dfrac{2 \times 210}{29}

= 14.4827\ldots

= 14.5

cm (3 s.f.)

(b)(i)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

(b)(ii)Answer: No[2]

(x + y)(x - y) = x^2 - y^2

, so

x^2 - y^2 = 23

5^2 = 25

x^2 - y^2 = 23

, but

5^2 = 25

, so

x^2 \neq y^2 + 5^2

: triangle

LMN

is not right-angled.