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© State of New South Wales, Department of Education and Training 2004

Mathematics Stage 4

MS4.1 Perimeter and area

Part 2 Perimeter and Pythagoras

Part 2 Perimeter and Pythagoras 1

Contents – Part 2

Introduction – Part 2..........................................................3

Indicators ...................................................................................3

Preliminary quiz – Part 2 ...................................................5

Perimeters.........................................................................7

Perimeter of composite shapes.......................................11

Right-angled triangles .....................................................15

Pythagoras’s theorem .....................................................19

Pythagorean triads ..........................................................23

Suggested answers – Part 2 ...........................................27

Exercises – Part 2 ...........................................................29

Review quiz – Part 2 .......................................................43

Answers to exercises – Part 2.........................................47

2 MS4.1 Perimeter and area


Introduction – Part 2

This part deals with two concepts: perimeter and Pythagoras’s theorem.

Perimeters of plane shapes were introduced in Stages 2 and 3.

The first half of this part reviews those concepts and extends them to

calculating the perimeters of simple composite figures.

Pythagoras’s theorem is introduced as a notion that the three sides of a

right-angled triangle are linked by the relationship that the sum of the

squares of the two smaller sides equals the square of the largest side,

known as the hypotenuse.

Both these concepts will be extended and used in later parts.

Indicators

By the end of Part 2, you will have been given the opportunity to work

towards aspects of knowledge and skills including:

• finding the perimeter of simple composite figures

• identifying the hypotenuse as the longest side in any right-angled

triangle and also as the side opposite the right angle

• establishing the relationship between the lengths of the sides of a

right-angled triangle in practical ways, including the dissection of

areas

• identifying a Pythagorean triad as a set of three numbers such that

the sum of the squares of the first two equals the square of the third.


By the end of Part 2, you will have been given the opportunity to work

mathematically by:

• selecting and using appropriate devices to measure lengths and

distances

• describing the relationship between the sides of a right-angled

triangle.


Preliminary quiz – Part 2

Before you start this part, use this preliminary quiz to revise some skills

you will need.

Activity – Preliminary quiz

Try these.

1 Write out the full words for these abbreviations.

a mm _______________________________________________

b cm ________________________________________________

c m _________________________________________________

d km ________________________________________________

2 Some of these units measure length and some measure area.

km2, cm, ha, mm2, m, cm2, m2, km.

a Which units measure length? ___________________________

b Which units measure area? _____________________________

3 Convert:

a 32 mm to cm ________________________________________

b 48 cm to mm ________________________________________

c 2.5 m to cm _________________________________________

4 Use a ruler to measure the length of this line. Give answers both in

centimetres and millimetres.


5 A square has a side length of 15 cm. What is its perimeter? ______

6 A rectangle has length of 16.8 cm, and a width of 12.1 cm.

Calculate its perimeter. ___________________________________

7 Measure the perimeter of these plane shapes, in millimetres.

a ______________________ b _______________________

8 Use your calculator to calculate these correct to two decimal places.

a 6.22 _______________________________________________

b 7.252 ______________________________________________

c 75 ______________________________________________

d 31 ______________________________________________

Check your response by going to the suggested answers section.


Perimeters

A plane figure is any flat shape.

The perimeter of a plane figure is the sum of the lengths of the sides.

In other words, it is the length of the rim, or the distance around the edge

of the figure.

Perimeter comes from theGreek word .

means ‘around’, andmetros means ‘measure’.

Follow through the steps in this example. Do your own working in the

margin if you wish.

Determine the perimeter of this triangle by measurement.

A

B

C

Solution

Carefully measure each of the three sides. Length AB is

25 mm, length BC is 34 mm, and length AC is 55 mm.

∴ Perimeter = 25 + 34 + 55 = 114 mm

You could have measured the perimeter in centimetres and given the result

as 11.4 cm.


Activity – Perimeters

Try this.

1 Use a ruler to measure the perimeter of this plane shape.

_______________________________________________________

_______________________________________________________


Sometimes plane figures are not drawn to scale. In these cases they

would have measurements written on them.


margin if you wish.

Determine the perimeter of this quadrilateral.

Picture is not drawn to scale.

6.8

cm

19.1 cm

14.3 cm


Solution

Two of the sides are the same length. Sides that are the same

length are shown with one ( | ) or two ( || ) or more short

parallel markings. The perimeter of this plane shape isperimeter = 6.8 + 6.8 + 14.3 + 19.1 = 47.0 cm


Try this.

2 Determine the perimeter of this plane shape.

/

\

==

5.3 m

1.6 m

_______________________________________________________

_______________________________________________________


Sometimes when lengths are the same. They will be marked.

Other times you might know that certain shapes form squares, rectangles,

or isosceles triangles. You know that these shapes have some of their

sides the same. Use this information to make your task of calculating the

perimeter easier.

You should now practise what you have learned by doing the exercises.

Go to the exercises section and complete Exercise 2.1 – Perimeters.



Perimeter of composite shapes

A composite shape is one that is not a simple, one-figure shape.

It can be made up of two or more shapes.

The shape below represents the ground plan of

a large building. Can you see that this shape is

made up of three rectangles?

All angles are right angles and all measurements are in metres.

12

9

21

13 6

16

83

Suppose you needed to calculate its perimeter. You will notice that the

lengths of two sections are not given.

To find the perimeter of the figure, you must first find the unmarked

sections by using the fact that the opposite sides of a rectangle are equal.

21 m

overall length (21 + 8 = 29)

8 m

13 m ? m

(13 + ? = 29)

? m

(? +

3 =

15)

6 m

9 m

3 m


Here is one way to help you find the missing lengths.

Overall length = 21 + 8

= 29 m 21 m

8 m

Step 1

To find the side marked ‘?’

13 + ? = 29

13 + 16 = 29

Missing length = 16 m

Or you can see by inspection that this side

must be 16 m long.

13 m

? m

29 mStep 2

Overall width = 9 + 6

= 15 m

6 m

9 m

Step 3

So, to find this right side marked with a ‘?’,

3 + ? = 15

3 + 12 = 15

Missing side = 12 m

3 m

? m

15

m

Step 4

The diagram can now show all measurements.

∴ Perimeter = 9 + 21 + 3 + 8 + 12 + 16 + 6 + 13 = 88 m


Activity – Perimeter of composite shapes

Try this.

1 Find the perimeter of this shape. All measurements are in metres.

8

2

2 2

2 2

_______________________________________________________

_______________________________________________________


Sometimes it is not necessary to

work out the size of every length.

In this diagram, the steps on the right

have unequal heights, but the three

heights together add to 24 cm.

24 c

m

36 cm

Similarly, the three top lengths total 36 cm. Can you see why?

This gives a total perimeter for this figure of 2 × 24 + 2 × 36 = 120 cm.

Practise what you have learned by doing the exercises.

Go to the exercises section and complete Exercise 2.2 – Perimeter of

composite shapes.



Right-angled triangles

In your study of mathematics you have learned about different types of

triangles; those based on angles and those based on sides.

• Triangles named according to angles.

acute-angled triangle(all angles acute)

right-angled triangle(one right angle)

obtuse-angled triangle(one angle obtuse)

• Triangles are also named according to the nature of their sides.

scalene triangle(no sides equal)

isosceles triangle(two equal sides)

equilateral triangle(all sides equal)

In this section you will learn more about one of these types of triangle:

the right-angled triangle.

A right-angled triangle has one angle 90°. On diagrams a right angle is

shown as .

right angle

Right angle is shortfor upright angle.

hypotenuse

The longest side of a right-angled triangle is the side opposite the

right angle. This side is called the hypotenuse.


A hypotenuse only exists in a right-angled triangle. If there is no right

angle, there is no hypotenuse.

Triangles can be shown in different orientations. Once you locate the

right angle, look for the side opposite it to see the hypotenuse.

In this diagram the hypotenuse is the

side labelled AC.

hypotenuseA

B

C

Activity – Right-angled triangles

Try this.

1 Label the hypotenuse in each of these right-angled triangles by

writing the word alongside it.


The ancient Egyptians had long known how to get a right angle.

They made a triangle whose sides were 3, 4 and 5 units long, with rope

containing equally spaced knots. When these ropes were pulled taut, the

angle between the two short sides was 90°.


They used this to reposition land boundaries after the Nile floods, and in

building square corners. Carpenters still use this today to make sure

wooden frames have 90° angles.

right angle

You can check for a right angle by using a protractor.

908070

60

50

4030

2010

0

180170

160150

140

130

120110

100

90100110

120

130

140

150

160

170

180 0

1020

30

40

5060

7080

You can also use the four corners of an A4 sheet of paper, such as the

corners of this page. Place it in the angle you are checking and see that it

fits snugly.



Try these.

2 Here is a knotted rope similar to the ones used by ancient Egyptians.

a Check to see that the knots are equally spaced.

b Does this triangle contain a right angle? __________________


In the next exercise you will examine relationships between the lengths

of three sides of a right-angled triangle.

Go to the exercises section and complete Exercise 2.3 – Right-angled

triangles.


Pythagoras’s theorem

Pythagoras was a Greek mathematician and philosopher who lived about

55 BC. He founded a secret society, called the Pythagoreans, which is

credited with discovering Pythagoras’s theorem.

This theorem states that in any

right-angled triangle, the square on

the hypotenuse is equal to the sum of

the squares of the other two sides.

In other words: c2 = a2 + b2 a

bc

In the previous section you looked at

showing a relationship between the three

sides of a right-angled triangle by using

these sides as the sides of squares drawn

on them.

Did you discover a similar relationship to

the one Pythagoras found?

a2b2

c2

B

C

A

Check to see that a2 + b2 = c 2 in this case.

The remainder of this section will be devoted to a practical activity where

you dissect areas to verify Pythagoras’s theorem. For this you will need

some sheets of paper, a ruler and a pair of scissors.


Activity – Pythagoras’s theorem

Try these.

1 On your own paper draw any right-angled triangle ABC.

Here is an example, but you don’t have to

use this one and yours, of course, should

be larger.

B

c

A

b

a C

2 Cut it out. Then draw and cut out another three exactly the same.

(If you have access to a drawing program on a computer, you could

draw one triangle then copy it three times.)

You now have four congruent right-angled triangles.

3 Measure the two smaller sides of one triangle and add them together

(a + b). Use this total length for the sides of a square.

Draw your square on a blank sheet of paper.

a + b

a + b

a +

b

a +

b


4 Use your four triangles and place them inside the square in the two

ways as shown below.

b

a

b

a

a

a

a

b

b

b

b

area = ___

area = __

a b

cb

a

ab

a

b

c

cc

square area = c2

5 What is the uncovered area in each case?

For the left-hand square, uncovered area = ________ square units.

For the right-hand square, uncovered area = _____ + _____

=________ square units.

The two arrangements of the triangles inside the larger square give

two different ways of finding the uncovered area.

6 Do the uncovered parts in each square have the same area? _______

Explain why? ___________________________________________

_______________________________________________________

7 Does it follow that c2 = a2 + b2? _____________________________


Does this prove Pythagoras’s theorem for you?

This method will work no matter what lengths you choose for the sides

of the right-angled triangle. You might want to repeat this activity using

a different, but still right-angled, triangle.


Consolidate your learning by doing the exercises.

Go to the exercises section and complete Exercise 2.4 – Pythagoras’s

theorem.

There are quite a few interesting Internet sites about the life of

Pythagoras and Pythagoras’s theorem.

Access related sites on Pythagoras’s theorem by visiting the LMP

webpage below. Select Stage 4 and follow the links to resources for this

unit MS4.1 Measurement, Part 2.

<http://www.lmpc.edu.au/mathematics>

This photograph from the side of the Art Gallery of NSW shows Pythagoras’sname among many other famous names from the past.

Source © State of New South Wales, Department of Education and Training 2004,Janine Angove.


Pythagorean triads

In previous sessions you explored Pythagoras’s theorem.

This is a rule linking the three sides of any right-angled triangle.

The rule states that in any right-angled triangle,

a2 + b2 = c2, where a, b, and c are the lengths of the

sides of the triangle.B

c

A

b

a C

The opposite, or inverse, is true. That is, if the

lengths of the three sides of a triangle can be linked

by a2 + b2 = c2, then the triangle is right angled.

The triangle shown has side lengths 3 units, 4 units

and 5 units. Now 32 + 42 = 9 + 16 = 25.

This is equal to 52. Since 32 + 42 = 52 the triangle is

right angled and its hypotenuse is 5 units.3

45

This is an example where the whole numbers (3, 4, 5) can be linked by

the rule 32 + 42 = 52.

This triangle, too, is right angled. Its two shorter

sides are 2 units and 3 units, and 22 + 32 = 13.

The longest side is close to 3.6 units and

3.62 = 13. (Actually, 3.62 = 12.96, but for this

example assume it to be 13.)

In this case while two of the side lengths are

whole numbers, the third side is not. 2 units

3 un

its

3.6

units

In this section you will be concerned with whole number relationships

such as (3, 4, 5) which can be linked by the rule 32 + 42 = 52.

Mathematicians have long been interested in sets of three whole (integer)

numbers like these. They call these numbers triads (or triples).


The triad, or set, of positive integers (a, b, c) where a2 + b2 = c2 is called

the Pythagorean triad of numbers.

About fifteen such triads were known before Pythagoras’s time, but now

there are formulas that can generate an infinite number of such sets.

The 3–4–5 right triangle is the most famous.

It is the smallest right-angled triangle with

integer sides.

The numbers (3, 4, 5) is an example of a

Pythagorean triad. These numbers can also be

written as {3, 4, and 5}.3

45


margin if you wish.

Does the set (7, 24, 25) form a Pythagorean triad?

Solution

Squaring the first two numbers, then adding them gives

72 + 242 = 49 + 576 = 625.

Squaring the last number, 252 = 625.

Since both results are the same, (7, 24, 25) is a

Pythagorean triad.

Activity – Pythagorean triads

Try these.

1 Is (5, 12, 13) a Pythagorean triad?

_______________________________________________________

_______________________________________________________

_______________________________________________________

_______________________________________________________


2 Is (7, 9, 11) a Pythagorean triad?

_______________________________________________________

_______________________________________________________

_______________________________________________________

_______________________________________________________


3

45

Suppose you enlarge the 3-4-5 triangle. The resulting triangles will still

be right angled. If the sides double to 6-8-10 you can show that

62 + 82 = 102 and so (6, 8, 10) forms a Pythagorean triad.

Similarly, tripling each side to 9-12-15 also generates the

Pythagorean triad (9, 12, 15) since 92 + 122 = 152.

In fact, multiplying the numbers in a Pythagorean triad by any whole

number produces another Pythagorean triad.


Try this

3 Given that (3, 4, 5) is a Pythagorean triad, explain why (12, 16, 20)

is also a triad.

_______________________________________________________

_______________________________________________________

_______________________________________________________

_______________________________________________________

_______________________________________________________



Do the exercises to consolidate your learning.

Go to the exercises section and complete Exercise 2.5 – Pythagorean

triads.

Congratulations you have completed the learning for this part. It is now

time for you to complete the review quiz so you can show your teacher

how much you have learned.


Suggested answers – Part 2

Check your responses to the preliminary quiz and activities against these

suggested answers. Your answers should be similar. If your answers are

very different or if you do not understand an answer, contact your teacher.

Activity – Preliminary quiz

1 a millimetre b centimetre c metre

d kilometre

2 a cm, m, km b km2, ha, mm2, cm2, m2

3 a 3.2 cm b 480 mm c 250 cm

4 9 cm, 90 mm

5 Perimeter = 15 + 15 + 15 + 15 = 60 cm (or 4 ×15)

6 Perimeter = 16.8 + 16.8 + 12. 1 + 12.1 = 57.8 cm

7 a Perimeter = 56 + 56 + 29 + 29 = 170 mm

b Perimeter = 51 + 54 + 20 = 125 mm

8 a 38.44 b 52.56 c 8.66 d 5.57


1 (Your lengths measured should be one, or at most two, millimetres

out from these unless you are dealing with a resized photocopy.)

P = 66 + 42 + 33 + 59 = 200 mm

2 P = 2 ×1.6 + 2 × 5.3 = 13.8 m

Activity – Perimeter of composite shapes

1 The length of the left side is 2 + 2 + 2 = 6 m. The top side measures

8 + 2 + 2 = 12 m. Hence P = 8 + 2 + 2 + 2 + 2 + 2 + 12 + 6 = 36 m.



1

hypo

tenu

se

hypo

tenu

se

hypo

tenu

se

hypotenuse

2 All measurements are never exact, and diagrams or knotted ropes are

no exception. You should be able to see that the knots are equally

spaced and that there is a right angle at the bottom left of the triangle

(within reason).

Activity – Pythagoras’s theorem

(Individual answers will vary for other questions.)

6 The large square and triangles occupy the same areas. Therefore the

uncovered parts have a constant area. They are just arranged

differently. The relationship c2 = a2 + b2 holds.


1 52 +122 = 25 +144

= 169 (the sum of two smaller numbers squared)

132 = 169 (largest number squared)

(5,12,13) is a Pythagorean triad

2 72 + 92 = 49 + 81

= 130 (sum of squares of two smaller numbers)

112 = 121 ��(largest number squared)

Since both results are not the same then (7, 9, 11) is not a

Pythagorean triad.

3 (12, 16, 20) can be written as (4 × 3,4 × 4,4 × 5) .

This is just multiplying each of the numbers in (3, 4, 5) by 4.

So (12, 16, 20) is also a triad.


Exercises – Part 2

Exercises 2.1 to 2.5 Name ___________________________

Teacher ___________________________

Exercise 2.1 – Perimeters

1 Measure the lengths of each side of the following figures in

millimetres. Use these lengths to calculate the perimeters.

a

b


2 The following figures are not drawn to scale. Use the measurements

on them to calculate the perimeters.

a

40 m

24 m

32 m 36 m

b

125 mm

56 mm

c

18.4

cm

5.1 cm

d

0.8

m

1.7 m2.3 m


Exercise 2.2 – Perimeter of composite shapes

1 Find the perimeter of each of the following composite figures.

All measurements are in centimetres and all angles are right angles.

a

4

4

148

10

24

b

8

20

15

40

c 24

30

48

30

56

d 9

66

55

5

2 The perimeter of a square is 36 cm. What is the length of each side?

_______________________________________________________


3 The rectangle has a perimeter of

30.0 cm. What is its length?

_______________________________

_______________________________

6.4

cm

length

_______________________________________________________

_______________________________________________________

4 Gayle drew the following diagram (not drawn to scale).

3 cm 2.5 cm

3 cm

She was wondering what measurement she should write in the space

for the sloping ( / ) side. These are some suggestions from Kim,

Frank and Mary.

A value between0 and 3.

A value between3 and 5.5.

A value between5.5 and 8.5.

Kim Frank Mary

Who is correct, and why? _________________________________

_______________________________________________________

_______________________________________________________


Exercise 2.3 – Right-angled triangles

This activity shows some right-angled triangles whose sides are whole

numbers of units. Here you will explore whether relationships can be

formed between the lengths of three sides of each triangle.

1 By counting the squares, or otherwise, find the number of units of

area for each of the three squares drawn on the sides of the

right-angled triangle.

a

Area of largest square = _____ square units

Area of smallest square = _____ square units

Area of remaining square = _____ square units

Add the areas of the two smallest squares = _____ square units

Does this sum equal the area of the largest square? __________


b

Area of largest square = _____ square units

Area of smallest square = _____ square units

Area of remaining square = _____ square units

Add the areas of the two smallest squares = _____ square units

Does this sum equal the area of the largest square? __________


c Check that in ∆ABC, ∠C = 90� .

The side lengths of ∆ABC are a, b, and c.

a2

b2

c2

A

B

C

a

b

c

ii Which side of the triangle is the hypotenuse in ∆ABC?

________________________________________________

iii Complete:

a = _____ units __________________________ a2 = _____

b = _____ units __________________________ b2 = _____

c = _____ units __________________________ c2 = _____

a2 + b2 =_____+_____

=_____

From this example, can you conclude that a2 + b2 = c2? ___


Exercise 2.4 – Pythagoras’s theorem

Pythagoras’s theorem is a rule that holds only for right-angled triangles.

This exercise is an activity that should illustrate this point.

For this activity you will need a compass (or other instrument to draw

circles), a protractor (or a right angle) to compare angles, some sheets of

blank paper, and a ruler.

• Draw a number of identical circles on your sheet. Tracing the base

of a coffee mug should give you circles about reasonable size.

• On each diagram construct the same line segment AC, where A is a

point outside the circle and C is the centre of the circle.

(You can draw A anywhere. It does not have to be to the right of the

circle as is shown here.)

Here are some examples, but you do not have to follow these.

A

B

C A

B

C

A

B

C A

B

C

• Place point B at various positions on the rim of the circles, and

complete triangles ABC in each case. (Draw at least six circles.)

• For one of your circles place B at either the top or bottom of the

circle.

1 Which lengths always remain the same?

AB? _____ BC? _____ AC? _____


2 Measure the lengths of the three sides of the triangles. Give answers

in centimetres correct to one decimal place. Use these measurements

to complete the table of values on the next page. (Space has been

left for up to eight circle diagrams. If you drew more, then attach an

additional table.)

Figure AB AC BC AB2 AC2 BC2 AC2 + BC2

1

2

3

4

5

6

7

8

3 Calculate and enter the values on the right of the double vertical line.

4 Do any of your figures show the relationship AC2 + BC2 = AB2?

If so, what is the size of ∠ C? ______________________________

5 In which diagram (P, Q, or R) is

a AC2 + BC2 > AB2? ______________

b AC2 + BC2 < AB2? ______________

c AC2 + BC2 = AB2? ______________

P

A

B

C

6 In which diagram (P, Q, or R) is

a ∠ C < 90°? ____________________

b ∠ C = 90°? ____________________

c ∠ C > 90°? ____________________

Q

A

B

C


7 What name is given to the triangle side

AB in diagram R?

__________________________________

__________________________________

R

A

B

C

Can this name be used for side AB in diagrams P or Q? _________

Why or why not? ________________________________________

8 From this activity, when does the relationship AC2 + BC2 = AB2

hold? _________________________________________________


Exercise 2.5 – Pythagorean triads

1 Show that the following are Pythagorean triads.

a (8, 15, 17) __________________________________________

___________________________________________________

___________________________________________________

b (9, 40, 41) __________________________________________

___________________________________________________

___________________________________________________

c (15, 36, 39) _________________________________________

___________________________________________________

___________________________________________________

2 Which of these is not a Pythagorean triad?

a (8, 15, 17) b (20, 21, 29) c (10, 12, 15)

___________________________________________________

___________________________________________________

___________________________________________________

___________________________________________________

3 Given that (5, 12, 13) is a Pythagorean triad, give a reason for

(15, 36, 39) also being a triad.

_______________________________________________________

_______________________________________________________

4 (11, 60, 61) forms a Pythagorean triad. Write down two other triads

based on these numbers. __________________________________


5 Here is a simple formula that gives some of the Pythagorean triads.

Suppose that m is any positive integer. Then m2 – 1, 2m, and m2 + 1

is a Pythagorean triad. For example, if m = 5, then

m2 −1

= 52 −1

= 25 −1

= 24

2m

= 2 × 5

= 10

m2 +1

= 52 +1

= 25 +1

= 26

So (10, 24, 26) form a Pythagorean triad.

Find the Pythagorean triads when

a m = 3 ______________________________________________

___________________________________________________

___________________________________________________

___________________________________________________

b m = 8 ______________________________________________

___________________________________________________

___________________________________________________

___________________________________________________

6 The numbers (16, 63, 65), (25, 60, 65), (33, 56, 65), and (39, 52, 65)

are all Pythagorean triads. What can you say about triangles drawn

having these dimensions for sides?

_______________________________________________________

_______________________________________________________


7 The table gives the squares of numbers from 0 to 99.

The units digit is across the top, while the tens digit is down the side.

0 1 2 3 4 5 6 7 8 9

0 0 1 4 9 16 25 36 49 64 81

10 100 121 144 169 196 225 256 289 324 361

20 400 441 484 529 576 625 676 729 784 841

30 900 961 1024 1089 1156 1225 1296 1369 1444 1521

40 1600 1681 1764 1849 1936 2025 2116 2209 2304 2401

50 2500 2601 2704 2809 2916 3025 3136 3249 3364 3481

60 3600 3721 3844 3969 4096 4225 4356 4489 4624 4761

70 4900 5041 5184 5329 5476 5625 5776 5929 6084 6241

80 6400 6561 6724 6889 7056 7225 7396 7569 7744 7921

90 8100 8281 8464 8649 8836 9025 9216 9409 9604 9801

For example, 232 = 529.

Use the table to answer the following.

a (24, 32, h) form a Pythagorean triad. What is the value of h?

___________________________________________________

___________________________________________________

___________________________________________________

b Find the missing values in each of these triads.

i (33, 56, p) _______________________________________

________________________________________________

________________________________________________

ii (55, q, 73) _______________________________________

________________________________________________

________________________________________________

iii (r, 77, 85) _______________________________________

________________________________________________

________________________________________________



Review quiz – Part 2

Name ___________________________

Teacher ___________________________

1 Calculate the perimeter of this composite shape.

_______________________________________________________

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10 cm

2 cm

2 cm

2 cm

3 cm

3 cm

Figure not drawn to scale

2 Measure the perimeter of this figure correct to the nearest millimetre.

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3 A square has a perimeter of 18.4 cm. What is the length of one side

of the square? __________________________________________

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4 The perimeter of this rectangle is 50 m.

Calculate its width.

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__________________________________

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16 m

wid

th

5 Label the hypotenuse of this

right-angled triangle.

6 a State Pythagoras’s theorem in words. ____________________

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b Use the triangle shown to write a relationship between lengths

a, b, and c.

B

c

A

b

aC


7 Describe how you could use the following diagrams to prove

Pythagoras’s theorem.

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8 Why is (12, 35, 37) a Pythagorean triad? _____________________

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9 Which of the following is not a Pythagorean triad?

a (28, 45, 53)

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________________________

________________________

b (18, 80, 82)

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________________________

c (16, 21, 25)

________________________

________________________

________________________


10 In a Pythagorean triad, 25 is the largest number and one of the

other numbers is less than 10. What are the other two numbers in the

triad?


Answers to exercises – Part 2

This section provides answers to questions found in the exercises section.

Your answers should be similar to these. If your answers are very

different or if you do not understand an answer, contact your teacher.

Exercise 2.1 – Perimeters

1 a 233 mm b 279 mm (Your answers may vary slightly)

2 a 132 m b 556 mm c 167.6 cm d 14.6 m

Exercise 2.2 – Perimeter of composite shapes

1 a 96 cm b 120 cm c 208 cm d 72 cm

2 9 cm

3 8.6 cm

4 Frank is correct. The sloping ( / ) length must be greater than either

the vertical (3 cm) or horizontal (2.5 cm) lengths but shorter than

both of them added together.

Exercise 2.3 – Right-angled triangles

1 a largest square = 25 square units; smallest square = 9 square

units; remaining square = 16 square units. Sum of two smaller

areas = 25 square units, which equals the area of the largest

square.

b largest square = 169 square units;

smallest square = 25 square units;

remaining square = 144 square units.

Sum of two smaller areas = 169 square units, which equals the

area of the largest square.


c ii The hypotenuse is AB (or side c).

iii a = 9, a2 = 81; b = 12, b2 = 144; c = 15, c2 = 225.

t is true that a2 + b2 = c2.

Exercise 2.4 – Pythagoras’s theorem

1 AC is a fixed line and remains the same length. BC is the radius of

the circle. As all circles are the same, then BC remains constant.

Only AB changes in length as B moves around the circle.

2 (answers vary)

3 (answers vary)

4 Only when B is at the top or bottom of the circle (that is, ∠ C = 90°)

will AC2 + BC2 = AB2.

5 a P b Q c R

6 a P b R c Q

7 Hypotenuse. This name can only be used for the longest side of a

right-angled triangle, which occurs only in diagram R.

8 Relationship holds only for right-angled triangles.

Exercise 2.5 – Pythagorean triads

1 a 82 + 152 = 289; 172 = 289. Since both results (289) are the

same, then (8, 15, 17) form a Pythagorean triad.

b 92 + 402 = 1681; 412 = 168

c 152 + 362 = 1521; 392 = 1521

2 c (10, 12, 15). 102 + 122 ≠ 152

3 Multiplying each of the values in (5, 12, 13) by 3 gives (15, 36, 39)

4 Any whole number multiple of (11, 60, 61) such as (22, 120, 122)

and (33, 180, 183)

5 a (6, 8, 10) b (16, 63, 65)

6 The four right-angled triangles have whole number values for sides

and all four triangles have hypotenuse length 65 units.

7 a h = 40 (h2 = 1600)

b i p = 65 ii q = 48 iii r = 36

Documents

43583 MS4.1 Part 2 - Login Department of Educationlrr.cli.det.nsw.edu.au/legacy/Mathematics/43583_06_04_P2.pdf · Pythagoras’s theorem ... Answers to exercises – Part 2 ... Other