AA Pearson Maths 9 SB-02.fm Page 76 Monday, October 24...

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AA Pearson Maths 9 SB-02.fm Page 76 Monday, October 24, 2011 9:32 AM

77

2

Pythagoras’ Theorem

Did Pythagoras ‘steal’ the theorem that bears his name?

One of the world’s most famous mathematical statements is not named after its discoverer.

Several ancient civilisations, including the Babylonians, Chinese and Egyptians, knew about the special relationship between the sides of a right-angled triangle. Pythagoras of Samos (approx. 580–496

BC

) was a Greek philosopher who probably learnt about it from a group of Egyptian engineers known as the ‘rope stretchers’. They used a circle of rope with 12 knots tied at regular intervals around the circle. Pegging the rope to the ground at intervals of 3, 4 and 5 knots produced a right-angled triangle, which was used to ensure that building foundations or walls were constructed accurately. So, why does Pythagoras get all the credit? He was a notable mathematician who headed a secretive cult known as the Brotherhood, which was devoted to the study of

mathematics, with their motto being ‘all is number’. Although he did not discover the theorem, it is possible that Pythagoras was the first to produce a proof of the theorem using geometry. The photograph on the opposite page is of a statue of Pythagoras on the island of Samos, in Greece.

Forum

Why do you think Pythagoras and his followers held secret meetings and kept their discoveries to themselves?

Think of some real-life objects that can consist of, or be described by, right-angled triangles.

Why learn this?

Pythagoras’ Theorem is a useful mathematical tool applied by builders, carpenters, engineers and surveyors to determine the straight-line distance between two points or to check whether an angle is a right angle. It is also used in the audio-visual industry to determine the height and width of TV and projection screens.

After completing this chapter you will be able to:

•

use Pythagoras’ Theorem to determine whether a triangle is right-angled

•

use Pythagoras’ Theorem to find side lengths in right-angled triangles

•

apply Pythagoras’ Theorem to practical situations

•

complete simple calculations using surds

•

express irrational answers in exact (surd) form

•

identify Pythagorean triples.


78

PEARSON

mathematics 9

Recall

2

Equipment required:

calculator for Questions

2

–

4

and

6

.

Prepare for this chapter by attempting the following questions. If you have difficulty with a question, go to Pearson Places and download the Recall Worksheet from Pearson Reader.

1

Round each of the following numbers to two decimal places.

(a)

45.789

(b)

12.2311

(c)

4.549 567 835 6

2

Use your calculator to find the following, correct to two decimal places where appropriate.

(a)

12

2

(b)

55

2

(c)

37.5

2

(d)

3

2

+

4

2

(e)

9

2

+

13

2

(f)

2.5

2

+

7.1

2

3

Use your calculator to find the exact values of each of the following.

(a) (b) (c)

4

Use your calculator to find the following. Round your answers to two decimal places.

(a) (b) (c)

5

Solve the following equations.

(a)

x

+

5

=

11

(b)

144

+

x

=

225

(c)

45

−

x

=

12

6

Solve the following for positive values of

x

. Where necessary, express your answer as a decimal correct to two decimal places.

(a)

x

2

=

36

(b)

x

2

=

39

(c)

x

2

=

325

(d)

x

2

+

6

=

70

(e)

x

2

−

20

=

101

(f)

=

325

converse proof rational approximation theorem

hypotenuse Pythagoras’ Theorem right-angled triangle

irrational number Pythagorean triple surd

Worksheet R2.1

Worksheet R2.2

Worksheet R2.3 81 169 256

Worksheet R2.4 65 658 321.45

Worksheet R2.5

Worksheet R2.6

x2

4-----

Key Words


2


79

Pythagoras’ Theorem and right-angled triangles

A triangle that contains a 90

°

angle (a right angle) is called a

right-angled triangle

. The longest side of a right-angled triangle, which is opposite the right angle, is called the

hypotenuse

.

When labelling triangles, we use capital letters for the vertices and lower case letters for the side lengths.

So, triangle

ABC

on the right, which can be written as

Δ

ABC

, has vertices

A

,

B

and

C

with sides

a

,

b

and

c

opposite the vertices that have the same letter. It is usual to label the hypotenuse as side

c

and the two shorter sides as

a

and

b

. (It does not matter which side is

a

and which is

b

.)

What is a theorem?

A

theorem

is a mathematical statement that can be shown to be true. A

proof

is a step-by-step argument that demonstrates the truth of a mathematical theorem.

The area of a square of side

a

=

a

2

. In a right-angled triangle, the areas of the squares drawn onto the two shorter sides add to give the area of the square drawn onto the hypotenuse.


: For any right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the two shorter sides.

c

2

=

a

2

+

b

2

C B

A

a

cb

hypotenuse

a c

c2

b2

a2

b

a

bc

hypotenuse

a

Area = a2

b

Area = b2

c

Area = c2 + =

2

.

1


2

.

1

80

PEARSON

mathematics 9

Demonstrating Pythagoras’ Theorem

We can visually demonstrate the theorem as shown. If we take an isosceles right-angled triangle and place it onto square grid paper that has been divided into small, identical, triangular tiles, the combined area of the squares on the two shorter sides is eight small triangles, the same number of triangles covered by the square on the hypotenuse.

This shows only that the theorem works for one particular type of right-angled triangle. For the theorem to be shown as true, we must prove that it works for all right-angled triangles.

Proofs of Pythagoras’ Theorem by dissection

Mathematicians have established at least 85 different methods of proving that Pythagoras’ Theorem is true for every right-angled triangle. In 1830, a proof using areas was published by the mathematician Henry Perigal, who had the diagram of the proof carved on his gravestone!

Demonstrating Pythagoras’ Theorem using areasStep 1 On a piece of grid paper, draw any right-angled triangle, and label the sides a, b and c.

Step 2 Construct a square on each side of the triangle as shown below. Label the squares A, B and C to correspond to the sides a, b and c.

Step 3 Find the centre of square B. Rule a line PQthrough the centre, so that it is parallel tothe hypotenuse of the triangle.

Step 4 Rule a second line, RS, perpendicular(at right angles) to PQ that passes through the centre of the square.

Step 5 Cut square B into its four pieces. Then, cut out square A and square C.

Step 6 Rearrange the four pieces of square B as shown.

Step 7 Place square A in the centre. It should fit exactly. The total area of this square = area of square A + area of square B

= a2 + b2

Step 8 Compare the area of this square to the area of square C, c2. The two squares should be identical in area, demonstrating that c2 = a2 + b2.

square C

R

c

P

S

Qsquare Asquare B

a

b


2 Pythagoras’ Theorem

2.1

81

The converse of Pythagoras’ TheoremA converse is a reversal of a proved theorem.

Because Pythagoras’ Theorem is only true for right-angled triangles, the converse of Pythagoras’ Theorem can be used to check whether or not a triangle is right-angled. That is, if the sum of the squares of the two shorter sides is equal to the square of the longer side, the triangle is right-angled.

Consider a triangle of sides 3 cm, 4 cm, and 5 cm.

We can show this is a right-angled triangle using the converse of Pythagoras’ Theorem.

Here, a = 3, b = 4 and c = 5.

Now: c2 = 52

= 25 anda2 + b2 = 32 + 42

= 9 + 16= 25

As c2 = a2 + b2 holds for this triangle, we can say that this is a right-angled triangle.

Builders use the converse of Pythagoras’ Theorem to make sure building frames and walls are right-angled. They call it ‘the 3-4-5 principle’—can you see why?

If Pythagoras’ Theorem holds, then the triangle must be a right-angled triangle. This statement is known as the ‘converse’ of Pythagoras’ Theorem.

Worked Example 1Use the converse of Pythagoras’ Theorem to determine whether each of the following triangles is a right-angled triangle. If it is, state which angle is the right angle.

(a) (b)

Thinking(a) 1 Identify the longest side as c and the

other two sides as a and b.(a) a = 30 mm, b = 40 mm, c = 50 mm

2 Evaluate the left-hand side of Pythagoras’ Theorem, c2.

c2 = 502

= 2500

3 Evaluate the right-hand side of Pythagoras’ Theorem, a2 + b2.

a2 + b2 = 302 + 402

= 900 + 1600= 2500

4 Does c2 = a2 + b2? If so, the triangle is right-angled.

c2 = a2 + b2, so ΔABC is a right-angled triangle.

5 The right angle is the angle opposite the hypotenuse, c.

∠ACB is the right angle.

5

3

9 cm2

25 cm2

16 cm2

534

4

1

30 mm40 mm

50 mmB

A

C

7 m

11 m

9 m

A

B

C


2.1

82 PEARSON mathematics 9

Pythagoras’ Theorem and right-angled triangles

Fluency1 Use the converse of Pythagoras’ Theorem to determine whether each of the following

triangles is a right-angled triangle. If it is, state which angle is the right angle.

(a) (b) (c)

(d) (e) (f)

(g) (h) (i)

(b) 1 Identify the longest side as c and the other two sides as a and b.

(b) a = 7 m, b = 9 m, c = 11 m

2 Evaluate c2. c2 = 112

= 121

3 Evaluate a2 + b2. a2 + b2 = 72 + 92

= 49 + 81= 130

4 Does c2 = a2 + b2? If so, the triangle is right-angled. (Here, the triangle is not right angled, so we use the symbol ≠, which means ‘is not equal to’).

c2 ≠ a2 + b2, so ΔABC is not a right-angled triangle.

NavigatorQ1 Column 1, Q2, Q3, Q4, Q5 Column 1, Q7 (a), (b), (d), Q9, Q10, Q11

Q1 Column 2, Q2, Q3, Q4, Q5 Column 2, Q6, Q7 (a), (c), (e), Q8, Q9, Q10, Q11

Q1 Column 3, Q2, Q4, Q5 Column 2, Q6, Q7 (b), (e), Q8, Q9, Q10, Q11, Q12

2.1

Answerspage 603

1

B

CA

5 cm

4 cm

3 cm 4 cm

3 cm

2 cm

M

O

N 5 cm

3 cm

6 cm

K

L

J

6 m

9 m

7 mH

I

G

6 m

8 m 10 m

E

DF

7.5 m

4.5 m

6 m

FD

E

13 km

12 km

5 km

P

R

Q 10 km

12.5 km 7.5 km

R

Q

P

7.5 km4.5 km

9 kmT

U

S



2.1

83

2 For each of the following triangles, select the correct statement of Pythagoras’ Theorem from those provided.

(a) A p2 = q2 + r2 C r2 = p2 + q2

B q2 = p2 + r2 D q2 − p2 = r2

(b) A z2 + y2 = x2 C x2 − y2 = z2

B x2 + z2 = y2 D z2 = x2 + y2

(c) A n2 + p2 = m2 C n2 − m2 = p2

B p2 + m2 = n2 D n2 + m2 = p2

3 For each of the following right-angled triangles, identify which side length (AB, BC or AC) is the hypotenuse, and which angle is the right angle (∠ABC, ∠ACB or ∠CAB).

(a) (b) (c) (d)

4 Which one of these triangles is a right-angled triangle?

A B C D

Understanding5 Find which of these triangles with the following side lengths are right-angled triangles.

(a) 9 mm, 12 mm, 15 mm (b) 2 cm, 4 cm, 6 cm

(c) 2 m, 2 m, 4 m (d) 7 km, 24 km, 25 km

(e) 18 mm, 24 mm, 30 mm (f) 5 cm, 12 cm, 13 cm

6 Jack has constructed part of a building frame. He needs to check that it is ‘square’; that is, that the corners are right angles. The frame is 5 m wide and 3 m high. Jack has measured the diagonal to be 6 m.

(a) Use the converse of Pythagoras’ Theorem to check whether the frame is ‘square’.

(b) To make the frame ‘square’, should Jack increase or decrease the length of the diagonal?

r

q

p

R P

Q

yz

xY

X

Z

mn

p

P

M N

A

BC

B

A

C

C

B

A B

A

C

26

15

24

15

2521

10

24

2624

18

12

5 m

3 m6 m

AA Pearson Maths 9 SB-02.fm 3 Monday, October 24, 2011 9:32 AM

2.1


7 For each of the following, construct right-angled triangles using the side lengths given in the table.

(a) Find the length of the hypotenuse, c, by measurement, then complete the rest of the table.

(b) Which columns of the table would you check to see if Pythagoras’ Theorem was true?

(c) What would you look for in those columns?

Reasoning8 Pythagoras’ Theorem holds for the right-angled triangle whose sides are of length 5 cm,

12 cm and 13 cm.

(a) Triple each side length. Does Pythagoras’ Theorem still hold?

(b) Halve each original side length. Does Pythagoras’ Theorem still hold?

(c) What conclusion can you make from your answers to (a) and (b)?

9 Both a triangle whose sides are of length 3 units, 4 units and 5 units and a triangle whose sides are of lenth 5 units, 12 units and 13 units are right-angled triangles. Find another two right-angled triangles from the following list and justify your selection.

A 9 units, 12 units, 15 units B 4 units, 5 units, 6 units

C 10 units, 24 units, 26 units D 7 units, 14 units, 15 units

Open-ended10 Draw a right-angled isosceles triangle, measure the three sides and show that Pythagoras’

Theorem holds.

11 You have a 40 m long piece of rope with which to create a large right-angled triangle on the school oval. Determine the lengths of the sides of three different triangles that you can create. You do not have to use the whole length of the rope.

12 Show how you would explain Pythagoras’ Theorem to a friend.

a b c a2 b2 a2 + b2 c2

3 cm 4 cm

12 cm 16 cm

60 mm 80 mm

5 cm 12 cm

Problem solvingOlives and potsPythagoras loved the challenge of solving problems. One of his favourite problems involved olives and clay pots.

If nine olives are placed in each pot, two pots will remain empty.

If six olives are placed in each pot, there will be three olives left over.

How many olives and clay pots are there?

• Draw a diagram.

• Make a model.

• Guess and check.

• Test all possible combinations.

Strategy options


Technology Exploration GeoGebra

2 Pythagoras’ Theorem 85

Demonstrating Pythagoras’ Theorem

Versions of this Exploration for other technologies are available in Pearson Reader.

Equipment required: 1 brain, 1 computer with GeoGebra

Open the GeoGebra program. You will see seven menu options (File, Edit etc.) at the top of the screen. Below these are 11 icons called tools. We will refer to these by number, counting from the left. By clicking on the small arrow in the bottom right-hand corner of the tool icon, a drop-down list of more tools appears. (The arrow turns red when you hover the cursor over it.) If you hover over a tool, the tool’s name and how to use it will appear in the top right-hand corner of the screen. (Some versions of GeoGebra have different names for the tools; however, the icons remain the same.)

1 The GeoGebra screen is divided into two panels. The thinner panel on the left is called the ‘Algebra View’. The panel on the right is called the ‘Graphics View’. Click on the View menu and deselect ‘Axes’ to get rid of the Cartesian Plane. (Or, right click in the ‘Graphics View’ and deselect ‘Axes’.)

2 Click on the Options menu. Select ‘Point Capturing’ and ‘On (Grid)’.

3 From the Options menu, select ‘Labeling’ and ‘New Points Only’.

4 If a larger font is required, click on the options menu and select ‘Font Size’. Choose an appropriate size from the list provided.

Creating a right-angled triangle5 From the third icon, select the ‘Interval Between Two

Points’ tool . Draw an interval AB by clicking two points in the Graphics View.

6 From the fourth icon, select the ‘Perpendicular Line’

tool . Create a perpendicular line through A by clicking on the interval AB and then on Point A.

7 From the third icon, select the ‘Interval Between

Two Points’ tool . Create an interval AC by clicking on Point A and then anywhere along the perpendicular line you just created.

8 Still using the ‘Interval’ tool, join Points B and C with an interval.

9 Hide the perpendicular line by right clicking on the line outside the interval AC and deselecting ‘Show Object’.

10 From the fifth icon, select the ‘Regular Polygon’ tool

.

11 Click on Points B and C, heading in a clockwise direction. Enter ‘4’ under ‘Points’ in the box that appears on screen. If the square that appears overlaps the triangle, undo and reselect your points in the opposite order.

12 Repeat Step 11 for the other two sides of the triangle (AB and AC). If the resulting figure is too big to all be seen, rolling the mouse will make it smaller.

Analysing your construction13 Look at the Algebra View on the left of the screen.

You will see poly1, poly2 and poly3 at the bottom of a list of points and lines. Each one will have a number associated with it. This number is the area of the square. Poly1 is the area of the square on the hypotenuse because you created this square first.

14 In the Input bar (at the bottom of the screen), type: sum=poly2+poly3. Press ‘Enter’.

15 What do you notice about the values of poly1 and sum?

16 To label poly1 as the area of the square on the

hypotenuse, select the ‘Insert Text’ tool from the 10th icon.

17 Click in the Graphics View where you would like the text to appear and type the following into the box that appears: “The square on the hypotenuse = “ + poly1. (You must type the inverted commas, as shown.) Click OK.

Technology



18 With the ‘Insert Text’ tool still selected, click on a new position in the Graphics View. Type the following: ‘The sum of the squares on the other two sides = “ + sum’. Click ‘OK’.

19 Using the ‘Select’ tool under the first icon, , you can click on the text boxes and move them anywhere you like in the Graphics View. You can also edit colours and text sizes by right clicking on each and selecting ‘Object Properties’.

20 Using the ‘Select’ tool, click and drag either Point A or Point B, changing the lengths of the sides of the triangle.

21 What do you notice about the values of ‘The square on the hypotenuse’ and ‘The sum of the squares on the other two sides’?

22 Write a statement linking the ‘The area of the square on the hypotenuse’ and ‘The sum of the areas of the squares on the other two sides’.

Replacing the squares on the sides with triangles23 Right click inside each of the squares about the

triangle and select ‘Delete’.

24 From the fifth icon, select the ‘Regular Polygon’ tool

.

25 Click on Points B and C, heading in a clockwise direction. Enter ‘3’ under ‘Points’ in the box that appears on the screen. An equilateral triangle (a three-sided regular polygon) will appear, with BC as one of its sides. If the triangle you create goes over the original triangle, undo and reselect your points in the opposite order.

26 Repeat Step 25 for the other two sides of the right-angled triangle (AB and AC).

27 Poly1, poly2 and poly3 should appear in the Algebra View. This time they are the areas of the equilateral triangles.

28 In the Input bar (at the bottom of the screen), type: sum=poly2+poly3. Press ‘Enter’.

29 What do you notice about the values of poly1 and sum in the Algebra View?

30 Select the ‘Insert Text’ tool from the 10th icon.

31 Click in the Graphics View where you would like the text to appear and type the following into the box that appears: “The triangle on the hypotenuse = “ + poly1. (You must type the inverted commas, as shown.) Click ‘OK’.

32 With the ‘Insert Text’ tool still selected, click on a new position in the Graphics View. Type in the following: “The sum of the triangles on the other two sides = “ + sum. Click ‘OK’ when you are done.

33 Use the ‘Select’ tool , to click on the text boxes and move them anywhere you like in the Graphics View. You can also edit colours and text sizes by right clicking on each and selecting ‘Object Properties’.

34 Using the ‘Select’ tool, click and drag either Point A or Point B, changing the lengths of the sides of the triangle.

35 What do you notice about ‘The triangle on the hypotenuse’ and ‘The sum of the triangles on the other two sides’?

36 Write a statement linking the ‘The area of the triangle on the hypotenuse’ and ‘The sum of the areas of the triangles on the other two sides’.

Replacing the sides with semicircles37 Right click inside the triangles on the sides of the

triangle and select ‘Delete’.

38 From the sixth icon, select the ‘Semicircle Through

Two Points’ tool .

39 Starting with the interval BC ( the hypotenuse), click on pairs of points in a clockwise manner to create semicircles on each of the three sides of the triangle. If the semicircle you create overlaps the original triangle, undo and reselect your points in the opposite order.


2 Pythagoras’ Theorem 87

40 From the second icon, select the ‘New Point’ tool

. Starting with the arc (semicircle) on the hypotenuse, place a point on each of the three arcs.

41 From the sixth icon, select the ‘Circumcircular Sector

Through Three points’ tool . Click on the three points that make up the semicircle on the hypotenuse, starting with Point C, then the new Point D and then Point B (endpoint, arc, endpoint). Repeat for the other arcs about the triangle.

42 Notice that the areas of the semicircles are represented in the Algebra View by the letters h, k and p, where h is the area on the hypotenuse.

43 In the Input bar, calculate the sum of the smaller two areas by typing: sum=k+p. Press ‘Enter’.

44 What do you notice about the values of h and sum in the Algebra View?

45 Select the ‘Insert Text’ tool from the 10th icon.

46 Click in the Graphics View where you would like the text to appear and type the following into the box that appears: “The semicircle on the hypotenuse = “ + h. (You must type the inverted commas, as shown.) Click ‘OK’.

47 With the ‘Insert Text’ tool still selected, click on a new position in the Graphics View. Type, or copy and paste, the following: “The sum of the semicircles on the other two sides = “ + sum. Click ‘OK’.

48 From the first icon, click on the ‘Select’ tool . Then, by clicking and dragging either Point A or Point B, change the lengths of the sides of the triangle.

49 What do you notice about ‘The area of the semicircle on the hypotenuse’ and ‘The sum of the areas of the semicircles on the other two sides’?

50 Write a statement linking the ‘The semicircle on the hypotenuse’ and ‘The sum of the semicircles on the other two sides’.

Conclusion51 You have constructed squares, equilateral triangles

and semicircles on the sides of a right-angled triangle. What conclusion can you make about the sum of the areas on the shorter sides of the triangle? Make a general statement that links all the ideas into one. What feature is the important linking feature of the three examples we have constructed?

Taking it furtherCan this process be repeated with regular pentagons or hexagons on each side? Check this for yourself using GeoGebra and the techniques described above.



Finding the length of the hypotenuse

If we know the lengths of the two shorter sides of a right-angled triangle, we can use Pythagoras’ Theorem to find the length of the hypotenuse.

Surds and decimal approximationsThe length of the hypotenuse may not always be a whole number. It may have an exact decimal value, such as c = which gives an exact answer of c = 2.5. However, sometimes the length cannot be evaluated exactly by a decimal equivalent and is written as an approximation.

Consider a right-angled triangle with shorter side lengths of 1 m and 2 m:

c2 = a2 + b2

c2 = 12 + 22

= 1 + 4= 5

c = m

If we evaluate on the calculator, it is shown as 2.236 067 977. This is a decimal approximation. There is no exact decimal equivalent because the number does not terminate (it has an infinite number of decimal places) and it has no repeating pattern. These types of numbers are called irrational numbers. Irrational numbers such as and are known as surds. The only way to write these numbers accurately is in surd form; that is, as or However, this is not always practical when we are working with the side lengths of triangles. Depending on the type of answer and the degree of accuracy required, a calculator can be used to find a rational approximation. We will show this by using the ‘approximately equal to’ sign, ≈, then writing it equal to a decimal number correct to the required number of decimal places. At other times we may leave the answer in surd form (also known as ‘exact form’).

Worked Example 2Calculate the length of the hypotenuse in the following triangle.

Thinking1 State Pythagoras’ Theorem and define

the side lengths a, b, and c, letting c be the unknown length of the hypotenuse, x.

c2 = a2 + b2

a = 20, b = 21, c = x

2 Substitute the values for a, b and c into Pythagoras’ Theorem.

x2 = 202 + 212

3 Simplify the equation. x2 = 400 + 441= 841

4 Solve for x by finding the square root of both sides of the equation.

x =x = 29 units

2

21

20

x

841

6.25,

5

5

2, 5, 20 352 5.

2.2AA Pearson Maths 9 SB-02.fm Page 88 Monday, October 24, 2011 9:32 AM


2.2

89

For example, c = m≈ 2.236 067 977= 2.24 m (correct to two decimal places)

Pythagoras’ Theorem may be applied to other shapes provided they contain right-angled triangles. Sometimes, it is necessary to identify the required right-angled triangle by carefully adding a line or lines to the diagram.

Worked Example 3Find the length of the hypotenuse in the following right-angled triangle. Give your answer in exact form (surd form).

Thinking1 State Pythagoras’ Theorem, and define

the side lengths.c2 = a2 + b2

a = 6, b = 10, c = x

2 Substitute the values for a, b and c into Pythagoras’ Theorem.

x2 = 62 + 102

3 Simplify the equation. x2 = 36 + 100= 136

4 Solve for x by finding the square root of both sides of the equation. Leave the answer as a square root (surd form).

x =

Worked Example 4Use Pythagoras’ Theorem to find the value of the unknown length in each of the following diagrams. Give your answer as a rational approximation, correct to two decimal places, if necessary.

(a) (b)

Thinking(a) 1 Identify a right-angled triangle and

define the side lengths.(a) Let a = 2, b = 4, c = t.

5

3

10

x6

136

4

2 m

4 m

t m

B

DC

A

10 m

15 m

12 m

k m

2 m

4 m

t m


2.2


2 State Pythagoras’ Theorem, then subtitute in the values.

c2 = a2 + b2

t2 = 22 + 42

3 Simplify the equation. = 4 + 16 = 20

4 Solve for t by finding the square root of both sides. Use a calculator to find its rational approximation and write it correct to two decimal places, with the correct units.

t =≈ 4.472 135 955

t = 4.47m (2 d.p.)

(b) 1 Draw a line from point B down to the line CD to form a right-angled triangle BED.

(b)

2 Redraw the right-angled triangle and define the side lengths.

c = k, a = 10, b = ED= 15 − 12= 3

3 State Pythagoras’ Theorem, then substitute in the values.

c2 = a2 + b2

k2 = 102 + 32

4 Simplify the equation. = 100 + 9= 109

5 Solve for k by finding the square root of both sides of the equation. Use a calculator to find its rational approximation and write it correct to two decimal places, with the correct units.

k =≈ 10.440 306 51

k = 10.44m (2 d.p.)

Worked Example 5Find the unknown lengths in the following diagram. Give your answers in exact form (surd form).

Thinking1 Identify the right-angled triangle where

two of the three sides are known and define the side lengths.

a = 3b = 7c = x

20

B

DEC

A

10 m

12 m

15 m

k m

B

DE 3 m

10 m k m

109

5

y

3 4

7

x



2.2

91

Finding the length of the hypotenuse

Equipment required: calculator

Fluency1 Calculate the length of the hypotenuse in the following triangles.

(a) (b) (c)

(d) (e) (f)

2 State Pythagoras’ Theorem, then substitute in the values.

c2 = a2 + b2

x2 = 32 + 72

3 Simplify the equation. = 9 + 49= 58

4 Solve for x by finding the square root of both sides. Leave your answer as a square root (surd form).

x =

5 Identify the second right-angled triangle and define the side lengths using your answer for x.

a = x = b = 4c = y

6 State Pythagoras’ Theorem, then substitute in the values. Use the exact surd value determined previously.

c2 = a2 + b2

y2 = ( )2 + (4)2

y2 = 58 + 16

7 Simplify the equation. = 74

8 Solve for y by finding the square root of both sides. Leave your answer as a square root (surd form).

y =

NavigatorQ1, Q2, Q3 (a)–(f), Q4, Q5, Q8, Q10 (a) & (b), Q12, Q14, Q15, Q17

Q1, Q2, Q3 (d)−(i), Q4, Q6, Q7, Q8, Q9, Q10, Q11, Q15, Q16, Q17

Q1 (d)−(f), Q2, Q3 (g)−(l), Q4, Q7, Q8, Q9, Q10, Q11, Q12, Q13, Q14, Q15, Q16, Q17

58

58

58

74

2.2

Answerspage 603

2

d

12 m

5 m

k

6 cm

8 cm

v

9 mm

12 mm

x

36 m

15 m

24 cm

h10 cm

y

60 mm

25 mm


2

.

2

92

PEARSON

mathematics 9

2

Find the length of the hypotenuse in the following right-angled triangles. Give your answer in exact form (surd form).

(a) (b) (c)

(d) (e) (f)

3

Use Pythagoras’ Theorem to find the value of the unknown length in each of the following diagrams. Give your answers as rational approximations, correct to two decimal places if necessary.

(a) (b) (c)

(d) (e) (f)

(g) (h)

(i)

(j) (k) (l)

3

x13 m

4 m

n

12 m

6 m

l4.8 m

3.8 m

j

14.7 cm

8 cmg

3 cm

8 cm

f

16 mm

14 mm

4

8 cm

15 cm

p cm 30 m

16 m

d m

h cm

12 cm

6 cm

8 cm

4 m

4 m

7 m

r m

28 m

21 m g m

l cm

12 cm

8 cm

18 cm

t cm

7 cm

15 mmf mm

23 mm

v cm

17 cm

5 cm

15 cm

k m

14 m

10 m

10 m

c m

3.2 km t km

4.5 km

z m

12 m

5 m

12 m


2


2

.

2

93

4

Find the unknown lengths in the following diagrams. Give your answers in exact form (surd form).

(a) (b)

5

The length of the hypotenuse in this right-angled triangle, correct to two decimal places, is:

A

39.80 mm

B

63.51 mm

C

88.00 mm

D

4034.00 mm

6

The value of

y

in the following diagram is closest to:

A

7.33

B

7.34

C

10.21

D

53.87

7

The value of

x

in the diagram is:

A

cm

B

cm

C

cm

D

cm

Understanding

8

A ladder is leaning against a built framework and just reaches the roof, a height of 4.27 m above the ground. The foot of the ladder is 1.41 m from the base of the framework.

(a)

Draw a diagram of a right-angled triangle that represents the ladder, the ground and the framework. Label the diagram with the known measurements, and label the unknown side

x

.

(b)

Use Pythagoras’ Theorem to calculate the length of the ladder, correct to two decimal places.

9

Corporal Vince is in charge of designing a ropes course for an army training camp. In the final part of the course, recruits must slide down a rope connected from the top of a tower to a point on the ground 35 m from the base of the tower. The top of the tower is 14 m high and at right angles to the ground.

(a)

Draw a diagram of a right-angled triangle that shows the tower, the rope and the ground. Label the diagram with the known measurements and label the unknown side

x

.

5

y

8 m15 m

7 m

x

y

12 cm

5 cm

5 cm

x

53 mm

35 mm

y

4.17

6.04

x cm

25 cm

12 cm

15 cm

x cm

244

288

369

769


2

.

2

94

PEARSON

mathematics 9

(b)

Use Pythagoras’ Theorem to calculate the length of rope Corporal Vince needs for this part of the course, correct to the nearest metre.

10

Find the exact value of the variables in the following diagrams. For any surd values, also give the answer correct to two decimal places.

(a) (b)

(c) (d)

11

The values of

x

and

y

in the diagram are closest to:

A

x

= 9.80 cm,

y

= 8 cm

B

x

= 9.80 cm,

y

= 5.66 cm

C

x

= 13.86 cm,

y

= 8 cm

D

x

= 13.86 cm,

y

= 5.66 cm

Reasoning

12 (a)

Find the length of the hypotenuse of a right-angled isosceles triangle with a side length of 10 cm. Now divide the length of the hypotenuse by the length of the shorter side. Compare your answer to the value of rounded to four decimal places.

(b)

Find the length of the hypotenuse for several other right-angled isosceles triangles. Then, divide the length of the hypotenuse by the length of the shorter side. What do you notice about the result?

13

A Pythagorean spiral has the shape and side lengths shown.

(a)

Use Pythagoras’ Theorem to calculate the values of the hypotenuses of

A

,

B

,

C

,

D

and

E

as exact values and correct to two decimal places where necessary.

(b)

More triangles are added until the letter

Z

is used so that the side lengths are

A

,

B

,

C

,

D

,

E

,

F

,

G

, …,

X

,

Y

,

Z

. How many side lengths cannot be written as exact decimal values?

a cm

10 m

b cm

5 m

18 m

4 m

60 mm

16 mm

9 mm

15 mm

f mme mmd

mm

3 cm

5 cm

a cma cm

b cm b cm

a cm

4 cm7 cm

c cm

4 cm

4 cm

x cm

12 cm 4 cm

y cm√48 cm

2,

1

1

1

11

1

1A

B

C D

E

F


2


2

.

2

95

14 (a)

Use the letters shown to describe the shortest path possible to go from point

A

to point

K

if you can only follow the straight lines shown.

(b)

If each shorter side of a triangle is 1 cm in length, what is the total length of the path described in

(a)

? State your answer as an exact value.

15

Alex wants to buy a new flat screen TV. He has allocated a rectangular space on his lounge wall that is 98 cm in length and 54 cm in width. At the TV store, Alex learns that TVs come in the following screen sizes: 55 cm, 66 cm, 81 cm, 94 cm, 106 cm, 116 cm and 127 cm. He also learns that these ‘screen sizes’ are the diagonal distance from one corner of the screen to the other.

Which is the largest screen size that could fit in the space on Alex’s wall?

Open-ended

16

A swimming pool is 50 m long and 25 m wide. Anastasia is trying to swim a length of the pool, starting from a corner at the shallow end. However, she has difficulty swimming in a straight line, and, instead, heads diagonally across the pool, usually ending up somewhere between the two corners at the other end of the pool. Draw a diagram and calculate four possible distances, correct to two decimal places, that Anastasia may actually end up swimming when completing a lap of the 50 m pool.

17

The city council needs to build a wheelchair ramp from the ground to the top of a step 30 cm off the ground.

(a)

Design three different ramps by first choosing the horizontal length, then calculating the sloping ramp length.

(b)

Sketch your three ramps roughly to scale.

(c)

Which ramp design would you choose if the ramp were to be used everyday?

AB

CD

E F G H J

I K

GamePerfect square root bingoEquipment required: 2 brains, 2 dice

How to win: Cross three numbers in a row off your grid. Play the best of three games.

How to play:Each player should draw up a 3 × 3 grid and fill each grid space with any one of the following square roots (used only once):

Roll both dice. The sum of the dice is the number both players can cross off their board.

Sample Roll: 2 and a 3 are rolled 2 + 3 = 5, so you could cross off your bingo board.

The aim is to get 3 in a row.

Consider (the following before your next game): Which square root is likely to come up more times than the others? Where would be a good location on your board to place this number?4, 9, 16, 25, 36, 49, 64, 81, 100,

121, 144

25



Finding the length of a

shorter sideThe length of one of the shorter sides of a right-angled triangle can be found if we are given the lengths of the hypotenuse and the other side.

It may be necessary to draw a right-angled triangle from a diagram in order to apply Pythagoras’ Theorem.

3

Worked Example 6Find the value of the unknown shorter side in this right-angled triangle.Give your answer as an exact value.

Thinking1 State Pythagoras’ Theorem and define

the side lengths.c2 = a2 + b2

Let a = y, b = 14, c = 19

2 Substitute the values into Pythagoras’ Theorem.

192 = y2 + 142

3 Rewrite so that the unknown is isolated on one side of the equation. (Here, 142 has been subtracted from both sides.)

y2 = 192 − 142

4 Simplify the equation. = 361 − 196= 165

5 Solve for y by finding the square root of both sides of the equation. Leave your answer as a surd (exact form).

y =

Worked Example 7Find the value of the variable, correct to two decimal places, in the following diagram.

6

y

1419

165

7

10 cm

12 cm

x cm

2.3AA Pearson Maths 9 SB-02.fm Page 96 Monday, October 24, 2011 9:32 AM


2.3

97

Finding the length of a shorter side

Equipment required: calculator

Fluency1 Find the value of the unknown shorter side in these right-angled triangles. Give your

answer as an exact value.

(a) (b) (c)

Thinking1 Draw a diagram of the appropriate

right-angled triangle.

2 State Pythagoras’ Theorem and define the side lengths.

c2 = a2 + b2

Let a = 10, b = x, c = 12

3 Substitute the values into Pythagoras’ Theorem.

122 = 102 + x2

4 Rewrite so that the unknown is isolated on one side of the equation. (Here, 102 has been subtracted from both sides).

x2 = 122 − 102

5 Simplify the equation. = 144 – 100= 44

6 Solve for x by finding the square root of both sides of the equation. Use a calculator to find its rational approximation and write it correct to two decimal places.

x =≈ 6.6 332 495

x = 6.63 m (2 d.p.)

NavigatorQ1 (a)–(i), Q2 (a)–(c), Q3 (a), Q4, Q5, Q6, Q8

Q1 (d)–(l), Q2, Q3, Q4, Q5, Q6, Q7, Q8

Q1 (g)–(l), Q2, Q3 (b), Q4, Q5, Q6, Q7, Q8, Q9

10 cm12 cm

x cm

44

2.3

Answerspage 604

6

1237

25

7

12

15


2

.

3

98

PEARSON

mathematics 9

(d) (e) (f)

(g) (h) (i)

(j) (k) (l)

2

Find the value of the variable, correct to two decimal places, in each of the following diagrams.

(a) (b) (c)

(d) (e) (f)

3 (a)

The value of

x

in this right-angled triangle is closest to:

A

7.30

B

12.99

C

17.13

D

23.10

(b)

The hypotenuse of a right-angled triangle has a length of 55 cm. One of the shorter sides has a length of 53 cm. The length of the third side is closest to:

A

2 cm

B

14.6 cm

C

14.7 cm

D

216 cm

41 m

9 m

20 cm52 cm

85 mm36 mm

y

1429

y

821.3

y

326

127

45

y15 m

y

11.2 m20.4 m

y

7

x m

12 m

15 m

x m14 m

x m

10 m

6 m

5 m

x mm

28 mm

34 mm

10 mm

x cm

y cm

10 cm

15 cm

13 cm

y m

20 m12 m

x m

10 m

x

15.27.9


2


2

.

3

99

Understanding

4

A playground slide has a length of 4.76 m, while it covers a horizontal distance of 4.1 m on the ground. What is the height of the slide, correct to two decimal places?

5

A ‘guy rope’ is attached to the top of a tent pole and pegged into the ground so that the rope is tight. If the rope is 3.8 m long, and the pole is 2.9 m long, how far from the base of the pole should the rope be pegged if it is to be tight?

Reasoning

6

Find the lengths of the shorter sides, correct to two decimal places, if the length of the hypotenuse of a right-angled isosceles triangle is:

(a)

20 cm

(b)

12 m

(c)

100 cm

7

Matilda uses a piece of rope 12 m long to form an equilateral triangle.

(a)

How long are the sides of the triangle?

(b)

Find the height of the triangle, correct to two decimal places.

Open-ended

8

Safety regulations for ladders state that the distance that a ladder can be placed from a wall is no closer than one-quarter of its length and no further than half its length. If a ladder is 3.2 m long, give three heights that it can reach up the wall.

9

Find two sets of possible lengths of the shorter sides, correct to two decimal places, if the length of the hypotenuse of a right-angled scalene triangle is as given.

(a)

20 cm

(b)

35 m

4.76 m

4.1 m

3.8 m2.9 m

PuzzleRoots, but no plantsThe square root of a number multiplied by itself gives the number. The cube root of a number is multiplied by itself twice to give the number. The fourth root is multiplied by itself three times.

“I know a number of which the cube root is ten times the fourth root”, said Sue.

“Not possible!” said Alec.

“Is too!” said Sue.

Who was right?


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